IGCSE Physics Topic 1: General Physics

Hey Tara! Welcome to your Topic 1 study guide. This topic is the foundation of ALL of physics, and I promise you it is not as scary as it looks. We will take it step by step, with examples from everyday life in Bangalore that you already understand. Every time you see a formula, I will show you exactly how to use it with worked examples. You have got this!

1.1 Physical Quantities & Measurement Techniques ▼

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Measurements & Physical Quantities - IGCSE Physics

Measuring Length and Volume

Using a Ruler

A ruler is the simplest instrument you use every day. It measures length - the distance from one point to another. Here is how to use it properly:

Place the ruler right next to the object - do not hold it above or at an angle. Place it flat on the surface so there is no gap between the ruler and the object.
Read to the nearest millimetre (mm) - a standard ruler is marked in cm and mm. The small divisions are 1 mm each. So if a pencil ends between 15.2 cm and 15.3 cm, estimate which mm line it is closest to.
Avoid parallax error - your eye must be directly above the marking you are reading, not looking at an angle. If you look from the side, you will read the wrong value.
Use the markings, not the end of the ruler - some rulers have a worn or chipped end. It is better to measure from, say, the 1.0 cm mark to the end of the object, then subtract 1.0 cm from your reading.

Indian example: Imagine you are measuring the length of your NCERT Physics textbook. You place the ruler alongside it. The book starts at 0.0 cm and ends at 24.8 cm. The length of the book is 24.8 cm, or 248 mm, or 0.248 m.

⚠ Exam Tip

Always state your measurement with the correct unit. A number without a unit is meaningless in physics. "24.8" means nothing. "24.8 cm" is a proper measurement.

Using a Measuring Cylinder

A measuring cylinder measures the volume of a liquid. You have seen your amma measure water for cooking - a measuring cylinder is the same idea, but more precise.

Place the cylinder on a flat, level surface - do not hold it in your hand; it will tilt and give a wrong reading.
Pour in the liquid gently to avoid splashing.
Read at the meniscus - this is the key part! Water curves upward at the edges of the glass (because water molecules are attracted to glass). The bottom of this curve is the meniscus. Always read the volume at the bottom of the meniscus, not at the edges where the water climbs up.
Keep your eye level with the liquid surface - if you look from above, the reading will be too high; from below, too low. Bend down so your eye is at the same height as the liquid.

Always read the volume at the bottom of the meniscus, with your eye level with the liquid surface.

Indian example: Suppose you want to measure the volume of water that fits in a stainless steel tumbler (the kind you drink water from at home). You pour the water from the tumbler into a measuring cylinder. The bottom of the meniscus sits at the 250 cm³ mark. So the tumbler holds 250 cm³ (which is the same as 250 mL) of water.

Measuring Volume of an Irregular Object (Displacement Method)

What if you need to find the volume of a small stone or a metal idol? You cannot use a ruler because the shape is irregular. Instead, you use the displacement method:

Fill a measuring cylinder partway with water. Record the initial volume (say, 50 cm³).
Gently lower the object into the water using a thread (do not drop it - that causes splashing!).
The water level rises. Record the new volume (say, 62 cm³).
Volume of the object = new volume - initial volume = 62 - 50 = 12 cm³.

Indian example: You want to find the volume of a small brass Ganesha idol. You lower it into a measuring cylinder. The water rises from 40 cm³ to 53 cm³. The volume of the idol is 53 - 40 = 13 cm³.

⚠ Exam Tip

The object must be completely submerged (fully underwater) for the displacement method to work. If there are air bubbles stuck to the object, tap the cylinder gently to release them before reading.

Measuring Time Intervals

Time is something we measure constantly in everyday life. In physics, we need to be precise about it.

Instruments for Measuring Time

Analogue clocks - the regular clocks with hour, minute, and second hands. Good for longer time intervals (hours, minutes) but not precise for short intervals.
Digital stopwatches - these measure to the nearest 0.01 s (one hundredth of a second). You press start when the event begins and stop when it ends.
Digital timers with sensors - these use light gates or pressure pads to start and stop automatically. They are more accurate because there is no human reaction time error.

Indian example: Think about Sports Day at your school. When the PE teacher times the 100 m race, they use a stopwatch. They press start when they see the starting gun fire (not when they hear it - light travels faster than sound!). They press stop when the runner crosses the finish line.

Another example: Timing how long your pressure cooker whistles. If it whistles 3 times in 4 minutes, each whistle interval is about 80 seconds. But if you wanted to measure this precisely, you would use a digital stopwatch.

⚠ Exam Tip

Human reaction time is about 0.2 to 0.5 seconds. This means every time you start or stop a stopwatch by hand, your measurement could be off by up to 0.5 s. This is why we use the "multiples" method for short time intervals (see next section).

Measuring Small Distances and Short Times by Averaging

Here is a brilliant trick that scientists use: if a single measurement is too small to measure accurately, measure MANY of them together and divide.

Why Do We Measure Multiples?

Let us say you want to measure the thickness of a single page of your textbook. If you try to measure one page with a ruler, you might get "about 0.1 mm" - but that is very imprecise. The ruler is not fine enough.

Instead:

Measure the thickness of 100 pages together: say, 8.4 mm.
Divide by 100: thickness of one page = 8.4 / 100 = 0.084 mm.

Now you have a much more precise answer!

The Pendulum: A Classic Example

A pendulum is a weight hanging from a string that swings back and forth. One complete back-and-forth swing is called one oscillation. The time for one complete oscillation is called the period (T).

A single swing might take about 1 second. If you time just one swing with a stopwatch, your reaction time error (about 0.3 s) is a huge fraction of the measurement - that is a 30% error! Terrible!

Instead, you time 20 swings. If 20 swings take 18.4 s, then one swing takes 18.4 / 20 = 0.92 s. Now your reaction time error of 0.3 s is only 0.3/18.4 = 1.6% of the total time. Much better!

T = total time / number of oscillations

T = period (time for one complete oscillation), in seconds (s) total time = time measured for all oscillations, in seconds (s) number of oscillations = how many complete swings you counted

Worked Example A student times 20 swings of a pendulum and records 18.4 s. Calculate the period T of the pendulum.

Step 1: Write the formula

T = total time / number of oscillations

Step 2: Substitute the values

T = 18.4 s / 20

Step 3: Calculate

T = 0.92 s

Answer: T = 0.92 s

Worked Example A student measures 25 oscillations of a pendulum three times and gets: 22.5 s, 23.1 s, and 22.8 s. Calculate the average period T.

Step 1: Find the average total time

Average time = (22.5 + 23.1 + 22.8) / 3 = 68.4 / 3 = 22.8 s

Step 2: Calculate the period

T = 22.8 s / 25 = 0.912 s

Answer: T = 0.912 s (or 0.91 s to 2 significant figures)

Worked Example At a temple in Bangalore, a brass bell hangs from a rope. A student times 15 complete swings of the bell as 27.0 s. What is the period of the bell?

Step 1: Write the formula

T = total time / number of oscillations

Step 2: Substitute

T = 27.0 s / 15

Step 3: Calculate

T = 1.80 s

Answer: T = 1.80 s (Each complete swing of the bell takes 1.80 seconds)

⚠ Exam Tip

When counting oscillations, start your count from zero, not one! When you release the pendulum and start the stopwatch, the pendulum is at position 0. When it comes back to the same position going in the same direction, that is oscillation 1.

Also, always start timing from the middle (equilibrium) position of the swing, not from the highest point. The pendulum moves fastest at the middle, so it is easier to judge the exact moment it passes through.

Supplement

Scalars and Vectors

This is one of those ideas in physics that sounds complicated but is actually really simple once you get it.

What is a Scalar?

A scalar is a quantity that has only magnitude (size). That is it. Just a number with a unit.

For example: "The temperature in Bangalore today is 28 degrees Celsius." You do not need to say in which direction the temperature is 28 degrees - that would not make sense! Temperature is just a number.

What is a Vector?

A vector is a quantity that has both magnitude AND direction. You need to state which way it is pointing for it to make full sense.

For example: "The auto-rickshaw is travelling at 30 km/h towards Majestic." The "towards Majestic" part is the direction. Without it, you only know how fast the auto is going, not where it is going.

The Cricket Analogy

Think about Jasprit Bumrah bowling in a cricket match:

Speed of the ball: 145 km/h - this is a scalar. It tells you how fast the ball is moving, but not in which direction.
Velocity of the ball: 145 km/h towards the off stump - this is a vector. It tells you both the speed AND the direction. The batsman cares about the direction!

The Scalar Quantities You Must Know

Scalar Quantity	What It Measures	Example
Distance	How far something has travelled (total path)	The walk from your classroom to the canteen is 200 m
Speed	How fast something is going	An auto-rickshaw going at 40 km/h
Time	Duration of an event	The school assembly lasted 30 minutes
Mass	Amount of matter in an object	A bag of rice has a mass of 5 kg
Energy	Ability to do work	A cup of chai gives you about 300 kJ of energy
Temperature	How hot or cold something is	Bangalore in April: 34 degrees C

💡 Memory Trick

Remember the scalars with: "D-S-T-M-E-T" = "Dosas Served To Me Every Tuesday"

Distance, Speed, Time, Mass, Energy, Temperature - all scalars!

The Vector Quantities You Must Know

Vector Quantity	Why It Needs Direction	Example
Force	A push or pull acts in a specific direction	You push a door with 10 N to the right
Weight	Force of gravity always acts downwards	Your weight is 500 N downwards
Velocity	Speed in a specific direction	A train moving at 80 km/h northwards
Acceleration	Rate of change of velocity has a direction	The bus accelerates at 2 m/s² forwards
Momentum	Mass x velocity, so it inherits direction from velocity	A cricket ball has momentum towards the boundary
Electric field strength	Force per unit charge acts in a direction	Electric field points from positive to negative
Gravitational field strength	Force per unit mass, acts towards centre of Earth	g = 9.8 N/kg directed downward

💡 Memory Trick

Remember the vectors with: "F-W-V-A-M-E-G" = "Five Wickets! Virat And MS Earn Glory"

Force, Weight, Velocity, Acceleration, Momentum, Electric field strength, Gravitational field strength - all vectors!

⚠ Exam Tip

A very common exam question asks: "What is the difference between speed and velocity?" The answer: Speed is a scalar (magnitude only); velocity is a vector (magnitude AND direction). Similarly for distance (scalar) vs displacement (vector).

Adding Vectors at Right Angles (Resultant)

When two vectors act at right angles (90 degrees) to each other, we can find the single vector that has the same overall effect. This single vector is called the resultant.

Think about it this way: if you walk 3 km east and then 4 km north, you have not ended up 7 km from where you started. You have ended up at a diagonal distance that is less than 7 km. The resultant is that diagonal.

R = √(A² + B²)

R = resultant (the combined effect of both vectors) A = magnitude of the first vector B = magnitude of the second vector (at right angles to A) This is just Pythagoras' theorem! The resultant is the hypotenuse.

θ = tan⁻¹(B / A)

θ = angle of the resultant measured from the direction of vector A B = the side opposite to the angle A = the side adjacent to the angle

Two vectors at right angles (3 N east + 4 N north) combine to give a resultant of 5 N at 53.1 degrees from east.

Worked Example Two forces act on an object: 3 N towards the east and 4 N towards the north. Find the magnitude and direction of the resultant force.

Step 1: Sketch the vectors

Draw the 3 N vector pointing east (horizontal). From its tip, draw the 4 N vector pointing north (vertical). The resultant goes from the start of the first vector to the tip of the second.

Step 2: Use Pythagoras to find the magnitude

R = √(3² + 4²) = √(9 + 16) = √25 = 5 N

Step 3: Use trigonometry to find the direction

θ = tan⁻¹(4 / 3) = tan⁻¹(1.333) = 53.1 degrees from east (towards north)

Answer: The resultant force is 5 N at 53.1 degrees north of east.

Worked Example A boat tries to cross the Ganges river. The boat moves at 4 m/s directly towards the opposite bank. The river current pushes the boat downstream at 3 m/s. Find the resultant velocity of the boat.

Step 1: Identify the two velocities at right angles

Boat speed across river = 4 m/s (let us call this the "forward" direction)
River current speed = 3 m/s (this is sideways, at 90 degrees to the boat's intended direction)

Step 2: Calculate the resultant speed

R = √(4² + 3²) = √(16 + 9) = √25 = 5 m/s

Step 3: Calculate the direction

θ = tan⁻¹(3 / 4) = tan⁻¹(0.75) = 36.9 degrees from the intended direction (towards downstream)

Answer: The boat actually travels at 5 m/s, at an angle of 36.9 degrees downstream from its intended straight-across path. The boatman will end up further downstream than planned!

Worked Example An aeroplane flies at 200 km/h due north. A crosswind blows from the west at 50 km/h (pushing the plane eastward). Find the resultant velocity of the aeroplane.

Step 1: Identify the perpendicular vectors

Plane speed = 200 km/h northward
Wind speed = 50 km/h eastward (perpendicular to north)

Step 2: Calculate resultant magnitude

R = √(200² + 50²) = √(40000 + 2500) = √42500 = 206.2 km/h

Step 3: Calculate direction

θ = tan⁻¹(50 / 200) = tan⁻¹(0.25) = 14.0 degrees from north (towards east)

Answer: The aeroplane's resultant velocity is 206.2 km/h at 14.0 degrees east of north.

⚠ Exam Tip

When finding the resultant of two perpendicular vectors, you will ALWAYS need: (1) Pythagoras for the magnitude, (2) tan⁻¹ for the angle. Always draw a diagram first - it helps you see which side is opposite and which is adjacent for the angle.

Remember: the resultant of two perpendicular vectors is ALWAYS less than their arithmetic sum. If forces are 3 N and 4 N, the resultant is 5 N, not 7 N!

🔍 Apply It: Real-World Physics

Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.

A BMTC bus driver in Bangalore needs to check if a new bus fits under a heritage arch that is exactly 3.50 m tall. The bus manufacturer says the bus is 3450 mm.

Does the bus fit? What instrument should the driver use to verify the arch height, and why?

▼

Identify the Physics

This is about unit conversion (mm to m) and choosing the right measuring instrument for a large distance.

Work It Out

Step 1: Convert the bus height to metres.
1 m = 1000 mm, so 3450 mm ÷ 1000 = 3.450 m = 3.45 m

Step 2: Compare.
Arch height = 3.50 m. Bus height = 3.45 m.
3.45 m < 3.50 m, so yes, the bus fits — with 0.05 m (5 cm) to spare.

Step 3: Choose the instrument.
The arch is about 3.5 m tall — far too large for a ruler (30 cm) or even a metre ruler. The driver should use a measuring tape (e.g., a 5 m steel tape). It can measure large lengths accurately to the nearest mm, it’s portable, and it’s flexible enough to reach the top of the arch.

💡 The Aha! Moment

5 cm of clearance sounds safe, but buses bounce on speed bumps! In real life, engineers add a safety margin. In physics exams, always convert to the same unit before comparing — never compare mm with m directly.

A chai wallah near Majestic bus station measures out milk for chai using a small measuring cylinder. He reads the meniscus at eye level and gets 48 mL. His helper, who is taller, reads the same cylinder from above and gets 46 mL.

Who is correct and why?

▼

Identify the Physics

This is about parallax error — the reading changes depending on the angle you look at the scale from.

Work It Out

The chai wallah (48 mL) is correct.

When you read a measuring cylinder, your eye must be level with the bottom of the meniscus. The chai wallah does this.

His taller helper looks down at the cylinder. From above, the line of sight crosses the scale at a lower reading than the true value — this is why he reads 46 mL instead of 48 mL.

This mistake is called a parallax error. It happens whenever you read a scale from the wrong angle.

💡 The Aha! Moment

Parallax error is one of the most common mistakes in practical physics — and one of the easiest marks to pick up in exams! The fix is simple: always position your eye perpendicular to the scale (at the same level as the measurement). This applies to rulers, thermometers, ammeters, and any analogue instrument.

A scientist at ISRO needs to measure the thickness of the heat shield tiles on a spacecraft. Each tile is approximately 5 cm thick, but she needs to measure it to the nearest 0.1 mm.

Should she use a ruler, vernier caliper, or micrometer? Why?

▼

Identify the Physics

This is about choosing an instrument based on the precision needed and the size of the object being measured.

Work It Out

Let’s compare the three instruments:

Ruler: precision ≈ ±1 mm. Can measure up to 30 cm or 100 cm. But ±1 mm is far too imprecise — she needs ±0.1 mm. ❌

Micrometer screw gauge: precision = ±0.01 mm (excellent!). But its maximum opening is typically only about 25 mm. The tile is 5 cm = 50 mm — it won’t fit! ❌

Vernier caliper: precision = ±0.1 mm. Can measure objects up to about 150 mm. The tile (50 mm) fits easily, and the precision matches exactly what she needs. ✅

Answer: Vernier caliper. It gives ±0.1 mm precision and can accommodate a 50 mm tile.

💡 The Aha! Moment

The micrometer is MORE precise than the vernier caliper, but it can’t measure objects thicker than ~25 mm. Precision isn’t the only factor — you must also check that the instrument’s range fits the object. This is a classic IGCSE trap question!

Tara’s PE teacher at school asks her to estimate the average speed of students running the 100 m dash. The stopwatch shows times of 14.2 s, 13.8 s, 14.1 s, 13.9 s, and 14.5 s for five students. But each timing has a reaction time error of about ±0.3 s.

What is the average time, and how significant is the reaction time error as a percentage?

▼

Identify the Physics

This is about averaging repeated measurements to improve reliability, and percentage error to judge how significant an error is.

Work It Out

Step 1: Find the average time.
Sum = 14.2 + 13.8 + 14.1 + 13.9 + 14.5 = 70.5 s
Average = 70.5 ÷ 5 = 14.1 s

Step 2: Calculate the percentage error.
Percentage error = (error ÷ measured value) × 100%
= (0.3 ÷ 14.1) × 100%
= 2.1%

Step 3: Is this significant?
A 2.1% error is relatively small but not negligible. For a school PE lesson, it’s acceptable. For an Olympic sprint (where races are decided by 0.01 s), this error would be huge!

Average speed = distance ÷ time = 100 ÷ 14.1 = 7.1 m/s (about 25.5 km/h).

💡 The Aha! Moment

Taking multiple readings and averaging doesn’t remove systematic error (like reaction time), but it does reduce the effect of random errors. To fix the reaction time problem, you could use light gates instead of a manual stopwatch — that’s a common exam answer!

An engineer at Hindustan Aeronautics Limited (HAL) in Bangalore is testing a turbine blade. She needs to find the volume of the irregularly shaped blade. She cannot use length × width × height.

What method should she use, and what readings does she need?

▼

Identify the Physics

This is about the displacement (Eureka can) method for finding the volume of an irregularly shaped object.

Work It Out

Since the blade is irregularly shaped, she cannot calculate the volume using a formula. Instead, she should use the displacement method:

Method 1 — Measuring cylinder (if the blade fits):
1. Fill a measuring cylinder with water and record the initial volume, V₁.
2. Carefully lower the blade into the water (it must be fully submerged).
3. Record the new volume, V₂.
4. Volume of blade = V₂ − V₁.

Method 2 — Displacement (Eureka) can (if the blade is too large for a measuring cylinder):
1. Fill the displacement can until water flows out of the spout. Wait until dripping stops.
2. Place a measuring cylinder under the spout.
3. Gently lower the blade into the can until fully submerged.
4. Collect all displaced water in the measuring cylinder.
5. The volume of displaced water = volume of the blade.

Key readings: initial water level (V₁) and final water level (V₂), OR the volume of displaced water collected.

💡 The Aha! Moment

This works because of Archimedes’ principle — the submerged object pushes aside (displaces) a volume of water exactly equal to its own volume. The object MUST be fully submerged, and you must make sure no air bubbles are trapped on it. This method works for ANY shape — even something as complex as a turbine blade!

Practice Questions: 1.1

20 multiple choice questions. Tap an option to check your answer.

Score 0 / 20

Question 1

When reading the volume of water in a measuring cylinder, where should you take the reading?

A At the top of the meniscus

B At the bottom of the meniscus

C At the highest point where water touches the glass

D Halfway between the top and bottom of the meniscus

The meniscus is the curved surface of water. Water curves up at the edges because it is attracted to glass. You always read the volume at the bottom of the meniscus (the lowest point of the curve) with your eye level with the liquid surface.

Question 2

A student measures the thickness of 200 sheets of paper and finds it to be 18.0 mm. What is the thickness of one sheet?

A 0.18 mm

B 0.090 mm

C 0.90 mm

D 9.0 mm

Thickness of one sheet = total thickness / number of sheets = 18.0 mm / 200 = 0.090 mm. This is the "measuring multiples" technique - measuring many together and dividing to get a precise value for one.

Question 3

A student times 20 oscillations of a pendulum and records 24.0 s. What is the period of the pendulum?

A 2.40 s

B 1.20 s

C 0.83 s

D 480 s

T = total time / number of oscillations = 24.0 / 20 = 1.20 s. Option A (2.40 s) would be wrong because that uses 10 oscillations. Option D multiplies instead of dividing.

Question 4

Which of the following is a reason for timing many oscillations of a pendulum instead of just one?

A To make the pendulum swing faster

B To increase the period of the pendulum

C To reduce the effect of human reaction time on the result

D To make the experiment quicker to complete

Human reaction time (about 0.2-0.5 s) causes error when starting and stopping the stopwatch. If you time 20 oscillations (about 20 s total), the reaction time error is a tiny percentage of the total. If you only time 1 oscillation (about 1 s), the error is a huge percentage. Timing multiples reduces the percentage error.

Question 5

A stone is lowered into a measuring cylinder containing 45 cm³ of water. The water level rises to 52 cm³. What is the volume of the stone?

A 52 cm³

B 45 cm³

C 7 cm³

D 97 cm³

Volume of stone = final water level - initial water level = 52 - 45 = 7 cm³. This is the displacement method - the stone pushes the water up by exactly its own volume.

Question 6

Which of the following is a vector quantity?

A Speed

B Mass

C Velocity

D Distance

Velocity is a vector because it has both magnitude (how fast) and direction (which way). Speed, mass, and distance are all scalars - they only have magnitude.

Question 7

Which of the following is a scalar quantity?

A Force

B Acceleration

C Momentum

D Energy

Energy is a scalar - it has magnitude only. You would not say "I have 500 joules of energy going north." That does not make sense! Force, acceleration, and momentum are all vectors - they all need a direction.

Question 8

Two forces act on an object at right angles: 6 N east and 8 N north. What is the magnitude of the resultant force?

A 14 N

B 10 N

C 2 N

D 48 N

R = √(6² + 8²) = √(36 + 64) = √100 = 10 N. This is Pythagoras! 6, 8, 10 is a Pythagorean triple (like 3, 4, 5 but doubled). The answer is NOT 14 N (that would be adding them as if they were in the same direction).

Question 9

A student looks at the ruler from an angle instead of directly above when taking a length measurement. This error is called:

A Zero error

B Parallax error

C Systematic error

D Calibration error

Parallax error occurs when you read a scale from an angle instead of directly in front of or above the marking. Your eye must be perpendicular to the scale for an accurate reading. This applies to rulers, measuring cylinders, thermometers, and all analogue instruments.

Question 10

A swimmer crosses a river at 2 m/s. The river current flows at 1.5 m/s perpendicular to the swimmer. What is the swimmer's resultant speed?

A 3.5 m/s

B 2.5 m/s

C 0.5 m/s

D 2.0 m/s

R = √(2² + 1.5²) = √(4 + 2.25) = √6.25 = 2.5 m/s. The velocities are perpendicular, so you use Pythagoras. You cannot just add them (that would give 3.5 m/s, which is wrong for perpendicular vectors).

Question 11

What is the period of a pendulum if 30 complete oscillations take 36.0 seconds?

A 1.20 s

B 0.83 s

C 1.80 s

D 1080 s

T = total time / number of oscillations = 36.0 / 30 = 1.20 s. Always divide the total time by the number of oscillations. Never multiply!

Question 12

Which of the following correctly lists ONLY vector quantities?

A Force, mass, velocity

B Force, velocity, acceleration

C Speed, velocity, momentum

D Weight, distance, force

Force, velocity, and acceleration are all vectors (they all have magnitude and direction). Option A includes mass (scalar). Option C includes speed (scalar). Option D includes distance (scalar).

Question 13

Why is it important to place a measuring cylinder on a flat, level surface before reading?

A To make the water evaporate faster

B To prevent the cylinder from breaking

C To ensure the water surface is level so the reading is accurate

D To make the water warmer

If the cylinder is tilted, the water surface is no longer horizontal and the scale reading will not match the true volume. A flat, level surface keeps the water surface level and gives an accurate reading.

Question 14

Two forces of 5 N and 12 N act at right angles on an object. What is the resultant force?

A 7 N

B 17 N

C 13 N

D 60 N

R = √(5² + 12²) = √(25 + 144) = √169 = 13 N. Another Pythagorean triple! 5, 12, 13. The answer is NOT 17 (simple addition) or 7 (simple subtraction).

Question 15

Temperature is classified as:

A A scalar quantity

B A vector quantity

C Both scalar and vector

D Neither scalar nor vector

Temperature is a scalar. It has magnitude only (e.g., 35 degrees C). It makes no sense to say "35 degrees C heading north." Temperature does not have a direction.

Question 16

The smallest division on a standard 30 cm ruler is:

A 1 cm

B 1 mm

C 0.1 mm

D 10 mm

A standard ruler is marked in millimetres (mm). The small divisions are 1 mm apart. This means a ruler can measure to the nearest 1 mm (or 0.1 cm). For anything smaller than 1 mm, you would need a more precise instrument like a micrometer.

Question 17

What is the angle of the resultant when a 3 N force acts east and a 3 N force acts north?

A 30 degrees from east

B 45 degrees from east

C 60 degrees from east

D 90 degrees from east

θ = tan⁻¹(3/3) = tan⁻¹(1) = 45 degrees. When the two perpendicular vectors are equal in magnitude, the resultant always makes a 45-degree angle with either one. This makes sense - it goes exactly diagonally between them.

Question 18

A student records three measurements for 10 oscillations: 12.3 s, 12.1 s, and 12.5 s. What is the best estimate for the period?

A 12.3 s

B 1.23 s

C 3.69 s

D 36.9 s

First, find the average time for 10 oscillations: (12.3 + 12.1 + 12.5) / 3 = 36.9 / 3 = 12.3 s. Then, period T = 12.3 / 10 = 1.23 s. Option A gives the average time for 10 oscillations, not the period of one oscillation.

Question 19

Weight is classified as a vector quantity because:

A It is measured in kilograms

B It can change on different planets

C It has both magnitude and direction (always acts downwards)

D It is the same as mass

Weight is a force (the force of gravity on an object). Like all forces, it has a magnitude (measured in newtons, NOT kilograms) and a direction (always towards the centre of the Earth, i.e., downwards). That is what makes it a vector. Weight is NOT the same as mass!

Question 20

An object is pulled by a force of 8 N to the east and 6 N to the north. The magnitude of the resultant force is closest to:

A 14 N

B 2 N

C 10 N

D 48 N

R = √(8² + 6²) = √(64 + 36) = √100 = 10 N. Yet another Pythagorean triple: 6, 8, 10 (which is 3, 4, 5 multiplied by 2). Start recognising these triples - they save time in exams!

1.2 Motion ▼

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Speed, Velocity & Acceleration - IGCSE Physics

Speed

Speed is one of the first things you learn in physics, and you already understand it intuitively. When your parents say "the auto is going too fast," they are talking about speed.

Definition

Speed is the distance travelled per unit time. In simple terms, it tells you how much distance an object covers in each second (or each hour, or any unit of time).

v = s / t

v = speed, in metres per second (m/s) or km/h s = distance travelled, in metres (m) or kilometres (km) t = time taken, in seconds (s) or hours (h)

You can rearrange this formula into three forms. Remember the triangle trick: write s at the top, v and t at the bottom. Cover the one you want to find:

v = s / t (cover v: s over t)
s = v x t (cover s: v times t)
t = s / v (cover t: s over v)

⚠ Exam Tip

Unit consistency is crucial! If speed is in m/s, distance must be in metres and time in seconds. If speed is in km/h, distance must be in km and time in hours. Do NOT mix units!

To convert km/h to m/s: divide by 3.6 (because 1 km/h = 1000m / 3600s = 1/3.6 m/s).

To convert m/s to km/h: multiply by 3.6.

Worked Example An auto-rickshaw in Bangalore travels 6 km from Koramangala to Majestic in 15 minutes. Calculate its average speed in (a) km/h and (b) m/s.

Part (a): Speed in km/h

First, convert 15 minutes to hours: 15 min = 15/60 = 0.25 h
v = s / t = 6 km / 0.25 h = 24 km/h

Part (b): Speed in m/s

Convert: 6 km = 6000 m, 15 min = 15 x 60 = 900 s
v = s / t = 6000 m / 900 s = 6.67 m/s

OR simply: 24 km/h / 3.6 = 6.67 m/s

Answer: (a) 24 km/h (b) 6.67 m/s

Worked Example The Rajdhani Express covers the 2,400 km journey from New Delhi to Mumbai in 16 hours. Calculate its average speed in km/h and m/s.

Step 1: Speed in km/h

v = s / t = 2400 km / 16 h = 150 km/h

Step 2: Convert to m/s

150 km/h / 3.6 = 41.7 m/s

Answer: 150 km/h or 41.7 m/s. That is about the speed of a fast bowler's delivery in cricket!

Worked Example A Bangalore city bus travels at a steady speed of 36 km/h. How far does it travel in 5 minutes?

Step 1: Convert units to match

Speed = 36 km/h = 36/3.6 = 10 m/s
Time = 5 min = 5 x 60 = 300 s

Step 2: Use s = v x t

s = 10 m/s x 300 s = 3000 m = 3 km

Answer: The bus travels 3000 m (3 km) in 5 minutes.

Velocity

You have already seen this in Section 1.1 when we talked about scalars and vectors. Now let us make it formal.

Definition

Velocity is speed in a given direction. It is a vector quantity.

The difference between speed and velocity:

Speed = 50 km/h (scalar - just tells you how fast)
Velocity = 50 km/h due north (vector - tells you how fast AND which way)

Why does direction matter? Imagine a car driving around a circular roundabout at a constant speed of 30 km/h. Its speed never changes, but its velocity is constantly changing because the direction is constantly changing! This is an important concept for understanding acceleration later.

Cricket analogy: When a batsman hits the ball, the ball comes towards the bat at, say, 140 km/h. After the shot, the ball goes away at maybe 120 km/h. The speed did not change much, but the velocity changed dramatically because the direction completely reversed!

Average Speed

average speed = total distance / total time

The total distance is the ENTIRE path length, not just the straight-line distance The total time includes ALL the time, including stops and rests

This is a very important concept. Average speed takes into account the whole journey, including the fast bits, the slow bits, and even stops.

Worked Example Tara walks 800 m from her house to school in 10 minutes, rests for 2 minutes, then walks 200 m to the school canteen in 3 minutes. Calculate her average speed for the entire journey.

Step 1: Find total distance

Total distance = 800 m + 200 m = 1000 m

Step 2: Find total time

Total time = 10 min + 2 min + 3 min = 15 min = 15 x 60 = 900 s

Step 3: Calculate average speed

Average speed = 1000 m / 900 s = 1.11 m/s

Answer: Average speed = 1.11 m/s. Notice that the rest time of 2 minutes IS included in the total time, even though Tara was not moving. That is what makes it an average.

Worked Example A BMTC bus travels 12 km in 30 minutes, then stops for 10 minutes, then travels another 8 km in 20 minutes. Calculate the average speed for the entire journey.

Step 1: Total distance

12 km + 8 km = 20 km

Step 2: Total time

30 min + 10 min + 20 min = 60 min = 1 hour

Step 3: Average speed

Average speed = 20 km / 1 h = 20 km/h

Answer: Average speed = 20 km/h. The 10-minute stop is included in the total time. If we had only used the moving time (50 min), we would get 24 km/h - that would be wrong for average speed.

⚠ Exam Tip

When a question asks for "average speed," you MUST include all time - including time spent stationary. The total distance is the distance actually travelled along the path, not the straight-line displacement. This catches many students out!

Distance-Time Graphs

Graphs are one of the most powerful tools in physics. They tell a story about how an object moves, and once you learn to read them, you can extract all kinds of useful information.

What does a distance-time graph show?

The horizontal axis (x-axis) shows time. The vertical axis (y-axis) shows distance from the starting point. As time increases (you move right), the distance changes depending on how the object is moving.

Three key shapes on a distance-time graph: horizontal (stationary), straight line (constant speed), and upward curve (accelerating).

Graph Shape	What It Means	Example
Horizontal line	Object is stationary (at rest) - distance is not changing	An auto-rickshaw waiting at a red light
Straight line going up	Object moving at constant speed - distance increases steadily	A train cruising at 100 km/h on a straight track
Curve getting steeper	Object is accelerating - covering more distance each second	A bus pulling away from a stop
Curve getting flatter	Object is decelerating - covering less distance each second	A bus approaching a stop and slowing down
Steeper straight line	Faster constant speed (steeper = faster)	Comparing a cycle vs a car on the same road

Calculating Speed from a Distance-Time Graph

Here is the key idea: the speed of an object is the gradient (slope) of the distance-time graph.

speed = gradient = rise / run = change in distance / change in time

rise = how much the distance changes (read off y-axis) run = how much the time changes (read off x-axis)

If the line is straight, the gradient (and therefore speed) is constant. If the line is curved, the gradient is changing, meaning the speed is changing.

Speed = rise/run = 60 m / 4 s = 15 m/s. The gradient of a straight-line d-t graph gives constant speed.

Worked Example From the distance-time graph above, an object travels from 30 m at t = 2 s to 90 m at t = 6 s. Calculate its speed.

Step 1: Read the values from the graph

At t = 2 s, distance = 30 m
At t = 6 s, distance = 90 m

Step 2: Calculate the gradient

Speed = (change in distance) / (change in time) = (90 - 30) / (6 - 2) = 60 / 4 = 15 m/s

Answer: Speed = 15 m/s

Speed-Time Graphs

Now let us move to speed-time graphs. These are even more useful than distance-time graphs because they can tell you about acceleration AND distance.

Key shapes: horizontal line (constant speed), straight line going up (constant acceleration), straight line going down (deceleration).

Graph Shape	What It Means	Example
Horizontal line	Constant speed - speed is not changing	A Namma Metro train cruising between stations
Straight line going up	Constant acceleration - speed increasing at a steady rate	A car accelerating smoothly from a traffic light
Straight line going down	Constant deceleration - speed decreasing steadily	An auto-rickshaw braking to a stop
Curve getting steeper (upward)	Increasing acceleration	A motorcycle whose engine gets more powerful at higher revs
Curve getting flatter (upward)	Decreasing acceleration	A car approaching its top speed
Line at zero	Object is stationary	A parked scooter

Distance from a Speed-Time Graph

Here is the second powerful thing about speed-time graphs: the area under the graph equals the distance travelled.

Why? Because distance = speed x time. And "speed x time" is exactly what area means when the x-axis is time and the y-axis is speed!

distance = area under the speed-time graph

For a rectangle (constant speed): area = base x height = time x speed For a triangle (constant acceleration from/to zero): area = 1/2 x base x height For combined shapes, split into rectangles and triangles, then add areas

The area under a speed-time graph gives the distance travelled. Split complex shapes into triangles and rectangles.

Worked Example A car accelerates uniformly from rest to 20 m/s in 4 seconds, then travels at constant speed for 8 more seconds. Calculate the total distance travelled.

Step 1: Distance during acceleration (triangle)

Area of triangle = 1/2 x base x height = 1/2 x 4 s x 20 m/s = 40 m

Step 2: Distance during constant speed (rectangle)

Area of rectangle = base x height = 8 s x 20 m/s = 160 m

Step 3: Total distance

Total = 40 + 160 = 200 m

Answer: Total distance = 200 m

Worked Example A Namma Metro train accelerates from rest to 18 m/s in 6 s, travels at 18 m/s for 30 s, then decelerates uniformly to rest in 4 s. Calculate the total distance between two stations.

Step 1: Acceleration phase (triangle)

Distance = 1/2 x 6 x 18 = 54 m

Step 2: Constant speed phase (rectangle)

Distance = 30 x 18 = 540 m

Step 3: Deceleration phase (triangle)

Distance = 1/2 x 4 x 18 = 36 m

Step 4: Total distance

54 + 540 + 36 = 630 m

Answer: Total distance between the two stations = 630 m

⚠ Exam Tip

A very common mistake is confusing the two graph types. Remember:

Distance-time graph: gradient = speed

Speed-time graph: gradient = acceleration, area = distance

Never say "area under a distance-time graph" or "gradient of a speed-time graph gives speed" - that is wrong!

Acceleration of Free Fall (g)

When you drop something, it speeds up as it falls. Gravity pulls it downward, making it go faster and faster. This acceleration due to gravity is given the symbol g.

g ≈ 9.8 m/s²

This means: every second, a falling object's speed increases by about 9.8 m/s After 1 second of falling: speed = 9.8 m/s After 2 seconds: speed = 19.6 m/s After 3 seconds: speed = 29.4 m/s ... and so on (In many IGCSE calculations, you can approximate g as 10 m/s² for simplicity)

Indian example: Imagine a coconut falls from a tree. The moment it detaches from the tree, it is at rest (speed = 0 m/s). After 1 second of falling, it is going at about 10 m/s. After 2 seconds, about 20 m/s. Gravity does not care how heavy the coconut is - a heavy coconut and a small coconut both accelerate at the same rate (ignoring air resistance). This was famously demonstrated by Galileo!

Important: The value of g is the same for ALL objects, regardless of their mass (as long as we ignore air resistance). A cricket ball and a marble, if dropped from the same height at the same time, will hit the ground at the same time.

⚠ Exam Tip

On your IGCSE exam, g = 9.8 m/s² will usually be given in the question or on the data sheet. Some questions say "take g = 10 m/s²" for simpler calculations. Always use the value given in the question!

Supplement

Acceleration

Definition

Acceleration is the rate of change of velocity. In simpler terms, it tells you how quickly the speed is changing.

a = (v - u) / t

a = acceleration, in metres per second squared (m/s²) v = final velocity, in m/s u = initial velocity (starting velocity), in m/s t = time taken for the change, in seconds (s) (v - u) = change in velocity = Δv

The unit m/s² means "metres per second, per second." If a = 2 m/s², it means the object's speed increases by 2 m/s every second. So after 1 second, it is 2 m/s faster; after 2 seconds, 4 m/s faster; and so on.

Worked Example A Bangalore city bus accelerates from rest (0 m/s) to 15 m/s in 10 seconds. Calculate the acceleration.

Step 1: Write down what you know

u = 0 m/s (from rest), v = 15 m/s, t = 10 s

Step 2: Write the formula

a = (v - u) / t

Step 3: Substitute and calculate

a = (15 - 0) / 10 = 15 / 10 = 1.5 m/s²

Answer: a = 1.5 m/s². This means the bus gains 1.5 m/s of speed every second.

Worked Example A cricket ball is bowled at 40 m/s. After hitting the pitch, it slows to 30 m/s. The ball is in contact with the pitch for 0.5 s. Calculate the deceleration of the ball.

Step 1: Write down the values

u = 40 m/s (initial speed), v = 30 m/s (final speed), t = 0.5 s

Step 2: Apply the formula

a = (v - u) / t = (30 - 40) / 0.5 = (-10) / 0.5 = -20 m/s²

Step 3: Interpret the answer

The negative sign means the ball is slowing down (decelerating). The deceleration is 20 m/s².

Answer: Deceleration = 20 m/s² (or acceleration = -20 m/s²). The pitch slowed the ball down rapidly!

Worked Example A Namma Metro train has an acceleration of 1.2 m/s². If it starts from rest, what speed does it reach after 8 seconds?

Step 1: Write down the values

a = 1.2 m/s², u = 0 m/s (from rest), t = 8 s, v = ?

Step 2: Rearrange a = (v - u)/t to find v

v = u + at = 0 + (1.2 x 8) = 9.6 m/s

Answer: v = 9.6 m/s (about 34.6 km/h)

Acceleration from a Speed-Time Graph

Just like speed is the gradient of a distance-time graph, acceleration is the gradient of a speed-time graph.

acceleration = gradient of speed-time graph = Δv / Δt

Δv = change in speed (read from y-axis) Δt = change in time (read from x-axis)

Acceleration = gradient = Δv/Δt = 10 m/s / 4 s = 2.5 m/s²

Worked Example From the speed-time graph above, calculate the acceleration of the object between t = 2 s and t = 6 s.

Step 1: Read values from the graph

At t = 2 s, speed = 5 m/s
At t = 6 s, speed = 15 m/s

Step 2: Calculate the gradient

a = Δv / Δt = (15 - 5) / (6 - 2) = 10 / 4 = 2.5 m/s²

Answer: Acceleration = 2.5 m/s²

Constant vs Changing Acceleration

On a speed-time graph:

Straight line = constant acceleration (the gradient is the same everywhere)
Curved line = changing acceleration (the gradient changes from point to point)

If the curve gets steeper, the acceleration is increasing. If it gets flatter, the acceleration is decreasing.

Deceleration (Negative Acceleration)

Deceleration is simply acceleration in the opposite direction to the motion. It means the object is slowing down. In calculations, deceleration shows up as a negative value of acceleration.

There is nothing special about deceleration - it is just acceleration with a minus sign. The formulas work exactly the same way.

Worked Example An auto-rickshaw travelling at 12 m/s brakes and comes to a stop in 4 seconds. Calculate the deceleration.

Step 1: Write the known values

u = 12 m/s, v = 0 m/s (comes to a stop), t = 4 s

Step 2: Apply the formula

a = (v - u) / t = (0 - 12) / 4 = -12 / 4 = -3 m/s²

Step 3: State the deceleration

The deceleration is 3 m/s² (the magnitude of the acceleration). The negative sign tells us it is slowing down.

Answer: Deceleration = 3 m/s². The auto-rickshaw loses 3 m/s of speed every second until it stops.

Worked Example A train travelling at 30 m/s applies brakes with a deceleration of 2 m/s². How long does it take to stop?

Step 1: Write the known values

u = 30 m/s, v = 0 m/s, a = -2 m/s² (negative because decelerating)

Step 2: Rearrange a = (v-u)/t to find t

t = (v - u) / a = (0 - 30) / (-2) = -30 / -2 = 15 s

Answer: It takes 15 seconds for the train to stop.

⚠ Exam Tip

When a question says "deceleration of 5 m/s²," it means the object is slowing down by 5 m/s every second. In the formula, use a = -5 m/s² (with the negative sign). The negative sign is important because it indicates the direction opposite to motion.

If a question asks "what is the deceleration?" and your calculated acceleration is -3 m/s², the deceleration is 3 m/s² (positive number, because deceleration is the magnitude of the negative acceleration).

Falling Objects: Air Resistance and Terminal Velocity

This is one of the most interesting topics in motion. Let us think about what happens when you drop something from a great height.

Falling Without Air Resistance (in a vacuum)

In a vacuum (like on the Moon, where there is no air), ALL objects fall with the same acceleration of g = 9.8 m/s², regardless of their mass or shape. A feather and a hammer would hit the ground at the same time! (This was actually demonstrated on the Moon by astronaut David Scott in 1971.)

Falling With Air Resistance (in the real world)

In the real world, air resistance (also called drag) plays a big role. Here is what happens when you drop a coconut from a tall coconut tree:

At the moment of release: The coconut is not moving, so there is no air resistance (drag only acts on moving objects). The only force is weight (gravity) acting downward. The coconut accelerates at g = 9.8 m/s².
As it falls faster: Air resistance increases because the coconut is moving faster through the air. Now there are two forces: weight downward and air resistance upward. The net (resultant) force downward decreases, so the acceleration decreases (but the coconut is still speeding up, just more slowly).
Eventually: The air resistance grows until it equals the weight. Now the net force is zero. With no net force, there is no acceleration. The coconut falls at a constant speed. This constant speed is called the terminal velocity.

At terminal velocity: air resistance = weight

Net force = 0, so acceleration = 0 The object falls at constant speed (its maximum speed)

Top: The forces on a falling object at three stages. Bottom: The speed-time graph showing how speed increases then levels off at terminal velocity.

Why a coconut and a leaf fall differently

If you drop a coconut and a leaf from the same height, the coconut hits the ground first. Why?

The coconut is heavy (large weight) and compact (small air resistance relative to its weight). Air resistance takes a long time to match its weight, so it reaches a high terminal velocity and hits the ground quickly.
The leaf is light (small weight) and has a large flat surface area (large air resistance relative to its weight). Air resistance matches its tiny weight almost immediately, so it reaches a low terminal velocity quickly and drifts down slowly.

Falling in a Liquid

The same principle applies when an object falls through a liquid (like a marble dropped in a tall jar of oil). The liquid provides much more resistance (drag) than air, so:

Terminal velocity is reached much sooner
Terminal velocity is much lower than in air

⚠ Exam Tip

When drawing or describing a speed-time graph for a falling object reaching terminal velocity:

1. The graph starts with a steep gradient (high acceleration near g).

2. The gradient decreases (curve gets flatter) as air resistance increases.

3. The graph becomes horizontal (zero gradient) at terminal velocity.

The curve must be smooth - no sudden kinks or straight-line segments! And it must never go beyond the terminal velocity (it cannot overshoot).

🔍 Apply It: Real-World Physics

Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.

A peregrine falcon hunting pigeons over Cubbon Park in Bangalore can reach speeds of 390 km/h in a dive. The Namma Metro Purple Line train has a top speed of 80 km/h.

How many times faster is the falcon? Convert the falcon’s speed to m/s.

▼

Identify the Physics

This is about speed comparison and unit conversion (km/h to m/s).

Work It Out

Step 1: How many times faster?
Ratio = falcon speed ÷ metro speed = 390 ÷ 80 = 4.875 times faster
(Nearly 5 times faster!)

Step 2: Convert 390 km/h to m/s.
To convert km/h → m/s, divide by 3.6 (because 1 km = 1000 m and 1 h = 3600 s, so the factor is 1000 ÷ 3600 = 1 ÷ 3.6).

390 ÷ 3.6 = 108.3 m/s (to 1 d.p.)

That’s about 108 metres every second — roughly the length of a football pitch!

💡 The Aha! Moment

The quick conversion trick: km/h ÷ 3.6 = m/s and m/s × 3.6 = km/h. Write this on your formula sheet! The peregrine falcon is the fastest animal on Earth — it reaches these speeds because it tucks in its wings and dives nearly vertically, reducing air resistance to a minimum.

During the Bangalore Marathon, a runner completes the first 21 km in 2 hours. She then stops at a water station for 3 minutes before running the remaining 21 km in 2 hours 15 minutes.

Calculate her average speed for the entire marathon. Be careful — the rest time counts!

▼

Identify the Physics

This is about average speed, which uses total distance ÷ total time — including any time spent resting!

Work It Out

Step 1: Find total distance.
Total distance = 21 + 21 = 42 km

Step 2: Find total time (convert everything to hours).
First half: 2 h
Rest: 3 min = 3 ÷ 60 = 0.05 h
Second half: 2 h 15 min = 2.25 h
Total time = 2 + 0.05 + 2.25 = 4.30 h

Step 3: Calculate average speed.
Average speed = total distance ÷ total time
= 42 ÷ 4.30
= 9.77 km/h (or about 9.8 km/h)

In m/s: 9.77 ÷ 3.6 = 2.71 m/s

💡 The Aha! Moment

The biggest trap in average speed questions: students calculate the average of the two speeds for each half. That’s WRONG! Average speed is always total distance ÷ total time. And rest time is still time — the clock doesn’t stop when you stop moving. Cambridge examiners include rest periods specifically to test this.

An ISRO PSLV rocket accelerates from rest to 7,600 m/s in 20 minutes to reach orbit. A Namma Metro train accelerates from 0 to 80 km/h in 30 seconds.

What is the average acceleration of each? Which has the greater acceleration?

▼

Identify the Physics

This is about acceleration = change in velocity ÷ time, with unit conversions needed for both cases.

Work It Out

PSLV Rocket:
a = (v − u) ÷ t
v = 7600 m/s, u = 0, t = 20 min = 20 × 60 = 1200 s
a = (7600 − 0) ÷ 1200 = 6.33 m/s²

Namma Metro:
First convert 80 km/h to m/s: 80 ÷ 3.6 = 22.2 m/s
a = (v − u) ÷ t
v = 22.2 m/s, u = 0, t = 30 s
a = (22.2 − 0) ÷ 30 = 0.74 m/s²

The rocket has the greater acceleration — about 8.6 times greater than the Metro train.

💡 The Aha! Moment

Even though the rocket takes 20 minutes and the train takes only 30 seconds, the rocket’s acceleration is much greater because it reaches an enormously higher speed. Always convert ALL values to SI units (m/s and seconds) before calculating acceleration. If you leave time in minutes or speed in km/h, your answer will be wrong!

A coconut falls from a 20 m tall palm tree. Ignore air resistance. Use g = 10 m/s².

How fast is the coconut travelling when it hits the ground? How long does it take to fall? (Hint: you need v² = u² + 2as)

▼

Identify the Physics

This is a free fall problem. The coconut starts from rest (u = 0) and accelerates due to gravity (a = g = 10 m/s²).

Work It Out

Known values:
u = 0 (starts from rest), a = g = 10 m/s², s = 20 m

Part 1: Find the final velocity (v).
v² = u² + 2as
v² = 0² + 2 × 10 × 20
v² = 400
v = √400 = 20 m/s

That’s 72 km/h — as fast as a car on a city road!

Part 2: Find the time (t).
Use v = u + at
20 = 0 + 10 × t
t = 20 ÷ 10 = 2.0 s

The coconut takes just 2 seconds to fall 20 metres.

💡 The Aha! Moment

In free fall (no air resistance), ALL objects accelerate at the same rate regardless of mass. A coconut, a cricket ball, and a feather would all hit the ground at 20 m/s from 20 m — as long as there’s no air resistance. In real life, of course, air resistance matters — which brings us to terminal velocity!

A delivery drone over Electronic City in Bangalore drops two parcels from the same height: Parcel A is a light, large empty carton. Parcel B is a small, dense metal part. Both experience air resistance.

Explain why these two parcels reach different terminal velocities. Which one reaches terminal velocity first?

▼

Identify the Physics

This is about terminal velocity — the maximum speed a falling object reaches when air resistance equals its weight.

Work It Out

What happens as each parcel falls:

When an object falls, it accelerates due to gravity. As it speeds up, air resistance increases (air resistance depends on speed, shape, and surface area).

Terminal velocity is reached when: air resistance = weight. At this point, the resultant force is zero and the object stops accelerating.

Parcel A (light, large carton):
• Low weight (small gravitational force)
• Large surface area (high air resistance, even at low speeds)
• Air resistance matches its small weight at a LOW speed
• Reaches terminal velocity quickly
• Terminal velocity is LOW

Parcel B (small, dense metal part):
• High weight (large gravitational force)
• Small surface area (lower air resistance at the same speed)
• Must reach a much HIGHER speed before air resistance matches its weight
• Takes longer to reach terminal velocity
• Terminal velocity is HIGH

💡 The Aha! Moment

The light carton reaches terminal velocity first, but at a lower speed. The heavy metal part reaches terminal velocity later, but at a higher speed. This is why skydivers spread out their body (large area, lower terminal velocity) to slow down, and tuck in (small area, higher terminal velocity) to speed up — all without changing their weight!

Practice Questions: 1.2

20 multiple choice questions covering speed, velocity, acceleration, graphs, and terminal velocity.

Score 0 / 20

Question 1

An auto-rickshaw travels 9 km in 30 minutes. What is its average speed?

A 18 km/h

B 0.3 km/h

C 270 km/h

D 4.5 km/h

Speed = distance / time = 9 km / 0.5 h = 18 km/h. Remember to convert 30 minutes to 0.5 hours when using km/h. Option B divides time by distance. Option C multiplies.

Question 2

A car has a speed of 72 km/h. What is this in m/s?

A 7.2 m/s

B 20 m/s

C 259.2 m/s

D 1.2 m/s

To convert km/h to m/s, divide by 3.6: 72 / 3.6 = 20 m/s. Option A divides by 10 (wrong conversion). Option C multiplies by 3.6 (backwards!).

Question 3

What is the difference between speed and velocity?

A Speed is faster than velocity

B Velocity includes mass

C Velocity includes direction, speed does not

D There is no difference

Speed is a scalar (magnitude only). Velocity is a vector (magnitude AND direction). This is one of the most commonly asked definitions in IGCSE Physics.

Question 4

On a distance-time graph, a horizontal line represents:

A Constant speed

B Acceleration

C The object is stationary

D The object is moving backwards

On a distance-time graph, horizontal line means distance is not changing with time - the object is stationary (not moving). Constant speed would be a straight line going upward. Do not confuse with a speed-time graph where horizontal means constant speed!

Question 5

On a speed-time graph, the area under the line represents:

A Acceleration

B Distance travelled

C Speed

D Time taken

The area under a speed-time graph = distance travelled. This is because distance = speed x time, and "speed x time" is the area when speed is on the y-axis and time is on the x-axis. The gradient of a speed-time graph gives acceleration.

Question 6

A bus accelerates from 0 to 20 m/s in 8 seconds. What is its acceleration?

A 160 m/s²

B 2.5 m/s²

C 0.4 m/s²

D 28 m/s²

a = (v - u) / t = (20 - 0) / 8 = 2.5 m/s². Option A multiplies speed by time. Option C divides time by speed change (upside down). Option D adds them.

Question 7

A motorcycle travelling at 25 m/s brakes to a stop in 5 seconds. What is the deceleration?

A 5 m/s²

B 125 m/s²

C 0.2 m/s²

D 30 m/s²

a = (v - u)/t = (0 - 25)/5 = -25/5 = -5 m/s². The deceleration is 5 m/s² (the magnitude of the acceleration). The negative sign indicates slowing down, but deceleration is stated as a positive number.

Question 8

The acceleration of free fall g is approximately:

A 9.8 m/s

B 9.8 m/s²

C 9.8 km/h

D 98 m/s²

g ≈ 9.8 m/s². The unit is metres per second squared (m/s²) because acceleration is "change in speed per second." Option A has the wrong unit. Option D has the wrong value.

Question 9

A Shatabdi train travels 300 km at 100 km/h, then 200 km at 80 km/h. What is the average speed for the entire journey?

A 90 km/h

B 100 km/h

C Approximately 90.9 km/h

D 80 km/h

Total distance = 300 + 200 = 500 km. Time for first part = 300/100 = 3 h. Time for second part = 200/80 = 2.5 h. Total time = 5.5 h. Average speed = 500/5.5 = 90.9 km/h. Note: you cannot just average the speeds (90 km/h) because the distances were different!

Question 10

On a speed-time graph, a straight line going upwards represents:

A Constant (uniform) acceleration

B Constant speed

C Increasing acceleration

D Deceleration

A straight line going upward on a speed-time graph has a constant gradient, which means constant acceleration. If the line were curved upward, that would be increasing acceleration. Constant speed would be a horizontal line. Deceleration would be a line going downward.

Question 11

A car travels at a constant speed of 15 m/s for 20 seconds. What distance does it cover?

A 0.75 m

B 35 m

C 300 m

D 1.33 m

s = v x t = 15 x 20 = 300 m. On a speed-time graph, this would be a rectangle of height 15 and width 20, with area 300.

Question 12

An object accelerates uniformly from rest to 10 m/s in 5 seconds. What is the distance travelled?

A 25 m

B 50 m

C 2 m

D 10 m

On a speed-time graph, this is a triangle (starting from 0, going up to 10 m/s over 5 s). Area of triangle = 1/2 x base x height = 1/2 x 5 x 10 = 25 m. Option B would be the area of a rectangle (v x t without the 1/2) - that is wrong because the object was not at 10 m/s the whole time.

Question 13

An object reaches terminal velocity when:

A Gravity stops acting on it

B Air resistance equals its weight

C It stops moving

D Its acceleration equals g

Terminal velocity occurs when air resistance (drag) grows to equal the weight of the object. The net force becomes zero, so acceleration becomes zero, and the object falls at a constant (terminal) speed. Gravity never stops acting! The object does not stop moving - it keeps going at constant speed.

Question 14

The gradient of a distance-time graph gives:

A Speed

B Acceleration

C Distance

D Time

Gradient of d-t graph = change in distance / change in time = speed. This is the definition of speed! Acceleration comes from the gradient of a speed-time graph, not a distance-time graph.

Question 15

A falling object has an acceleration that decreases as it falls (in air). On a speed-time graph, this appears as:

A A straight line going up

B A straight horizontal line

C A curve that gets less steep (flattens out)

D A curve that gets steeper

If acceleration is decreasing, the gradient of the speed-time graph is decreasing. This means the line is curving and getting flatter (less steep). Eventually it becomes horizontal when the object reaches terminal velocity (acceleration = 0).

Question 16

A cricket ball is thrown vertically upwards. At the highest point, what is its velocity and acceleration?

A Velocity = 0, acceleration = 0

B Velocity = 0, acceleration = 9.8 m/s² downward

C Velocity = maximum, acceleration = 0

D Velocity = 9.8 m/s, acceleration = 9.8 m/s²

At the highest point, the ball momentarily stops, so velocity = 0. But gravity still acts on it! The acceleration is still 9.8 m/s² downward (this is what causes it to start falling back down). A common mistake is thinking acceleration is zero at the top - it is not!

Question 17

A car accelerates from 10 m/s to 30 m/s in 5 seconds. What is the acceleration?

A 6 m/s²

B 4 m/s²

C 8 m/s²

D 2 m/s²

a = (v - u)/t = (30 - 10)/5 = 20/5 = 4 m/s². Remember, it is the CHANGE in velocity (30-10 = 20), not the final velocity, that you divide by time. Option A uses 30/5 = 6, which incorrectly uses the final velocity instead of the change.

Question 18

Why does a leaf fall more slowly than a stone (in air)?

A Gravity is weaker on the leaf

B The leaf has less mass so gravity does not pull it

C The leaf reaches a much lower terminal velocity because air resistance matches its small weight quickly

D The stone is not affected by air resistance

The leaf has a large surface area relative to its small weight. Air resistance quickly equals its small weight, giving a very low terminal velocity. The stone has a small surface area relative to its large weight, so it reaches a much higher terminal velocity. Gravity acts on both objects equally (g = 9.8 m/s²), but air resistance makes the difference.

Question 19

A car accelerates uniformly from rest for 10 s, reaching 20 m/s. It then brakes uniformly to rest in 5 s. What is the total distance?

A 300 m

B 150 m

C 200 m

D 100 m

Acceleration phase: triangle area = 1/2 x 10 x 20 = 100 m. Braking phase: triangle area = 1/2 x 5 x 20 = 50 m. Total = 100 + 50 = 150 m. Both phases are triangles on the speed-time graph because the car starts or ends at rest.

Question 20

An object falls from rest. Ignoring air resistance, what is its speed after 3 seconds? (Take g = 10 m/s²)

A 3 m/s

B 10 m/s

C 30 m/s

D 90 m/s

Using v = u + at: v = 0 + (10)(3) = 30 m/s. Since the object starts from rest (u = 0) and accelerates at g = 10 m/s², after 3 seconds its speed is simply g x t = 10 x 3 = 30 m/s. It gains 10 m/s every second.

1.3 Mass and Weight ▼

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Mass and Weight — IGCSE Physics (0625)

Tara, have you ever wondered why astronauts float around inside the International Space Station, yet they still look the same size? Their mass has not changed one bit — but their weight has almost disappeared! Mass and weight sound like the same thing, but they are very different ideas in physics. Understanding this difference is one of the most important things in your IGCSE course, and examiners love to test it. Let us break it down step by step.

What is Mass?

Mass is a measure of the quantity of matter in an object at rest relative to the observer. In simpler words, mass tells you how much "stuff" is inside something.

Think about it this way: a 5 kg bag of Sona Masoori rice from Big Bazaar contains a certain amount of rice grains. Whether you carry that bag in Bangalore, take it on a train to Chennai, or even fly it to the Moon, the bag still contains the exact same rice grains. The mass stays at 5 kg no matter where you go in the universe.

Key facts about mass:

Mass is measured in kilograms (kg)
Mass is a scalar quantity (it has magnitude only, no direction)
Mass does not change with location — your mass is the same in Bangalore, on the Moon, or floating in deep space
Mass is measured using a beam balance (or electronic balance)
Mass tells you how much an object resists being accelerated (this is called inertia)

🧠 Memory Trick

"Mass stays, weight strays." Your mass is loyal — it never changes wherever you travel. Your weight is a wanderer — it changes depending on which planet or moon you are on!

What is Weight?

Weight is the gravitational force acting on an object that has mass. It is the force with which a planet, moon, or star pulls an object towards its centre.

Right now, the Earth is pulling you downwards towards its centre. That pull is your weight. If you stood on the Moon, the Moon would pull you much less strongly (because the Moon is smaller and less massive than the Earth), so your weight would be much less.

Key facts about weight:

Weight is measured in newtons (N) — because weight is a force!
Weight is a vector quantity (it has both magnitude and direction — always directed towards the centre of the planet)
Weight changes depending on where you are — it is different on the Earth, Moon, and Jupiter
Weight is measured using a spring balance (also called a newton meter or force meter)
Weight depends on two things: the object's mass and the gravitational field strength at that location

Mass vs Weight — The Big Comparison

Property	Mass	Weight
What is it?	Amount of matter in an object	Gravitational force on an object
Unit	Kilogram (kg)	Newton (N)
Type of quantity	Scalar (magnitude only)	Vector (magnitude + direction)
Changes with location?	No — same everywhere	Yes — depends on gravitational field strength
Measured with	Beam balance / electronic balance	Spring balance (newton meter)
Direction	None (scalar)	Always towards the centre of the planet
In zero gravity (deep space)	Still the same	Becomes zero

⚠ Exam Tip

A very common exam mistake: writing "mass is measured in newtons" or "weight is measured in kilograms." Remember — mass in kg, weight in N. The examiner will give zero marks if you swap the units. Also, never say weight is "the amount of gravity" — always say it is the gravitational force on an object.

Gravitational Field Strength (g)

Gravitational field strength is defined as the force per unit mass. It tells you how strong gravity is at a particular location. On the surface of the Earth, g = 9.8 N/kg. This means every kilogram of mass experiences a gravitational force of 9.8 N.

The equation linking weight, mass, and gravitational field strength is:

W = m × g

W = weight in newtons (N) m = mass in kilograms (kg) g = gravitational field strength in N/kg

You can rearrange this equation to find any of the three quantities:

g = W / m — to find gravitational field strength
m = W / g — to find mass

Here are the values of g you need to know:

Location	g (N/kg)	What it means
Earth	9.8 (use 10 if the exam says so)	Every 1 kg has a weight of 9.8 N
Moon	1.6	About 1/6 of Earth — you would feel very light!
Jupiter	24.8 (approximately 25)	About 2.5 times Earth — you would feel very heavy!
Deep space (far from any planet)	≈ 0	Weightless — you float!

Important: The gravitational field strength g is numerically equal to the acceleration of free fall. So g = 9.8 N/kg is the same as g = 9.8 m/s². The units are different because they describe different things (force per unit mass vs acceleration), but the number is the same. When a coconut falls from a tree in your backyard, it accelerates at 9.8 m/s² due to gravity!

⚠ Exam Tip

Some exam questions use g = 10 N/kg (or 10 m/s²) to make calculations easier. Always check the question — if it says "take g = 10 N/kg," use 10. If it says nothing, use 9.8 N/kg. Read the question carefully!

Worked Examples

Worked Example 1 A bag of Sona Masoori rice has a mass of 10 kg. Calculate its weight on Earth. (Take g = 9.8 N/kg)

Step 1: Write down what you know

m = 10 kg, g = 9.8 N/kg

Step 2: Choose the formula

W = m × g

Step 3: Substitute and calculate

W = 10 × 9.8 = 98 N

Answer: The weight of the rice bag is 98 N

Worked Example 2 Tara has a mass of 50 kg. Calculate her weight on (a) Earth (g = 9.8 N/kg), (b) the Moon (g = 1.6 N/kg), and (c) Jupiter (g = 24.8 N/kg).

Step 1: Write down what you know

m = 50 kg (this stays the same everywhere!)

Step 2: Use W = m × g for each location

(a) On Earth: W = 50 × 9.8 = 490 N
(b) On the Moon: W = 50 × 1.6 = 80 N
(c) On Jupiter: W = 50 × 24.8 = 1240 N

Step 3: Interpret the results

On the Moon, Tara would feel incredibly light — only 80 N compared to 490 N on Earth! She could jump much higher. On Jupiter, she would feel extremely heavy at 1240 N, and would struggle to even stand up.

Answer: (a) 490 N on Earth, (b) 80 N on the Moon, (c) 1240 N on Jupiter

Worked Example 3 A coconut weighs 14.7 N on Earth. What is its mass? (Take g = 9.8 N/kg)

Step 1: Write down what you know

W = 14.7 N, g = 9.8 N/kg

Step 2: Rearrange the formula

W = m × g, so m = W / g

Step 3: Substitute and calculate

m = 14.7 / 9.8 = 1.5 kg

Answer: The mass of the coconut is 1.5 kg

🧠 Memory Trick

Think of the formula triangle: put W on top, and m and g on the bottom side by side. Cover the quantity you want to find:
• Cover W → you see m × g (multiply)
• Cover m → you see W / g (divide)
• Cover g → you see W / m (divide)
Just like the dosa batter triangle — you need all three ingredients to work together!

Comparing Weights and Masses Using a Balance

There are two main types of balances you need to know about:

1. Beam Balance (for comparing masses): A beam balance works by comparing an unknown mass against a set of known masses. You place the object on one pan and add known masses to the other pan until the beam is level. When the beam is balanced, the unknown mass equals the sum of the known masses. A beam balance gives the same reading everywhere — on Earth, on the Moon, or on Jupiter — because both pans are affected equally by gravity.

2. Spring Balance / Newton Meter (for measuring weight): A spring balance works by stretching a spring. The heavier the object, the more the spring stretches. The scale is calibrated in newtons (N). A spring balance gives different readings on different planets because the weight of the object changes when g changes.

Here is an important scenario to understand: Imagine you use a beam balance to measure 1 kg of mangoes at a fruit stall in KR Market. If you take that beam balance and those mangoes to the Moon, the balance will still show 1 kg, because both the mangoes and the 1 kg weight are pulled less by the Moon's gravity — they are still equal. But if you use a spring balance, the mangoes would show only about 1.6 N on the Moon instead of 9.8 N on Earth!

A spring balance (left) measures weight in newtons and gives different readings on different planets. A beam balance (right) compares masses and gives the same reading everywhere.

⚠ Exam Tip

If a question asks "how would the reading change on the Moon?" — the answer depends on the instrument. A spring balance reading would decrease (because weight decreases). A beam balance reading stays the same (because both sides are affected equally by the weaker gravity).

Supplement (Extended)

Weight as the Effect of a Gravitational Field on a Mass

Every object that has mass creates a gravitational field around itself. A gravitational field is an invisible region of space where a mass experiences a force. The bigger the mass, the stronger the field.

Think of it like this: the Earth is like a giant magnet for mass. It creates an invisible "zone of pull" all around it. Any object with mass that enters this zone gets pulled towards the Earth's centre. That pull is what we call weight.

The concept works like this:

The Earth has mass, so it creates a gravitational field around it
When an object (like a cricket ball) is placed in this field, the field exerts a force on the ball
This force is the ball's weight
The strength of this field is called gravitational field strength (g)
Closer to the Earth's surface, the field is stronger; farther away, it weakens

Imagine you are standing at Lalbagh Botanical Garden. The Earth's gravitational field is pulling you downwards with a certain force. Now imagine you are on top of Mount Everest — you are farther from the Earth's centre, so the field is very slightly weaker, and your weight is very slightly less. But the change is tiny because even Everest's height is small compared to the Earth's radius.

Gravitational field strength g is defined as:

g = W / m

g = gravitational field strength (N/kg) W = weight (N) m = mass (kg)

This tells us that g is the force acting on each kilogram of mass. On Earth, g ≈ 9.8 N/kg, meaning every kilogram experiences a pull of 9.8 newtons.

Worked Example (Supplement) An auto-rickshaw has a mass of 300 kg. On Earth, its weight is 2940 N. Calculate the gravitational field strength and verify that it equals the acceleration of free fall.

Step 1: Write down what you know

m = 300 kg, W = 2940 N

Step 2: Use the formula

g = W / m = 2940 / 300 = 9.8 N/kg

Step 3: Interpret

g = 9.8 N/kg. This is the same as 9.8 m/s², which is the acceleration of free fall on Earth. So if you dropped a spanner from the auto-rickshaw, it would accelerate towards the ground at 9.8 m/s².

Answer: g = 9.8 N/kg, which is equivalent to the acceleration of free fall (9.8 m/s²)

🔍 Apply It: Real-World Physics

Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.

An ISRO astronaut has a mass of 70 kg on Earth (g = 10 N/kg). She travels to the Moon where g = 1.6 N/kg. Her bathroom scale (which actually measures weight in newtons but displays “kg”) shows a reading on the Moon.

What does the scale display on the Moon? Has her mass changed? If she tries to push a heavy 200 kg experiment module on the Moon, will it be easier to get it moving than on Earth?

▼

Identify the Physics

This is about the difference between mass (how much matter — stays the same everywhere) and weight (the gravitational force — changes with g). It also tests inertia, which depends on mass, not weight.

Work It Out

Step 1: Find her weight on the Moon.
W = m × g = 70 × 1.6 = 112 N

Step 2: What does the scale display?
On Earth, the scale is calibrated so that a weight of 700 N (70 × 10) displays “70 kg.” It divides the force by 10 to show “kg.”
On the Moon, the spring compresses with only 112 N of force. The scale divides by 10 and displays: 112 ÷ 10 = 11.2 kg.
This is WRONG — her mass is still 70 kg! The scale gives a false reading because it was calibrated for Earth’s gravity.

Step 3: Has her mass changed?
No. Mass is a property of matter — she has the same number of atoms whether she’s on Earth, the Moon, or floating in deep space. Her mass is always 70 kg.

Step 4: Pushing the 200 kg module on the Moon.
The module weighs less on the Moon (200 × 1.6 = 320 N vs 200 × 10 = 2000 N on Earth), so it’s easier to lift. But to push it horizontally on a frictionless surface, you still need to overcome its inertia, which depends on mass (200 kg), not weight. The resistance to getting it moving is exactly the same. In practice, if there’s friction, the lower weight means less friction, so it would be somewhat easier to slide.

💡 The Aha! Moment

A bathroom scale is a spring balance in disguise — it measures weight (force), not mass. That’s why it gives wrong readings on other planets. A beam balance would still show 70 kg on the Moon because it compares masses. And here’s the key insight: inertia depends on mass, not weight. A bowling ball is just as hard to get rolling on the Moon as on Earth!

At a vegetable market in KR Market, Bangalore, a vendor uses a beam balance to weigh tomatoes. His competitor next door uses a spring balance. Both give 2 kg for the same bag of tomatoes on Earth.

If both vendors somehow set up shop on Mars (g = 3.7 N/kg), which balance would still give the correct mass reading and which would give a wrong reading? Why?

▼

Identify the Physics

This is about how different instruments measure mass: a beam balance compares masses (so gravity cancels out), while a spring balance measures weight (force) and converts it to “mass” using Earth’s g.

Work It Out

Beam balance on Mars:
The beam balance compares the tomatoes on one side with standard 2 kg masses on the other. On Mars, BOTH sides experience the same reduced gravity. The tomatoes pull down with 2 × 3.7 = 7.4 N and the standard masses pull down with 2 × 3.7 = 7.4 N. They still balance! Reading: 2 kg — CORRECT.

Spring balance on Mars:
On Earth, 2 kg of tomatoes weigh 2 × 10 = 20 N, and the scale shows “2 kg.”
On Mars, the same tomatoes weigh 2 × 3.7 = 7.4 N. The spring stretches much less. The scale (calibrated for Earth) divides by 10 and shows: 7.4 ÷ 10 = 0.74 kg — WRONG.

The tomatoes still have a mass of 2 kg, but the spring balance thinks they’re lighter because gravity is weaker.

💡 The Aha! Moment

This is a favourite Cambridge question! The beam balance works anywhere in the universe because it compares masses — gravity affects both sides equally. The spring balance only works correctly where it was calibrated. Remember: beam balance = mass, spring balance = weight.

A cricket ball has a mass of 0.16 kg. When Jasprit Bumrah bowls it at 140 km/h, it follows a curved trajectory through the air. On the ground, the ball weighs 1.6 N (using g = 10 N/kg).

Does the ball’s weight change while it’s in the air? What is its weight at the highest point of its trajectory? What DOES change as it goes up and comes down?

▼

Identify the Physics

This tests whether you understand that weight is constant near Earth’s surface (g doesn’t change over a few metres), even though velocity and direction change during projectile motion.

Work It Out

Weight during flight:
W = m × g = 0.16 × 10 = 1.6 N throughout the entire flight.
Whether the ball is going up, at its highest point, or coming down, g is still 10 N/kg near Earth’s surface. The ball only rises a few metres — g doesn’t measurably change over such a small height.

What DOES change?
• Velocity changes — the ball slows down as it rises (gravity decelerates it) and speeds up as it falls (gravity accelerates it).
• Direction changes — the ball follows a curved path.
• Kinetic energy converts to gravitational potential energy and back.

But mass, weight, and gravitational field strength all stay the same.

💡 The Aha! Moment

Students often confuse “feeling lighter” with “weighing less.” The cricket ball doesn’t become lighter at the top of its arc — gravity pulls on it with exactly the same force the entire time. Weight only changes if you go to a different planet or move far from Earth’s surface. Near the ground, weight is constant.

The gravitational field strength on Jupiter is about 25 N/kg. Tara weighs 500 N on Earth (where g = 10 N/kg).

What is Tara’s mass? What would she weigh on Jupiter? Could she even stand up?

▼

Identify the Physics

This is about using W = m × g to calculate mass from weight on one planet, then using that mass to find weight on another planet.

Work It Out

Step 1: Find Tara’s mass.
W = m × g, so m = W ÷ g = 500 ÷ 10 = 50 kg

Step 2: Find her weight on Jupiter.
W = m × g = 50 × 25 = 1250 N

Step 3: Could she stand up?
On Jupiter, she would feel like she’s carrying an extra 75 kg on her back (total effective “felt mass” of 125 kg in Earth terms). Her muscles are used to supporting 500 N — now they must support 1250 N, which is 2.5 times her Earth weight. Standing would be extremely difficult. Walking would be nearly impossible. Even breathing would be much harder because her rib cage would feel 2.5 times heavier.

💡 The Aha! Moment

The two-step calculation (find mass first, then weight on new planet) is the most common exam pattern for mass and weight questions. Always remember: mass is the “bridge” between planets — it stays constant. Find it first, then multiply by the new g. And Jupiter really would crush you — astronauts train at just 3g (30 N/kg) and find it exhausting!

An astronaut on the International Space Station (ISS) is floating around, seemingly “weightless.” A common belief is that there’s no gravity up there. But the ISS orbits at only 408 km above Earth, where g is still about 8.7 N/kg — barely less than 10 N/kg on the surface.

If gravity is almost as strong at ISS altitude, why does the astronaut float? Is she truly weightless?

▼

Identify the Physics

This is about the difference between true weightlessness (no gravity at all) and apparent weightlessness (free fall — everything falls together so you feel no contact force).

Work It Out

Is there gravity at the ISS?
Yes! At 408 km altitude, g ≈ 8.7 N/kg. If the astronaut has a mass of 70 kg, her weight is 70 × 8.7 = 609 N. She definitely has weight.

So why does she float?
The ISS is in free fall around the Earth. It’s falling towards Earth constantly, but it’s moving sideways so fast (~28,000 km/h) that it keeps missing! The astronaut, the ISS, and everything inside are all falling at exactly the same rate.

Think of it this way: imagine you’re in a lift and the cable snaps. You and the lift both fall at the same rate. You’d float inside the lift — not because gravity disappeared, but because the floor is falling away beneath you just as fast as you’re falling. There’s no contact force pushing up on you, so you feel weightless.

This is apparent weightlessness, not true weightlessness.

💡 The Aha! Moment

This is one of the biggest misconceptions in physics! The ISS astronaut is NOT weightless — she has almost the same weight as on Earth. She’s in continuous free fall. The feeling of weightlessness comes from the absence of a contact force (nothing pushing on you), not the absence of gravity. If gravity actually disappeared, the ISS would fly off in a straight line into space!

Practice Questions: 1.3 Mass and Weight

20 multiple choice questions — tap an option to check your answer

Your Score 0 / 20

Question 1

What is the SI unit of mass?

A Newton (N)

B Kilogram (kg)

C Metre (m)

D Gram (g)

Mass is measured in kilograms (kg). Newton is the unit for force/weight. Gram is a sub-unit (1 kg = 1000 g), but the SI unit is kilogram.

Question 2

What is weight?

A The amount of matter in an object

B The size of an object

C The gravitational force acting on an object that has mass

D The density of an object

Weight is the gravitational force on an object. Option A describes mass, not weight.

Question 3

A cricket ball has a mass of 0.16 kg. What is its weight on Earth? (g = 10 N/kg)

A 0.16 N

B 1.6 N

C 16 N

D 160 N

W = m × g = 0.16 × 10 = 1.6 N. Remember to use the formula and check your decimal places.

Question 4

Which instrument is used to measure weight?

A Beam balance

B Ruler

C Spring balance (newton meter)

D Thermometer

A spring balance (newton meter) measures weight in newtons. A beam balance compares masses in kilograms.

Question 5

A bag of mangoes has a mass of 3 kg on Earth. What is its mass on the Moon?

A 0.5 kg

B 0.3 kg

C 3 kg

D 18 kg

Mass does not change with location. The mangoes contain the same amount of matter whether on Earth, the Moon, or anywhere else. Mass = 3 kg everywhere.

Question 6

Which of the following is a vector quantity?

A Mass

B Weight

C Temperature

D Volume

Weight is a force, and all forces are vectors (they have both magnitude and direction). Weight acts downwards towards the centre of the planet. Mass, temperature, and volume are all scalars.

Question 7

An object weighs 49 N on Earth (g = 9.8 N/kg). What is its mass?

A 49 kg

B 5 kg

C 480.2 kg

D 4.9 kg

m = W / g = 49 / 9.8 = 5 kg. Always rearrange the formula when finding mass from weight.

Question 8

On the Moon, g = 1.6 N/kg. An astronaut has a mass of 80 kg. What is the astronaut's weight on the Moon?

A 50 N

B 128 N

C 784 N

D 800 N

W = m × g = 80 × 1.6 = 128 N. On the Moon, the astronaut weighs much less than on Earth (where they would weigh 80 × 9.8 = 784 N).

Question 9

A beam balance shows that a gold chain has a mass of 50 g on Earth. If the same measurement is done on Mars, the beam balance will show:

A More than 50 g

B Less than 50 g

C Exactly 50 g

D Zero

A beam balance compares masses. Both pans are affected equally by gravity, so the reading stays 50 g regardless of the planet. If it balanced on Earth, it will balance on Mars too.

Question 10

Gravitational field strength is defined as:

A Mass per unit force

B Force per unit mass

C Force per unit volume

D Mass per unit volume

Gravitational field strength g = W/m, which is force per unit mass, measured in N/kg. Option D describes density, not gravitational field strength.

Question 11

The value of g on the surface of the Earth is approximately:

A 1.6 N/kg

B 6.7 N/kg

C 9.8 N/kg

D 24.8 N/kg

g on Earth is approximately 9.8 N/kg. 1.6 N/kg is the Moon, and 24.8 N/kg is Jupiter. You must memorise these values!

Question 12

A BMTC bus has a mass of 12 000 kg. What is its weight on Earth? (g = 10 N/kg)

A 1 200 N

B 12 000 N

C 120 000 N

D 1 200 000 N

W = m × g = 12 000 × 10 = 120 000 N. When working with large numbers, count your zeros carefully!

Question 13

The gravitational field strength g is numerically equal to:

A The mass of the planet

B The acceleration of free fall

C The weight of the planet

D The density of the planet

g = 9.8 N/kg is numerically equal to the acceleration of free fall, 9.8 m/s². They have different units but the same numerical value.

Question 14

An object has a weight of 40 N on a planet where g = 8 N/kg. What is the mass of the object?

A 320 kg

B 5 kg

C 48 kg

D 0.2 kg

m = W / g = 40 / 8 = 5 kg. Always rearrange the equation to solve for the unknown.

Question 15

A spring balance reads 20 N when a box hangs from it on Earth. The same box is taken to the Moon (g = 1.6 N/kg). The spring balance will read approximately:

A 20 N

B 12 N

C 3.3 N

D 0 N

First find the mass: m = W/g = 20/9.8 ≈ 2.04 kg. On the Moon: W = 2.04 × 1.6 ≈ 3.3 N. The spring balance reading decreases because weight depends on g.

Question 16

Mass is best described as:

A The force of gravity on an object

B A measure of the quantity of matter in an object at rest relative to the observer

C How heavy an object feels

D The volume of an object

The IGCSE definition of mass is: a measure of the quantity of matter in an object at rest relative to the observer. "How heavy something feels" describes weight, not mass.

Question 17

Two identical boxes are placed on a beam balance on Jupiter. The balance is level. If the same boxes are placed on the beam balance on Earth, the balance will be:

A Level (balanced)

B Tilted to the left

C Tilted to the right

D Impossible to determine

Identical boxes have identical masses. A beam balance compares masses, and since both sides are equal, the balance stays level on any planet. Gravity affects both sides equally.

Question 18

An astronaut floats freely inside the International Space Station. Which statement is correct?

A Both mass and weight are zero

B Mass is zero but weight is not zero

C Mass stays the same but weight is approximately zero

D Both mass and weight stay the same as on Earth

Mass never changes — the astronaut still has the same amount of matter. But in orbit, the astronaut is in free fall, so they experience apparent weightlessness (weight is approximately zero). Mass is always conserved.

Question 19

On Planet X, a 4 kg object has a weight of 36 N. What is the gravitational field strength on Planet X?

A 144 N/kg

B 9 N/kg

C 32 N/kg

D 40 N/kg

g = W / m = 36 / 4 = 9 N/kg. Planet X has a slightly lower gravitational field strength than Earth (9.8 N/kg).

Question 20

A Namma Metro train has a mass of 250 000 kg on Earth (g = 9.8 N/kg). If it could be transported to Jupiter (g = 25 N/kg), its weight would increase by a factor of approximately:

A 1 (no change)

B 2.5

C 6

D 25

The factor = g(Jupiter) / g(Earth) = 25 / 9.8 ≈ 2.55, so approximately 2.5 times. The weight would go from about 2 450 000 N on Earth to about 6 250 000 N on Jupiter. Mass stays at 250 000 kg on both planets.

1.4 Density ▼

Watch on YouTube

Density — IGCSE Physics (0625)

Tara, have you ever wondered why a small gold earring feels so heavy in your hand, but a big slab of thermocol (polystyrene) feels super light, even though the thermocol is much bigger? The answer is density! Gold packs a huge amount of mass into a tiny volume, while thermocol has very little mass spread out over a large volume. Density is one of the most useful ideas in physics, and it helps explain everything from why oil floats on water in your kitchen to why massive ships made of steel can float on the ocean.

What is Density?

Density is defined as mass per unit volume. It tells you how much mass is packed into each unit of volume. If a material has a high density, it means a lot of mass is squeezed into a small space. If a material has a low density, the mass is spread out over a large space.

ρ = m / V

ρ (rho) = density in kg/m³ or g/cm³ m = mass in kg or g V = volume in m³ or cm³

You can rearrange this equation to find mass or volume:

m = ρ × V — to find mass when you know density and volume
V = m / ρ — to find volume when you know mass and density

The Greek letter ρ (pronounced "rho") is used for density. It is not the letter "p"!

🧠 Memory Trick

Use the density triangle: put m on top, and ρ and V on the bottom. Cover what you want to find:
• Cover m → ρ × V
• Cover ρ → m / V
• Cover V → m / ρ
Think of it as: "Mangoes Rest on Vines" — M on top, R (ρ) and V on the bottom!

Common Density Values

Material	Density (kg/m³)	Density (g/cm³)	Everyday Example
Air (at sea level)	1.2	0.0012	The air you breathe
Cork	240	0.24	Bulletin board material
Wood (teak)	650	0.65	Furniture from your home
Coconut oil	920	0.92	Used for cooking in South India
Water	1 000	1.00	Your drinking water
Aluminium	2 700	2.70	Pressure cooker, idli plates
Iron / Steel	7 800	7.80	Dosa tawa, iron kadai
Gold	19 300	19.30	Gold jewellery from Tanishq

⚠ Exam Tip

Unit consistency is critical! If mass is in kg, volume must be in m³ to get density in kg/m³. If mass is in g, volume must be in cm³ to get density in g/cm³. A common mistake is mixing units. Remember: 1 m³ = 1 000 000 cm³ (that is 100 × 100 × 100). Also, the density of water is exactly 1 g/cm³ or 1000 kg/m³ — a very useful reference to remember!

Worked Examples

Worked Example 1 A block of ghee (clarified butter) has a mass of 460 g and a volume of 500 cm³. Calculate its density and determine whether it will float or sink in water.

Step 1: Write down what you know

m = 460 g, V = 500 cm³

Step 2: Use the density formula

ρ = m / V = 460 / 500 = 0.92 g/cm³

Step 3: Compare with water

Water has a density of 1.00 g/cm³. Since 0.92 g/cm³ < 1.00 g/cm³, the ghee is less dense than water.

Answer: The density of ghee is 0.92 g/cm³. It will float on water (which is why you see ghee floating in dal or sambar!).

Worked Example 2 A silver anklet (payal) has a density of 10 500 kg/m³ and a volume of 4.0 cm³. Calculate its mass in grams.

Step 1: Write down what you know

ρ = 10 500 kg/m³, V = 4.0 cm³. We need to make units consistent.

Step 2: Convert density to g/cm³

10 500 kg/m³ = 10.5 g/cm³ (divide by 1000 to convert kg/m³ to g/cm³)

Step 3: Use m = ρ × V

m = 10.5 × 4.0 = 42 g

Answer: The mass of the silver anklet is 42 g

Worked Example 3 A teak wood plank has a mass of 3.25 kg and a density of 650 kg/m³. Calculate its volume in cm³ and determine whether it will float in water.

Step 1: Write down what you know

m = 3.25 kg, ρ = 650 kg/m³

Step 2: Use V = m / ρ

V = 3.25 / 650 = 0.005 m³

Step 3: Convert to cm³

0.005 m³ × 1 000 000 = 5 000 cm³

Step 4: Will it float?

Density of teak = 650 kg/m³. Density of water = 1 000 kg/m³. Since 650 < 1 000, teak is less dense than water, so it will float.

Answer: Volume = 5 000 cm³. The teak wood will float on water because its density (650 kg/m³) is less than that of water (1 000 kg/m³).

How to Determine Density Experimentally

The IGCSE exam loves to ask about the practical methods for finding density. There are three scenarios you need to know, Tara. Let us go through each one carefully.

1. Density of a Regularly Shaped Solid (e.g., a metal cuboid)

If the solid has a regular shape (cube, cuboid, cylinder, sphere), you can calculate its volume using measurements and a formula.

Method:

Measure the mass of the solid using an electronic balance (in grams)
Measure the dimensions using a ruler or vernier calliper:
- For a cuboid: measure length (l), width (w), and height (h). Volume = l × w × h
- For a cylinder: measure radius (r) and height (h). Volume = πr²h
- For a sphere: measure radius (r). Volume = (4/3)πr³
Calculate density: ρ = m / V

Example: To find the density of an aluminium idli plate, you could measure its mass on a balance (say 180 g), measure its dimensions with a ruler and calculate the volume (say 72 cm³), then divide: ρ = 180 / 72 = 2.5 g/cm³.

2. Density of an Irregularly Shaped Solid that Sinks in Water (e.g., a stone idol)

If the solid has an irregular shape (like a stone Ganesha idol, a pebble, or a metal key), you cannot calculate its volume from measurements. Instead, you use the displacement method.

Method:

Measure the mass of the object using an electronic balance
Fill a measuring cylinder partially with water and record the initial water level (V₁)
Gently lower the object into the water using a thin string (so it does not splash). The water level rises.
Record the new water level (V₂)
Volume of the object = V₂ − V₁ (this is the volume of water displaced)
Calculate density: ρ = m / (V₂ − V₁)

Why does this work? When you submerge the object, it pushes aside (displaces) a volume of water exactly equal to its own volume. The rise in water level tells you the volume of the object. This is the same idea that Archimedes discovered in his bathtub — and shouted "Eureka!"

Displacement method: The water level rises from 80 cm³ to 100 cm³ when the stone is submerged, so the stone's volume is 20 cm³.

⚠ Exam Tip

When describing the displacement method in an exam, always mention: (1) use a measuring cylinder, (2) record the initial water level, (3) gently lower the object (to avoid splashing), (4) record the new water level, (5) subtract to find volume. If the object is too large for a measuring cylinder, you can use a displacement can (eureka can) instead — the overflow water is collected and its volume measured.

3. Density of a Liquid

Method:

Place an empty measuring cylinder on an electronic balance and record its mass (m₁). Alternatively, you can "tare" (zero) the balance with the empty cylinder on it.
Pour a known volume of the liquid into the measuring cylinder. Read the volume (V) at the bottom of the meniscus at eye level.
Record the new mass of the cylinder + liquid (m₂)
Mass of the liquid = m₂ − m₁
Calculate density: ρ = (m₂ − m₁) / V

Example: To find the density of coconut oil, you could measure an empty measuring cylinder (50 g), pour 100 cm³ of coconut oil into it, and weigh again (142 g). Mass of oil = 142 − 50 = 92 g. Density = 92 / 100 = 0.92 g/cm³.

⚠ Exam Tip

When reading the volume of a liquid in a measuring cylinder, always read from the bottom of the meniscus (the curved surface of the liquid). Your eye should be at the same level as the meniscus to avoid parallax error. This is a favourite question topic!

Floating and Sinking

Here is the golden rule for floating and sinking, Tara:

If the object's density is less than the liquid's density → the object floats
If the object's density is greater than the liquid's density → the object sinks
If the object's density is equal to the liquid's density → the object stays wherever you place it (neutral buoyancy)

Real-life examples:

A coconut floats in water because its overall density (including the air inside) is less than 1 g/cm³
An iron kadai sinks in water because iron has a density of 7.8 g/cm³, which is much greater than water's 1.0 g/cm³
Coconut oil floats on water because its density (0.92 g/cm³) is less than water (1.0 g/cm³) — you can see this when a drop of coconut oil sits on top of water in a glass
A wooden log floats in a river because wood (like teak at 0.65 g/cm³) is less dense than water

🧠 Memory Trick

"Less dense, goes up — More dense, goes down." Think of it like a crowded BMTC bus: if you are lighter (less dense), you get pushed upward by the crowd. If you are heavier (more dense), you sink to the bottom. The lighter material always floats on top of the denser material!

Supplement (Extended)

Liquids Floating on Liquids

The same floating and sinking rule applies to liquids! If you carefully pour different liquids into a tall glass, they will arrange themselves in layers based on density. The least dense liquid floats on top, and the most dense liquid sinks to the bottom.

Example from your kitchen: If you pour honey, water, and coconut oil into a glass, they will form three layers:

Top layer: Coconut oil (ρ = 0.92 g/cm³) — least dense, floats
Middle layer: Water (ρ = 1.00 g/cm³)
Bottom layer: Honey (ρ ≈ 1.42 g/cm³) — most dense, sinks

This also explains why when you mix oil and water, the oil always rises to the top — no matter how much you shake the mixture, the oil floats because it is less dense. You can see this in your amma's kitchen when she makes tadka (tempering)!

Worked Example (Supplement) Four liquids are poured into a tall glass. Their densities are: glycerine (1.26 g/cm³), kerosene (0.82 g/cm³), water (1.00 g/cm³), and mustard oil (0.91 g/cm³). List the liquids in order from bottom to top as they would settle in the glass. A small wooden cube with a density of 0.88 g/cm³ is dropped into the glass. Between which two layers will it float?

Step 1: Arrange liquids by density (highest at bottom)

Bottom: Glycerine (1.26 g/cm³)
Next: Water (1.00 g/cm³)
Next: Mustard oil (0.91 g/cm³)
Top: Kerosene (0.82 g/cm³)

Step 2: Find where the wooden cube sits

The wooden cube has density 0.88 g/cm³.
This is more dense than kerosene (0.82) — so it sinks through kerosene.
But it is less dense than mustard oil (0.91) — so it floats on mustard oil.

Answer: Bottom to top: glycerine, water, mustard oil, kerosene. The wooden cube will float between the kerosene layer and the mustard oil layer.

🔍 Apply It: Real-World Physics

Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.

The Dead Sea has water with a density of about 1240 kg/m³. Normal seawater is about 1025 kg/m³, and freshwater is 1000 kg/m³. The average density of the human body is about 1010 kg/m³.

Explain why people float effortlessly in the Dead Sea but struggle to float in a freshwater swimming pool.

▼

Identify the Physics

This is about the floating condition: an object floats when its density is less than the density of the liquid it’s in, and sinks when its density is greater.

Work It Out

In freshwater (ρ = 1000 kg/m³):
Human body density ≈ 1010 kg/m³. Since 1010 > 1000, the body is denser than freshwater — so you sink (or barely float if you fill your lungs with air, which lowers your average density).

In the Dead Sea (ρ = 1240 kg/m³):
Human body density ≈ 1010 kg/m³. Since 1010 < 1240, the body is less dense than the Dead Sea water — so you float easily. In fact, you float so high that a large portion of your body sticks out above the surface!

How high do you float?
The fraction submerged = body density ÷ liquid density = 1010 ÷ 1240 ≈ 0.81. So about 81% of your body is underwater and 19% sticks out — that’s why people can sit up and read a newspaper in the Dead Sea!

💡 The Aha! Moment

Whether you float or sink depends on one simple comparison: is your density greater or less than the liquid’s density? The Dead Sea is so salty (about 34% salt by mass) that its density is much higher than your body’s. The salt dissolves but adds mass without adding much volume — so the water becomes denser. This is also why it’s easier to float in the sea than in a swimming pool!

A goldsmith in Bangalore’s Avenue Road is given a “gold” chain by a customer who wants to sell it. Gold has a density of 19,300 kg/m³. The chain has a mass of 50 g. He measures its volume by water displacement: it displaces 4.5 cm³ of water.

Is it real gold? If not, what might it actually be made of?

▼

Identify the Physics

This is about using the density formula (ρ = m ÷ V) to identify an unknown material by comparing its density to known values.

Work It Out

Step 1: Convert units.
Mass = 50 g = 0.050 kg
Volume = 4.5 cm³ = 4.5 × 10⁻⁶ m³ (since 1 cm³ = 1 × 10⁻⁶ m³)

Step 2: Calculate density.
ρ = m ÷ V = 0.050 ÷ (4.5 × 10⁻⁶) = 11,111 kg/m³

Or more simply in g/cm³: 50 ÷ 4.5 = 11.1 g/cm³

Step 3: Compare to gold.
Gold: 19,300 kg/m³ (19.3 g/cm³)
This chain: 11,111 kg/m³ (11.1 g/cm³)
It is NOT real gold! The density is far too low.

Step 4: What could it be?
Lead has a density of about 11,340 kg/m³ (11.3 g/cm³) — very close to our calculated value. The chain is likely made of lead (possibly gold-plated to look real).

💡 The Aha! Moment

This is exactly how Archimedes caught the king’s dishonest goldsmith over 2000 years ago! Every pure material has a unique density — it’s like a fingerprint. If the density doesn’t match, the material is fake or impure. In exams, always calculate density and compare it to the table of known densities.

During Bangalore’s hot air balloon festival at Jakkur, a balloon pilot heats the air inside the balloon envelope. The envelope has a volume of 2800 m³. Cold air outside has a density of 1.225 kg/m³ and the hot air inside is heated to a density of 1.097 kg/m³. Use g = 10 N/kg.

Calculate the upward force (buoyancy minus weight of hot air) and determine how much payload the balloon can carry.

▼

Identify the Physics

This is about buoyancy created by a density difference. The balloon floats because the hot air inside is less dense than the cold air outside, creating a net upward force.

Work It Out

Step 1: Find the mass of cold air displaced (this creates the buoyancy force).
Mass of cold air = ρ × V = 1.225 × 2800 = 3430 kg

Step 2: Find the buoyancy force (weight of displaced cold air).
Buoyancy = m × g = 3430 × 10 = 34,300 N

Step 3: Find the weight of the hot air inside.
Mass of hot air = 1.097 × 2800 = 3071.6 kg
Weight of hot air = 3071.6 × 10 = 30,716 N

Step 4: Find the net upward force.
Net force = Buoyancy − Weight of hot air = 34,300 − 30,716 = 3584 N

Step 5: Maximum payload.
This 3584 N of net lift must support the basket, burner, envelope fabric, AND passengers.
Maximum payload mass = 3584 ÷ 10 = 358.4 kg
That’s enough for about 4–5 people plus equipment.

💡 The Aha! Moment

A hot air balloon works because of a tiny density difference — just 0.128 kg/m³! But multiplied by the huge volume (2800 m³), that tiny difference creates over 3500 N of lift. This is why balloons need to be so big. The pilot controls altitude by heating the air more (less dense = more lift) or letting it cool (denser = less lift).

A steel ship floats even though steel has a density of 7800 kg/m³ — much higher than water (1000 kg/m³). A solid steel ball sinks immediately.

Explain this apparent contradiction. If the ship has a total mass of 50,000 tonnes, what volume of water must it displace to float?

▼

Identify the Physics

This is about the difference between the density of the material (steel) and the effective (average) density of the whole object (the ship, which is mostly hollow air space).

Work It Out

Why does a solid steel ball sink?
A solid ball is 100% steel. Its density is 7800 kg/m³, which is much greater than water (1000 kg/m³). Since object density > liquid density, it sinks.

Why does a steel ship float?
A ship is a hollow shell of steel filled with air. Air has a density of only about 1.2 kg/m³. The average density of the ship (steel + air + cargo + everything inside) is calculated using the total mass divided by the total volume of the hull. This average density works out to be less than 1000 kg/m³, so the ship floats.

Volume of water displaced:
For the ship to float, the weight of water displaced must equal the weight of the ship.
Mass of ship = 50,000 tonnes = 50,000,000 kg
Mass of water displaced = 50,000,000 kg
Volume = mass ÷ density = 50,000,000 ÷ 1000 = 50,000 m³

💡 The Aha! Moment

The trick is that floating depends on average density, not material density. By shaping steel into a hollow hull, engineers spread a relatively small mass of steel over a huge volume. The air inside brings the average density well below water’s density. If you crushed the ship into a solid block of steel, it would sink instantly — same mass, much smaller volume, higher density!

In Tara’s chemistry lab, she pours honey (density ≈ 1400 kg/m³), water (1000 kg/m³), vegetable oil (920 kg/m³), and rubbing alcohol (790 kg/m³) into a tall glass. She then drops in a grape (density ≈ 1100 kg/m³), a cherry tomato (density ≈ 950 kg/m³), and a piece of cork (density ≈ 120 kg/m³).

Predict where each liquid layer settles and where each solid object ends up.

▼

Identify the Physics

This is about density layering: liquids that don’t mix will arrange themselves with the densest at the bottom. Solid objects sink through any liquid denser than them and float on any liquid less dense than them — they settle at the boundary between the right layers.

Work It Out

Liquid layers (bottom to top — densest first):
1. Honey (1400 kg/m³) — bottom
2. Water (1000 kg/m³)
3. Vegetable oil (920 kg/m³)
4. Rubbing alcohol (790 kg/m³) — top

Now for each solid object:

Grape (1100 kg/m³): Denser than water (1000) but less dense than honey (1400). It sinks through alcohol, oil, and water, but floats on honey. It settles at the water–honey boundary.

Cherry tomato (950 kg/m³): Denser than oil (920) but less dense than water (1000). It sinks through alcohol and oil, but floats on water. It settles at the oil–water boundary.

Cork (120 kg/m³): Less dense than all four liquids (even rubbing alcohol at 790). It floats right on top of the rubbing alcohol at the very surface.

💡 The Aha! Moment

Each object finds its “Goldilocks zone” — it sinks through liquids less dense than itself and floats on liquids denser than itself. It’s like a density-based elevator: the object stops at the floor where the liquid below is denser and the liquid above is less dense. This is exactly how a hydrometer works to measure liquid density!

Practice Questions: 1.4 Density

20 multiple choice questions — tap an option to check your answer

Your Score 0 / 20

Question 1

Density is defined as:

A Volume per unit mass

B Mass per unit volume

C Weight per unit volume

D Force per unit mass

Density = mass per unit volume (ρ = m/V). Option D is the definition of gravitational field strength, not density.

Question 2

What is the density of water?

A 1 g/m³

B 1 g/cm³

C 10 g/cm³

D 100 g/cm³

The density of water is 1 g/cm³ (or equivalently 1000 kg/m³). This is a key reference value you should memorise.

Question 3

A brass Ganesha idol has a mass of 540 g and a volume of 60 cm³. What is its density?

A 0.11 g/cm³

B 32 400 g/cm³

C 9.0 g/cm³

D 480 g/cm³

ρ = m/V = 540/60 = 9.0 g/cm³. Always divide mass by volume, never multiply.

Question 4

Which of the following materials will float on water?

A Iron (7 800 kg/m³)

B Aluminium (2 700 kg/m³)

C Cork (240 kg/m³)

D Gold (19 300 kg/m³)

An object floats if its density is less than water (1 000 kg/m³). Cork (240 kg/m³) is less dense than water, so it floats. All the other materials are denser than water.

Question 5

To find the volume of an irregularly shaped stone, you should use:

A A ruler to measure its length, width, and height

B A measuring cylinder filled with water and the displacement method

C A spring balance

D A thermometer

For irregularly shaped solids, you cannot use a ruler because the shape has no simple formula. The displacement method (measuring the rise in water level in a measuring cylinder) gives the volume.

Question 6

A measuring cylinder contains 45 cm³ of water. When a metal key is lowered in, the level rises to 52 cm³. What is the volume of the key?

A 45 cm³

B 52 cm³

C 7 cm³

D 97 cm³

Volume of the key = final level − initial level = 52 − 45 = 7 cm³. The key displaces 7 cm³ of water.

Question 7

A gold chain has a mass of 50 g and a density of 19.3 g/cm³. What is its volume?

A 965 cm³

B 0.386 cm³

C 2.59 cm³

D 50 cm³

V = m/ρ = 50/19.3 = 2.59 cm³. Gold is so dense that even 50 g of gold takes up a very small volume!

Question 8

To determine the density of a liquid, you need to measure:

A Only its volume

B Only its mass

C Its mass and its volume

D Its weight and its temperature

Density = mass/volume, so you need both mass (using a balance) and volume (using a measuring cylinder). Temperature is not needed for the basic density calculation.

Question 9

An aluminium cuboid measures 10 cm × 5 cm × 2 cm and has a mass of 270 g. What is its density?

A 0.37 g/cm³

B 2.7 g/cm³

C 27 g/cm³

D 270 g/cm³

Volume = 10 × 5 × 2 = 100 cm³. Density = 270/100 = 2.7 g/cm³. This matches the known density of aluminium.

Question 10

1 m³ is equal to how many cm³?

A 100

B 1 000

C 10 000

D 1 000 000

1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1 000 000 cm³. This is a common conversion that catches students off guard!

Question 11

A steel spoon (density 7 800 kg/m³) is dropped into a pot of honey (density 1 420 kg/m³). The spoon will:

A Float on the surface

B Sink to the bottom

C Stay in the middle

D Dissolve

The steel spoon (7 800 kg/m³) is much denser than honey (1 420 kg/m³), so it will sink. An object sinks in a liquid if its density is greater than the liquid's density.

Question 12

An empty measuring cylinder has a mass of 120 g. When 50 cm³ of a mystery liquid is poured in, the total mass becomes 160 g. What is the density of the liquid?

A 3.2 g/cm³

B 2.4 g/cm³

C 0.8 g/cm³

D 1.2 g/cm³

Mass of liquid = 160 − 120 = 40 g. Density = 40/50 = 0.8 g/cm³. Always subtract the mass of the container first!

Question 13

Which of these is the correct unit for density?

A kg/m²

B kg/m³

C N/m³

D kg × m³

Density = mass/volume. Mass in kg, volume in m³, so density is in kg/m³. The unit kg/m² would be mass per unit area, which is not density.

Question 14

A wooden toy boat (density 0.6 g/cm³) is placed in a tub of water. A lead fishing weight (density 11.3 g/cm³) is then placed in a separate tub of water. Which statement is correct?

A Both float

B Both sink

C The boat floats, the weight sinks

D The boat sinks, the weight floats

Wood (0.6 g/cm³) is less dense than water (1.0 g/cm³), so it floats. Lead (11.3 g/cm³) is much denser than water, so it sinks.

Question 15

A solid cube has sides of 4 cm and a mass of 128 g. What is the density of the material?

A 2.0 g/cm³

B 8.0 g/cm³

C 32 g/cm³

D 0.5 g/cm³

Volume = 4 × 4 × 4 = 64 cm³. Density = 128/64 = 2.0 g/cm³. Remember: for a cube, V = side³.

Question 16

Three liquids are poured into a tall glass and allowed to settle. Liquid P has density 0.8 g/cm³, liquid Q has density 1.2 g/cm³, and liquid R has density 1.0 g/cm³. From top to bottom, the order is:

A P, R, Q

B Q, R, P

C R, P, Q

D P, Q, R

The least dense liquid floats on top, and the most dense sinks to the bottom. Top to bottom: P (0.8) → R (1.0) → Q (1.2).

Question 17

A stone has a mass of 150 g. When placed in a measuring cylinder, the water level rises from 60 cm³ to 110 cm³. What is the density of the stone?

A 1.36 g/cm³

B 2.5 g/cm³

C 3.0 g/cm³

D 0.33 g/cm³

Volume of stone = 110 − 60 = 50 cm³. Density = 150/50 = 3.0 g/cm³.

Question 18

A solid object has a density of exactly 1.0 g/cm³. When placed in water, it will:

A Definitely float

B Definitely sink

C Remain at whatever depth it is placed (neutral buoyancy)

D Dissolve in the water

When an object's density exactly equals the liquid's density, it neither sinks nor floats. It stays at whatever position you place it — this is called neutral buoyancy.

Question 19

A block of iron has a mass of 3.9 kg and a volume of 500 cm³. What is its density in kg/m³?

A 0.0078 kg/m³

B 78 kg/m³

C 7 800 kg/m³

D 780 kg/m³

First, convert volume: 500 cm³ = 500 / 1 000 000 m³ = 0.0005 m³. Then ρ = 3.9 / 0.0005 = 7 800 kg/m³. Alternatively, find density in g/cm³ first: 3900g / 500cm³ = 7.8 g/cm³, then multiply by 1000 to get kg/m³ = 7 800 kg/m³.

Question 20

A scientist has four liquids with the following densities: A = 0.79 g/cm³, B = 1.26 g/cm³, C = 0.92 g/cm³, D = 1.05 g/cm³. A small plastic bead with density 0.95 g/cm³ is dropped into a container holding all four liquids in layers. The bead will settle between:

A Liquid A and Liquid C

B Liquid C and Liquid D

C Liquid D and Liquid B

D It floats on top of all liquids

Arrange layers from top (least dense) to bottom (most dense): A (0.79) → C (0.92) → D (1.05) → B (1.26). The bead has density 0.95. It is denser than C (0.92), so it sinks through C. But it is less dense than D (1.05), so it floats on D. Therefore, it settles between liquid C and liquid D.

1.5 Forces ▼

Watch on YouTube

Forces - IGCSE Physics (0625)

Forces are everywhere in your life, Tara! Every time a BMTC bus brakes, every time a cricket ball is hit for a six, every time you open a door or sit on a see-saw, forces are at work. In this section, you will learn what forces do, how to calculate them, and how objects stay balanced. Let us start!

1.5.1 Effects of Forces

What Can Forces Do?

A force is a push or a pull that acts on an object. Forces are measured in newtons (N). You cannot see a force directly, but you can always see what it does. A force can cause three effects on an object:

Change the speed of an object — speed it up or slow it down. When a BMTC bus driver presses the accelerator, the engine force speeds the bus up. When the driver hits the brakes, friction slows it down.
Change the direction of an object — even if the speed stays the same. When a cricket batsman hits the ball, the bat pushes the ball in a completely new direction. The ball was coming towards the batsman but now flies away towards the boundary.
Change the shape of an object — squash it, stretch it, bend it, or twist it. When you sit on a sofa cushion at home, your weight force compresses the cushion and changes its shape. When you stretch a rubber band, the pulling force changes its shape.

⚠ Exam Tip

When the exam asks "describe the effects of a force," always mention all three: change in speed, change in direction, and change in shape. If you write just one, you lose marks!

Types of Forces

There are many different types of forces. You need to know each one and be able to identify them in diagrams and real-life situations. Let us go through them one by one:

Force	What It Is	Indian Example	Contact or Non-Contact?
Friction	A force that opposes (resists) motion between two surfaces that are touching and sliding past each other	The brake pads on a BMTC bus grip the wheel — friction between the pads and the wheel slows the bus down	Contact
Air Resistance (Drag)	Friction between an object and the air it moves through. It opposes the direction of motion through air.	When you stick your hand out of a moving auto-rickshaw, you feel the air pushing your hand back — that is air resistance	Contact
Tension	The pulling force in a stretched rope, wire, cable, or string	During tug of war at your school sports day, both teams pull the rope. The rope is under tension.	Contact
Normal Contact Force	The support force that a surface pushes back with when an object sits on it. It acts perpendicular (at 90°) to the surface.	Your physics textbook sits on your desk. The desk pushes up on the book with a normal contact force, preventing it from falling through.	Contact
Upthrust	The upward force that a liquid or gas exerts on an object submerged (or partially submerged) in it	When you float in a swimming pool, the water pushes you upward. That upward push is upthrust.	Contact
Weight (Gravitational Force)	The force of gravity pulling an object towards the centre of the Earth. Weight = mass × gravitational field strength (W = mg).	A 1 kg bag of rice weighs about 10 N on Earth, because the Earth pulls it downward with a gravitational force	Non-contact
Electric Force	The force between electrically charged objects. Like charges repel, opposite charges attract.	After rubbing a plastic comb on your hair, the comb can pick up small pieces of paper — that is the electric force	Non-contact
Magnetic Force	The force between magnets or between a magnet and a magnetic material (iron, cobalt, nickel)	Fridge magnets stick to the steel fridge door without touching — magnetic force pulls them	Non-contact
Nuclear Force	The strong force that holds protons and neutrons together inside the nucleus of an atom. It only acts over extremely tiny distances.	The reason the Sun shines and nuclear power plants work — nuclear forces hold the nuclei of atoms together	Non-contact

Contact Forces vs Non-Contact Forces

The forces in the table above fall into two categories:

Contact forces require the objects to be physically touching. Friction, air resistance, tension, normal contact force, and upthrust are all contact forces. The objects must be in physical contact for these forces to act.
Non-contact forces can act across a distance, even through empty space. Gravitational, electric, magnetic, and nuclear forces are non-contact forces. They do not need the objects to be touching.

🧠 Memory Trick

To remember the non-contact forces, think: G.E.M.N. — Gravitational, Electric, Magnetic, Nuclear. Everything else is a contact force.

⚠ Exam Tip

Weight is NOT the same as mass! Mass is the amount of matter in an object (measured in kg). Weight is the gravitational force pulling that mass down (measured in N). A 50 kg person has a weight of about 500 N on Earth.

More everyday Indian examples of each force type:

Friction: When you rub your hands together on a cold Bangalore winter morning to warm them up, friction between your palms generates heat. When a BMTC bus brakes suddenly, friction between the tyres and the road is what actually stops the bus.
Air resistance: When you cycle to school, you feel the wind pushing against your face. The faster you pedal, the stronger the air resistance becomes. This is why professional cyclists crouch low — to reduce air resistance.
Tension: When your mother hangs wet clothes on a clothesline, the weight of the wet clothes pulls the line down. The line stretches slightly and the pulling force inside the rope is tension. The heavier the clothes, the greater the tension in the line.
Normal contact force: Right now, the chair you are sitting on pushes you upward with a normal contact force. If it did not, you would fall straight through the chair! This force is always perpendicular (at 90°) to the surface.
Upthrust: If you have ever tried to push a beach ball underwater in a swimming pool, you felt the water pushing it back up very strongly. That upward push from the water is upthrust. It is also what keeps boats floating.

Free-Body Diagrams

A free-body diagram shows all the forces acting on a single object, drawn as arrows. Each arrow starts from the object and points in the direction the force acts. The length of the arrow shows the size (magnitude) of the force — a longer arrow means a bigger force.

Here is an example: a book sitting on a desk. There are exactly two forces acting on it:

Free-body diagram of a book resting on a desk. The weight pulls down, the normal contact force pushes up. Since the book is stationary, both forces are equal in size (same arrow lengths).

Rules for drawing free-body diagrams:

Draw the object as a simple box or dot.
Draw each force as an arrow starting from the object.
Label every arrow with the name of the force and, if known, its value in newtons.
Make the arrow length proportional to the force size — bigger forces get longer arrows.
Only include forces acting on the object, not forces the object exerts on other things.

Here is a more complex example: a Namma Metro train accelerating along the track.

Free-body diagram of a Namma Metro train accelerating. The driving force (right) is bigger than friction + air resistance (left), so the train speeds up. Weight and normal force balance each other vertically.

Resultant Force

When more than one force acts on an object, we can combine them into a single force called the resultant force. The resultant force is the overall (net) effect of all the individual forces acting on the object.

Forces in the same direction: add them together.

Forces in opposite directions: subtract the smaller from the larger. The resultant force points in the direction of the larger force.

Resultant Force = Sum of forces (considering direction)

Forces in the same direction: add them Forces in opposite directions: subtract them The resultant is measured in newtons (N)

Left: forces in the same direction are added. Right: forces in opposite directions are subtracted. The resultant is in the direction of the larger force.

Worked Example Two boys push a broken-down auto-rickshaw. One pushes with 200 N and the other pushes with 150 N, both in the same direction. What is the resultant force?

Step 1: Identify

Both forces act in the same direction (forward).

Step 2: Add the forces

Resultant = 200 N + 150 N = 350 N

Step 3: State the direction

The resultant force is 350 N in the forward direction (the direction both boys are pushing).

Answer: Resultant force = 350 N forward

Worked Example In a tug of war at Tara's school sports day, Team A pulls with 800 N to the left and Team B pulls with 650 N to the right. What is the resultant force?

Step 1: Identify

The forces act in opposite directions: 800 N to the left, 650 N to the right.

Step 2: Subtract

Resultant = 800 N − 650 N = 150 N

Step 3: State the direction

The resultant is in the direction of the larger force, which is to the left (Team A's direction).

Answer: Resultant force = 150 N to the left. Team A wins!

Worked Example A BMTC bus has a driving force of 5000 N forward, air resistance of 1200 N backward, and road friction of 800 N backward. Find the resultant force.

Step 1: Identify

Forward force: 5000 N. Backward forces: 1200 N (air resistance) + 800 N (friction) = 2000 N backward.

Step 2: Find resultant

Resultant = 5000 N − 2000 N = 3000 N

Step 3: State the direction

The resultant is 3000 N in the forward direction (same as the driving force).

Answer: Resultant force = 3000 N forward (so the bus accelerates)

🧠 Memory Trick

Think of resultant force like a tug of war: if one side pulls harder, that side wins. The resultant is the difference, and it acts in the direction of the stronger side.

Supplement (Extended)

Resultant Force and Acceleration

If the resultant force on an object is not zero, the object will accelerate (change its velocity). The acceleration happens in the direction of the resultant force.

If the resultant force is in the direction of motion, the object speeds up.
If the resultant force is opposite to the direction of motion, the object slows down (decelerates).
If the resultant force is at an angle to the direction of motion, the object changes direction.

This is described by Newton's Second Law, which gives us one of the most important equations in all of physics:

F = m × a

F = resultant force in newtons (N) m = mass in kilograms (kg) a = acceleration in metres per second squared (m/s²)

This equation tells us: the bigger the resultant force, the bigger the acceleration. And the bigger the mass, the smaller the acceleration (heavier objects are harder to speed up).

You can rearrange this equation three ways:

F = m × a — to find the force
a = F / m — to find the acceleration
m = F / a — to find the mass

Worked Example A Namma Metro train has a mass of 40,000 kg and accelerates at 1.2 m/s². Calculate the resultant force acting on the train.

Step 1: Write what you know

m = 40,000 kg, a = 1.2 m/s²

Step 2: Write the formula

F = m × a

Step 3: Substitute and calculate

F = 40,000 × 1.2 = 48,000 N

Answer: F = 48,000 N (or 48 kN)

Worked Example A cricket ball has a mass of 0.16 kg. A bowler applies a force of 40 N to the ball. What is the acceleration of the ball?

Step 1: Write what you know

F = 40 N, m = 0.16 kg

Step 2: Rearrange for acceleration

a = F / m

Step 3: Substitute and calculate

a = 40 / 0.16 = 250 m/s²

Answer: a = 250 m/s² (that is an enormous acceleration — this is why cricket balls fly so fast when bowled!)

Worked Example An auto-rickshaw accelerates at 2.5 m/s² when the engine provides a resultant force of 1000 N. Calculate the mass of the auto-rickshaw.

Step 1: Write what you know

F = 1000 N, a = 2.5 m/s²

Step 2: Rearrange for mass

m = F / a

Step 3: Substitute and calculate

m = 1000 / 2.5 = 400 kg

Answer: m = 400 kg

⚠ Exam Tip

The F in F = ma is the resultant force, not just any single force! If you are given multiple forces, you must find the resultant first, then use F = ma. Also, make sure mass is in kg and acceleration is in m/s² before you calculate.

1.5.2 Turning Effect of Forces

What Is a Moment?

Have you ever tried to open a heavy door? You probably pushed near the handle, far from the hinges. If you tried pushing near the hinges instead, you would need a much bigger force — the door barely moves! This is because of the turning effect of a force, which we call the moment of the force.

The moment of a force is a measure of its ability to make an object rotate (turn) about a point called the pivot (or fulcrum). The moment depends on two things:

The size of the force — a bigger force creates a bigger turning effect.
The perpendicular distance from the pivot to the line of action of the force — the farther from the pivot, the bigger the turning effect.

Moment = Force × Perpendicular Distance from Pivot

Moment is measured in newton-metres (Nm) Force (F) is measured in newtons (N) Perpendicular distance (d) is measured in metres (m)

A lever with a pivot (fulcrum) in the middle. The moment of the force about the pivot depends on the force AND the distance from the pivot.

Important: The distance must be the perpendicular (at right angles) distance from the pivot to the line along which the force acts. If the force is applied at an angle, only the perpendicular component counts.

Worked Example Tara pushes a classroom door with a force of 15 N. The door handle is 0.8 m from the hinges (pivot). Calculate the moment about the hinges.

Step 1: Write what you know

Force = 15 N, Distance from pivot = 0.8 m

Step 2: Write the formula

Moment = Force × perpendicular distance

Step 3: Calculate

Moment = 15 × 0.8 = 12 Nm

Answer: Moment = 12 Nm

Worked Example A mechanic uses a spanner to tighten a bolt on a Bajaj auto-rickshaw. He applies a force of 60 N at the end of the spanner, which is 0.25 m long. What is the moment?

Step 1: Write what you know

Force = 60 N, Distance = 0.25 m

Step 2: Write the formula

Moment = Force × perpendicular distance

Step 3: Calculate

Moment = 60 × 0.25 = 15 Nm

Answer: Moment = 15 Nm

Worked Example A see-saw at Cubbon Park needs a moment of 200 Nm to turn. If a child sits 1.6 m from the pivot, what force (weight) must the child have?

Step 1: Write what you know

Moment = 200 Nm, Distance = 1.6 m

Step 2: Rearrange the formula

Force = Moment / distance = 200 / 1.6

Step 3: Calculate

Force = 125 N

Answer: The child must weigh 125 N (that is a mass of about 12.5 kg)

The Principle of Moments

The principle of moments states:

For a body in equilibrium (balanced), the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about that same point.

In simpler words: if a see-saw is balanced, the turning effect pushing it clockwise equals the turning effect pushing it anticlockwise. They cancel each other out exactly.

Sum of Clockwise Moments = Sum of Anticlockwise Moments

This is the condition for rotational balance All moments are calculated about the same pivot point

A see-saw with two children of different weights. For balance, the heavier child sits closer to the pivot and the lighter child sits farther away.

Worked Example Two children sit on a see-saw. Child A weighs 400 N and sits 2.0 m from the pivot. Child B weighs 500 N. How far from the pivot must Child B sit for the see-saw to balance?

Step 1: Write the principle of moments

Clockwise moment = Anticlockwise moment

Step 2: Identify moments

Child A creates an anticlockwise moment: 400 × 2.0 = 800 Nm
Child B creates a clockwise moment: 500 × d

Step 3: Set them equal and solve

500 × d = 800
d = 800 / 500 = 1.6 m

Answer: Child B must sit 1.6 m from the pivot

Worked Example A uniform metre ruler is balanced on a pivot at the 50 cm mark. A 2 N weight is hung at the 20 cm mark and a 1 N weight is hung at the 30 cm mark. A single weight W is hung at the 80 cm mark. Find W if the ruler is balanced.

Step 1: Find distances from pivot (50 cm mark)

2 N weight is at 20 cm, so distance = 50 − 20 = 30 cm = 0.30 m (left side, anticlockwise)
1 N weight is at 30 cm, so distance = 50 − 30 = 20 cm = 0.20 m (left side, anticlockwise)
W is at 80 cm, so distance = 80 − 50 = 30 cm = 0.30 m (right side, clockwise)

Step 2: Apply principle of moments

Clockwise moments = Anticlockwise moments
W × 0.30 = (2 × 0.30) + (1 × 0.20)
W × 0.30 = 0.60 + 0.20 = 0.80

Step 3: Solve for W

W = 0.80 / 0.30 = 2.67 N (to 3 significant figures)

Answer: W = 2.67 N

Worked Example During Diwali preparations, Tara and her friend use a 3 m wooden plank as a lever to lift a heavy box. The pivot is 0.5 m from the box end. If the box weighs 600 N, what force must Tara apply at the other end to lift it?

Step 1: Find distances from pivot

Box is 0.5 m from pivot (creates clockwise moment).
Tara pushes at the other end: 3.0 − 0.5 = 2.5 m from pivot (creates anticlockwise moment).

Step 2: Apply principle of moments

Tara's force × 2.5 = 600 × 0.5
F × 2.5 = 300

Step 3: Solve

F = 300 / 2.5 = 120 N

Answer: Tara needs to push with 120 N (much less than the 600 N weight — that is the beauty of levers!)

⚠ Exam Tip

Always convert distances to metres before calculating moments! If you are given 30 cm, convert it to 0.30 m. Moments must be in Nm (newton-metres), not N×cm.

Supplement (Extended)

Moments with Forces in More Than One Direction

Sometimes forces do not all act vertically. In these cases, you must use the perpendicular distance from the pivot to the line of action of the force.

The line of action is an imaginary line extending along the direction of the force. The perpendicular distance is the shortest distance from the pivot to this line, measured at right angles (90°).

For example, if you push a door at an angle instead of straight on, the effective moment is reduced because the perpendicular distance from the hinge to the line of push is shorter.

Worked Example A horizontal beam is 2.0 m long and pivoted at one end. A vertical force of 50 N acts downward at the other end, and a horizontal force of 30 N acts to the right at the free end. Calculate the total moment about the pivot.

Step 1: Identify each force and its perpendicular distance

The 50 N vertical force acts at 2.0 m from the pivot. Since the beam is horizontal and the force is vertical, the perpendicular distance is 2.0 m.
The 30 N horizontal force acts at the end of the beam. Since the beam is horizontal and this force is horizontal, its line of action passes through all points at the same height. The perpendicular distance from the pivot to this horizontal line is 0 m (the force acts along the beam direction, but since the force is horizontal and the beam is horizontal, the perpendicular distance is actually the vertical distance, which is 0 m — so this force creates no moment about the pivot if the pivot and the point of application are at the same height).

Step 2: Calculate moments

Moment of 50 N force = 50 × 2.0 = 100 Nm (clockwise)
Moment of 30 N force = 30 × 0 = 0 Nm (since perpendicular distance from pivot is zero for this horizontal force acting along the line of the beam)

Step 3: Find total moment

Total moment = 100 Nm clockwise

Answer: Total moment about the pivot = 100 Nm clockwise. The horizontal force creates no turning effect because its line of action passes through the pivot (zero perpendicular distance).

Worked Example A vertical signboard of weight 200 N is attached to a wall by a horizontal pole 1.5 m long. A wire attached to the end of the pole makes an angle with the wall and exerts a tension of 300 N. The perpendicular distance from the wall to the line of action of the tension is 1.2 m. Is the system in equilibrium?

Step 1: Clockwise moment (about the wall joint)

The weight of 200 N acts downward at the end of the pole.
Perpendicular distance = 1.5 m
Clockwise moment = 200 × 1.5 = 300 Nm

Step 2: Anticlockwise moment

The tension of 300 N has a perpendicular distance of 1.2 m from the pivot.
Anticlockwise moment = 300 × 1.2 = 360 Nm

Step 3: Compare

Clockwise moment = 300 Nm, Anticlockwise moment = 360 Nm. These are not equal.

Answer: No, the system is not in equilibrium. The anticlockwise moment (360 Nm) is greater than the clockwise moment (300 Nm), so the pole would rotate anticlockwise.

1.5.3 Conditions for Equilibrium

What Does Equilibrium Mean?

An object is in equilibrium when it is completely balanced — it is not accelerating and it is not rotating. Think of a book sitting still on your study desk, or a balanced see-saw with nobody going up or down.

For an object to be in equilibrium, two conditions must be met at the same time:

The resultant force must be zero. All the forces cancel each other out — the total force in every direction adds up to zero. This means the object will not start moving or speed up.
The resultant moment about any point must be zero. All the clockwise moments equal all the anticlockwise moments. This means the object will not start rotating.

⚠ Exam Tip

Both conditions must be true at the same time! If the forces balance but the moments do not, the object will spin. If the moments balance but the forces do not, the object will move. For true equilibrium, you need both.

Centre of Gravity

The centre of gravity of an object is the single point where all of its weight appears to act. You can think of it as the "balance point" of the object.

For a regular, uniform object (like a ruler, a brick, or a ball), the centre of gravity is at the geometric centre.
For an irregular object (like an oddly shaped piece of cardboard), the centre of gravity might not be at an obvious point — you need to find it experimentally.
The centre of gravity does not have to be inside the object! For example, the centre of gravity of a ring-shaped bangle is at the centre of the ring, where there is no material.

Finding the Centre of Gravity of an Irregular Shape (Plumb Line Method)

Here is a simple experiment to find the centre of gravity of a thin, flat, irregular piece of card:

Make a small hole near one edge of the card.
Hang the card from a pin through the hole so it can swing freely.
Hang a plumb line (a string with a small weight, like a heavy nut) from the same pin. Wait for it to stop swinging.
Draw a line on the card along the plumb line string. The centre of gravity lies somewhere on this line.
Repeat from a different hole — make another hole in a different edge, hang the card again, and draw a second line along the plumb line.
The centre of gravity is at the point where the two lines cross (intersect). You can do it a third time to check — all three lines should meet at the same point.

Finding the centre of gravity using the plumb line method. Hang the card from different holes, draw lines along the plumb line each time. The lines intersect at the centre of gravity.

🧠 Memory Trick

Think of balancing a cardboard cutout on your fingertip. The one point where it balances perfectly — that is the centre of gravity! The plumb line method just finds this point using gravity itself.

Centre of Gravity and Stability

The position of the centre of gravity affects how stable an object is — that is, how easily it topples over.

Three rules of stability:

A lower centre of gravity makes an object more stable. This is why a BMTC bus (heavy engine low down) is more stable than an auto-rickshaw (higher centre of gravity). The bus is harder to tip over.
A wider base makes an object more stable. A pyramid shape is very stable because it has a wide base. A tall, thin object like a cricket stump is easy to knock over because it has a narrow base.
An object topples when its centre of gravity moves beyond the edge of its base. As long as a vertical line from the centre of gravity falls within the base, the object is stable. The moment the line falls outside the base, the object tips over.

Real-life Indian examples:

A BMTC bus has a low centre of gravity (heavy engine underneath) and a wide wheelbase — very stable, rarely tips over.
An auto-rickshaw is narrower and has a higher centre of gravity — it can feel tippy when turning sharp corners in Bangalore traffic!
A matka (clay water pot) is round at the bottom and must be placed on a ring stand. Without the stand, it rolls and topples because its centre of gravity is above a tiny contact point.
Stacking steel tiffin boxes — when you stack too many, the centre of gravity rises and eventually the stack topples. If you put the heaviest box at the bottom, the centre of gravity stays lower and the stack is more stable.
During Dasara/Dussehra celebrations, tall Ravan effigies are very unstable because they have a high centre of gravity. They need thick, heavy bases and guy ropes to stay upright.

Three objects with different stability. The wider the base and the lower the centre of gravity (CG), the more stable the object is.

Supplement (Extended)

Explaining Stability Using Centre of Gravity and Pivot Point

When you tilt an object, you are effectively rotating it about the edge of its base (which becomes the pivot point). Here is what happens:

When you start tilting, the centre of gravity rises slightly. The weight of the object still acts downward through the centre of gravity.
As long as the vertical line from the centre of gravity falls within the base, the weight creates a restoring moment that pushes the object back to its upright position. The object is stable.
If you tilt it far enough that the vertical line from the centre of gravity falls beyond the edge of the base (beyond the pivot point), the weight now creates a toppling moment that tips the object further over. The object is unstable and will fall.

Why is a low centre of gravity more stable? Because you need to tilt the object through a larger angle before the centre of gravity passes over the edge of the base. With a higher centre of gravity, even a small tilt can move the line of weight outside the base.

Why is a wider base more stable? Because the edge of the base is farther from the centre, so the centre of gravity needs to move a greater horizontal distance to get beyond the base edge. A wider base gives more room.

Indian example: Imagine a loaded bullock cart. If the load is stacked very high, the centre of gravity is high and the cart might topple on a sloping road. If the load is kept low and spread wide, the cart is much more stable. This is exactly why Indian truck drivers are told to distribute heavy loads low and evenly!

Worked Example A rectangular block is 20 cm tall and has a base of 10 cm × 10 cm. Its centre of gravity is at its geometric centre (10 cm up from the base). Through what angle can it be tilted before it topples?

Step 1: Identify the pivot

When tilted, the block rotates about the bottom edge. The centre of gravity is 10 cm up and 5 cm in from the edge (half the base width).

Step 2: Find the critical angle

The block topples when the vertical from the CG passes over the edge. This happens when: tan(angle) = half-base-width / height-of-CG = 5 cm / 10 cm = 0.5

Step 3: Calculate

Angle = arctan(0.5) ≈ 26.6°

Answer: The block can be tilted up to about 26.6° before it topples. A block with a wider base (say 20 cm × 20 cm) would be tilted to arctan(10/10) = 45° before toppling — much more stable!

Worked Example Explain why a double-decker BMTC bus is fitted with heavy batteries under the lower deck floor rather than on the upper deck.

Step 1: Identify the physics concept

This is about stability and centre of gravity.

Step 2: Explain

Placing heavy batteries low down lowers the centre of gravity of the entire bus. A lower centre of gravity means that when the bus tilts (going around corners or on uneven roads), the vertical line from its centre of gravity is less likely to fall outside the base (the distance between the wheels). The weight continues to provide a restoring moment that brings the bus back upright.

Step 3: State the conclusion

With batteries on the upper deck, the centre of gravity would be much higher, and the bus could topple more easily on turns. Keeping them low makes the bus more stable and safer.

Answer: Placing heavy batteries under the lower deck lowers the centre of gravity of the bus, making it more stable. This means the bus can tilt further before the line of action of its weight falls outside its base (wheelbase), reducing the risk of toppling.

⚠ Exam Tip

When answering stability questions in the exam, always mention three things: (1) the position of the centre of gravity, (2) the width of the base, and (3) whether the vertical line from the CG falls within or outside the base. This structure gets you full marks.

🧠 Memory Trick

Remember "Low and Wide = Stable Pride": a low centre of gravity and a wide base make an object stable and hard to topple. Think of a pyramid — it has been standing for thousands of years!

Key Formulas Summary for 1.5

Formula	What It Calculates	Units
Resultant = sum of forces (with direction)	The overall net force on an object	Newtons (N)
F = m × a [Supplement]	Resultant force from mass and acceleration	N = kg × m/s²
Moment = F × d	Turning effect of a force about a pivot	Nm = N × m
Clockwise moments = Anticlockwise moments	Condition for rotational equilibrium (balance)	Nm = Nm

⚠ Exam Tip

Unit conversions to remember: Always convert cm to m before using in moment calculations (divide by 100). Always convert g to kg before using in F = ma (divide by 1000). Forgetting unit conversions is the most common reason students lose marks in forces questions!

🔍 Apply It: Real-World Physics

Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.

On Bangalore’s Outer Ring Road, a BMTC bus (mass 12,000 kg) is moving at a constant speed. The engine provides a driving force of 8000 N.

Since the bus is moving at constant speed, what is the total friction force? If the bus driver increases the engine force to 10,000 N, what is the resultant force and the acceleration?

▼

Identify the Physics

This is about balanced and unbalanced forces. When an object moves at constant speed, the forces are balanced (resultant force = 0). When the driving force increases, the forces become unbalanced and we use F = ma to find the acceleration.

Work It Out

Part 1: Find the friction force.
Constant speed means the resultant force is zero.
So: driving force = friction force
Friction force = 8000 N

Part 2: Find the resultant force when the engine force increases.
The friction force doesn’t change suddenly — it stays at 8000 N.
Resultant force = driving force − friction force
Resultant force = 10,000 − 8000 = 2000 N (forward)

Part 3: Find the acceleration.
F = ma, so a = F ÷ m
a = 2000 ÷ 12,000 = 0.167 m/s² (to 3 s.f.)

💡 The Aha! Moment

“Constant speed” is the magic phrase that tells you forces are balanced. The moment you see it, you know resultant force = 0, which means friction = driving force. This is one of the most common IGCSE force questions — don’t overthink it!

A construction worker at a building site in Whitefield, Bangalore, is using a wheelbarrow. The load weighs 600 N and sits 0.4 m from the wheel (pivot). The worker lifts the handles, which are 1.2 m from the wheel.

What force must the worker apply to lift the wheelbarrow? If he moves the load closer to the wheel (0.3 m away), how does the required force change?

▼

Identify the Physics

This is about the principle of moments (turning effect of a force). The wheel acts as the pivot. For the wheelbarrow to be balanced, clockwise moments must equal anticlockwise moments.

Work It Out

Part 1: Original position (load 0.4 m from pivot).
Clockwise moment (load) = Anticlockwise moment (worker)
600 × 0.4 = F × 1.2
240 = 1.2F
F = 240 ÷ 1.2 = 200 N

Part 2: Load moved closer (0.3 m from pivot).
600 × 0.3 = F × 1.2
180 = 1.2F
F = 180 ÷ 1.2 = 150 N

The required force decreases from 200 N to 150 N when the load is moved closer to the pivot.

💡 The Aha! Moment

Moving the load closer to the pivot reduces the moment it creates, so the worker needs less effort to balance it. This is exactly why wheelbarrows are designed with the wheel (pivot) close to the load — it’s a force multiplier! The same principle explains why door handles are placed far from the hinge.

An engineer at TVS Motor Company in Hosur (near Bangalore) is testing a motorcycle suspension spring. She hangs masses from the spring and records the extension: 2 kg → 4 cm, 4 kg → 8 cm, 6 kg → 12 cm, 8 kg → 16 cm, 10 kg → 22 cm.

Up to what load does the spring obey Hooke’s Law? What is the spring constant? What happened at 10 kg?

▼

Identify the Physics

This is about Hooke’s Law (F = kx). A spring obeys Hooke’s Law when the extension is directly proportional to the force — meaning equal increases in force produce equal increases in extension. When this stops being true, the spring has passed its limit of proportionality.

Work It Out

Step 1: Check the pattern.
2 kg → 4 cm (extension per 2 kg = 4 cm)
4 kg → 8 cm (extension per 2 kg = 4 cm) ✅
6 kg → 12 cm (extension per 2 kg = 4 cm) ✅
8 kg → 16 cm (extension per 2 kg = 4 cm) ✅
10 kg → 22 cm (extension per 2 kg = 6 cm) ❌

The spring obeys Hooke’s Law up to 8 kg (up to a force of 8 × 10 = 80 N).

Step 2: Find the spring constant (using data within Hooke’s Law range).
Convert units: at 2 kg, F = 2 × 10 = 20 N, extension = 4 cm = 0.04 m
k = F ÷ x = 20 ÷ 0.04 = 500 N/m

Step 3: What happened at 10 kg?
The spring extended more than expected (22 cm instead of the predicted 20 cm). This means the spring has passed its limit of proportionality — the spring is being permanently deformed and will not return to its original length.

💡 The Aha! Moment

The trick to spotting when Hooke’s Law breaks down is to check if equal force increases give equal extension increases. As soon as the extension “jumps” (6 cm instead of 4 cm here), you’ve gone past the limit. In an exam, plot the data — the point where the graph curves away from a straight line is the limit of proportionality.

A painter is standing on a uniform plank that rests on two supports (trestles). The plank is 4 m long and weighs 200 N. Support A is at the left end, support B is at the right end. The painter (weight 700 N) stands 1 m from support A.

Calculate the force on each support.

▼

Identify the Physics

This is about the principle of moments with two supports. The plank is uniform, so its weight acts at its centre (2 m from each end). We take moments about one support to find the force on the other.

Work It Out

Step 1: Draw it out mentally.
Support A is at position 0 m. Support B is at position 4 m.
Plank weight (200 N) acts at centre = 2 m from A.
Painter (700 N) stands at 1 m from A.

Step 2: Take moments about support A (to find force at B).
Clockwise moments = Anticlockwise moments
(Weight of plank × distance from A) + (Weight of painter × distance from A) = F_B × 4
(200 × 2) + (700 × 1) = F_B × 4
400 + 700 = 4F_B
1100 = 4F_B
F_B = 1100 ÷ 4 = 275 N

Step 3: Find force at A using equilibrium.
Total upward forces = Total downward forces
F_A + F_B = 200 + 700
F_A + 275 = 900
F_A = 900 − 275 = 625 N

💡 The Aha! Moment

The painter stands closer to support A, so support A takes more of the load (625 N vs 275 N). This makes intuitive sense — if you stand on one end of a seesaw, the support on that end bears most of the weight. Always check: F_A + F_B should equal the total downward force (625 + 275 = 900 N = 200 + 700). If it doesn’t add up, go back and check your working!

In a tug-of-war at Tara’s school Sports Day, Team A pulls with a total force of 2500 N to the left, and Team B pulls with 2400 N to the right. The rope’s mass is 5 kg.

What is the resultant force and in which direction does the rope move? What is the acceleration of the rope?

▼

Identify the Physics

This is about resultant force (two forces in opposite directions) and Newton’s second law (F = ma) to find acceleration.

Work It Out

Step 1: Find the resultant force.
The forces act in opposite directions, so we subtract.
Resultant force = 2500 − 2400 = 100 N to the left (towards Team A)

Step 2: Find the acceleration.
F = ma, so a = F ÷ m
a = 100 ÷ 5 = 20 m/s² to the left

💡 The Aha! Moment

The acceleration seems huge (20 m/s²!) because the rope has such a small mass (5 kg) compared to the unbalanced force (100 N). In reality, we should consider the mass of the students too — but in an exam, use only the mass you’re given. Always state the direction of the resultant force and acceleration — examiners award marks for direction!

Practice Questions: 1.5 Forces

20 multiple choice questions — tap an option to check your answer

Your Score 0 / 20

Question 1

A cricket batsman hits a ball. Which of the following effects does the bat have on the ball?

A It changes only the speed of the ball

B It changes only the direction of the ball

C It changes both the speed and the direction of the ball

D It changes the mass of the ball

When a batsman hits a cricket ball, the force of the bat changes the ball's speed (it was slowing down, now it flies away fast) AND its direction (it was coming towards the batsman, now it goes towards the boundary). Forces cannot change mass.

Question 2

Which of the following is a non-contact force?

A Friction

B Air resistance

C Tension

D Gravitational force

Gravitational force (weight) is a non-contact force — the Earth pulls you downward without touching you. Friction, air resistance, and tension all require physical contact between objects.

Question 3

A physics textbook sits on a desk. The desk exerts an upward force on the book. What is this force called?

A Tension

B Normal contact force

C Upthrust

D Friction

The desk pushes up on the book with a normal contact force. This force acts perpendicular (at 90°) to the surface. Upthrust is the upward force in a fluid (liquid or gas), not from a solid surface.

Question 4

Two forces of 50 N and 30 N act on an object in the same direction. What is the resultant force?

A 20 N

B 30 N

C 50 N

D 80 N

When forces act in the same direction, you add them: 50 + 30 = 80 N. You only subtract when forces are in opposite directions.

Question 5

A BMTC bus has a driving force of 4000 N forward and a total friction force of 4000 N backward. What happens to the bus?

A It accelerates forward

B It moves at a constant speed (or stays stationary)

C It decelerates

D It stops immediately

When the driving force equals the friction force, the resultant force is zero (4000 − 4000 = 0 N). With zero resultant force, the bus either stays stationary or continues at constant speed (Newton's First Law). It does not necessarily stop — it just does not change speed.

Question 6

In a free-body diagram of a falling cricket ball (ignoring air resistance), which forces should be shown?

A Only weight (downward)

B Weight (downward) and normal contact force (upward)

C Weight (downward) and a forward force

D No forces, because it is in the air

A falling ball in free fall (ignoring air resistance) has only one force acting on it: its weight pulling it downward. There is no normal contact force because it is not touching any surface. There is no forward force — objects do not need a force to keep moving (Newton's First Law).

Question 7

A fridge magnet sticks to a steel refrigerator door. What type of force holds it in place against gravity?

A Electric force

B Magnetic force

C Tension

D Upthrust

The magnet is attracted to the steel door by a magnetic force. This magnetic force pulls the magnet towards the door, and friction between the magnet and door surface prevents it from sliding down. The magnetic force is a non-contact force (it works through a small gap too).

Question 8 [Supplement]

A resultant force of 600 N acts on a Namma Metro train car of mass 12,000 kg. What is the acceleration?

A 0.05 m/s²

B 0.5 m/s²

C 5 m/s²

D 50 m/s²

Using a = F / m = 600 / 12,000 = 0.05 m/s². The large mass of the train means even a 600 N force only produces a tiny acceleration.

Question 9

A spanner is 0.3 m long. A force of 40 N is applied at the end, perpendicular to the spanner. What is the moment about the bolt?

A 1.2 Nm

B 12 Nm

C 120 Nm

D 13.3 Nm

Moment = Force × perpendicular distance = 40 × 0.3 = 12 Nm. Always remember: the distance must be in metres and perpendicular to the force.

Question 10

A uniform see-saw is balanced. Child A (weight 300 N) sits 2.0 m from the pivot. Child B sits 1.5 m from the pivot on the other side. What is Child B's weight?

A 200 N

B 300 N

C 400 N

D 450 N

Principle of moments: clockwise moment = anticlockwise moment. So 300 × 2.0 = W × 1.5, which gives 600 = 1.5W, so W = 400 N. The heavier child sits closer to the pivot.

Question 11

For an object to be in equilibrium, which TWO conditions must be satisfied?

A Resultant force = 0 and the object must be stationary

B Resultant force = 0 and resultant moment = 0

C Resultant moment = 0 and the object must have zero speed

D There must be no forces acting on the object at all

Equilibrium requires both: (1) resultant force = 0 (no net push or pull) and (2) resultant moment = 0 (no net turning effect). An object in equilibrium does not have to be stationary — it could be moving at constant velocity.

Question 12

What is the centre of gravity of an object?

A The exact middle of the object

B The heaviest point of the object

C The point where all the weight appears to act

D The point at the base of the object

The centre of gravity is defined as the point where all the weight of the object appears to act. For a uniform object it is at the geometric centre, but for an irregular or non-uniform object it may be elsewhere. It is not always at the "middle" or the "base."

Question 13

To find the centre of gravity of a thin, irregular piece of cardboard, you hang it from a pin and draw a line along a plumb line. Why must you repeat this from a different hole?

A To make the experiment look more scientific

B To reduce parallax error

C Because the centre of gravity lies at the intersection of two (or more) lines

D To calculate an average position

One line tells you the centre of gravity lies somewhere along that line, but you do not know exactly where. A second line from a different hole crosses the first line, and the intersection point is the centre of gravity. A third line can be drawn as a check.

Question 14

Which design feature makes a BMTC bus more stable than an auto-rickshaw?

A The bus has a higher centre of gravity

B The bus has a narrower base

C The bus has a wider base and lower centre of gravity

D The bus is lighter

A BMTC bus has a wider wheelbase (wheels are further apart) and the heavy engine is placed low, lowering the centre of gravity. Both of these features increase stability. The auto-rickshaw is narrower with a relatively higher centre of gravity, making it less stable on turns.

Question 15 [Supplement]

An auto-rickshaw of mass 400 kg has a driving force of 2000 N. The total friction force is 800 N. What is the acceleration?

A 2 m/s²

B 3 m/s²

C 5 m/s²

D 7 m/s²

First find the resultant force: 2000 − 800 = 1200 N forward. Then use a = F / m = 1200 / 400 = 3 m/s². Remember: always use the resultant force in F = ma, not just the driving force!

Question 16

A uniform metre ruler is balanced at the 50 cm mark. A 4 N weight is hung at the 10 cm mark and a 2 N weight is hung at the 25 cm mark. A single weight W is hung at the 90 cm mark. Find W.

A 4.0 N

B 5.25 N

C 6.0 N

D 3.5 N

Anticlockwise moments: 4 × (50 − 10)/100 + 2 × (50 − 25)/100 = 4 × 0.40 + 2 × 0.25 = 1.60 + 0.50 = 2.10 Nm. Clockwise moment: W × (90 − 50)/100 = W × 0.40. Setting equal: W × 0.40 = 2.10, so W = 5.25 N.

Question 17

When you sit on a sofa cushion, the cushion gets compressed. Which effect of a force is this?

A Change in speed

B Change in direction

C Change in shape

D Change in mass

The cushion squashes (changes shape) under your weight. This is the "change in shape" effect of a force. The cushion is not moving or changing direction, so it is not a speed or direction change. Forces never change mass.

Question 18 [Supplement]

A car is moving at constant velocity on a straight road. What can you say about the forces acting on it?

A There are no forces acting on it

B The driving force is greater than friction

C The resultant force is zero

D The resultant force is in the direction of motion

Constant velocity means no acceleration. From F = ma, if a = 0 then F = 0. So the resultant force must be zero. This does NOT mean no forces are acting — there are forces (driving force, friction, weight, normal force), but they all balance out to give a resultant of zero.

Question 19 [Supplement]

A tall Dussehra effigy is tilted slightly. It returns to its upright position because:

A Its centre of gravity is very high

B The line of action of its weight still falls within its base, creating a restoring moment

C There is no moment acting on it

D The wind pushes it back

When tilted slightly, the vertical line from the centre of gravity still falls within the base. This means the weight creates a moment about the pivot (the edge of the base) that rotates the effigy back to its upright position — this is called a restoring moment. If tilted too far, the line of weight falls outside the base and the effigy topples.

Question 20 [Supplement]

An Indian Railways train of mass 500,000 kg decelerates from 20 m/s to 10 m/s in 50 s. What is the resultant braking force?

A 10,000 N

B 50,000 N

C 100,000 N

D 200,000 N

First find the acceleration: a = (v − u) / t = (10 − 20) / 50 = −10 / 50 = −0.2 m/s² (negative because it is slowing down). Then F = ma = 500,000 × 0.2 = 100,000 N. The force is 100,000 N acting opposite to the direction of motion (braking force).

1.6 Momentum ▼

Watch on YouTube

Momentum & Collisions — IGCSE Physics Explained

Supplement (Extended)

Hey Tara! This entire section on momentum is Supplement (Extended) content, which means it appears on your Paper 4 exam. Momentum is one of the most satisfying topics in physics because once you understand it, you can predict what happens in collisions — from cricket balls hitting bats to cars crashing on Bangalore roads. Let us dive in!

What is Momentum?

Imagine two things coming towards you: a cricket ball bowled by Jasprit Bumrah at 140 km/h, and a heavy BMTC bus moving slowly at 5 km/h. Both feel dangerous, right? That is because both have a lot of momentum. Momentum depends on two things: how heavy something is (mass) and how fast it is moving (velocity).

Momentum is a measure of how hard it is to stop a moving object. A heavier object or a faster object has more momentum. Think of it as the "unstoppability" of an object.

p = m × v

p = momentum in kilogram metres per second (kg m/s) m = mass in kilograms (kg) v = velocity in metres per second (m/s)

Key point: Momentum is a vector quantity — it has both size and direction. If an object moves to the right, its momentum is positive. If it moves to the left, its momentum is negative. This becomes very important in collision problems!

🧠 Memory Trick

"Please Move the Van" — p = m × v. The letter p stands for momentum (because m was already taken by mass!).

Worked Example A cricket ball has a mass of 0.16 kg and is bowled at 40 m/s. What is its momentum?

Step 1: Write down what you know

m = 0.16 kg, v = 40 m/s

Step 2: Write the formula

p = m × v

Step 3: Substitute and calculate

p = 0.16 × 40 = 6.4 kg m/s

Answer: The cricket ball has a momentum of 6.4 kg m/s

Worked Example A Rajdhani Express train of mass 500,000 kg moves at 30 m/s. An auto-rickshaw of mass 400 kg moves at 15 m/s. Which has more momentum?

Step 1: Calculate train momentum

p(train) = 500,000 × 30 = 15,000,000 kg m/s

Step 2: Calculate auto momentum

p(auto) = 400 × 15 = 6,000 kg m/s

Step 3: Compare

15,000,000 is much larger than 6,000. The train has about 2,500 times more momentum than the auto-rickshaw!

Answer: The train has far more momentum (15,000,000 kg m/s vs 6,000 kg m/s). This is why trains take such a long distance to stop!

Worked Example A BMTC bus of mass 8,000 kg has a momentum of 120,000 kg m/s. What is its velocity?

Step 1: Write down what you know

m = 8,000 kg, p = 120,000 kg m/s, v = ?

Step 2: Rearrange the formula

p = m × v, so v = p ÷ m

Step 3: Substitute and calculate

v = 120,000 ÷ 8,000 = 15 m/s

Answer: The bus is moving at 15 m/s (about 54 km/h — typical speed on Outer Ring Road!)

Impulse

When a force acts on an object for some time, it changes the object's momentum. This change is called impulse. Think about a cricket batsman — the longer the bat stays in contact with the ball (bigger Δt), and the harder the batsman hits (bigger F), the more the ball's momentum changes.

Impulse = F × Δt = Δ(mv)

Impulse = change in momentum in kg m/s (or N s) F = force in newtons (N) Δt = time for which force acts in seconds (s) Δ(mv) = change in momentum = final momentum − initial momentum

Why do we care about impulse? It explains why catching a cricket ball hurts less if you pull your hands back! By increasing the time (Δt) over which you stop the ball, you reduce the force (F) on your hands. Same change in momentum, but spread over more time = less force = less pain!

⚠ Exam Tip

Impulse is the SAME as change in momentum. The units kg m/s and N s are equivalent. In exams, you might see either one — they mean the same thing!

Worked Example A 0.16 kg cricket ball moving at 30 m/s is hit by a bat and moves back at 20 m/s. The bat is in contact with the ball for 0.02 s. Find (a) the impulse and (b) the force on the ball.

Step 1: Define directions

Let the initial direction (towards batsman) be positive. So initial velocity u = +30 m/s, final velocity v = −20 m/s (it reverses direction).

Step 2: Calculate impulse (change in momentum)

Impulse = Δ(mv) = m(v − u) = 0.16 × (−20 − 30) = 0.16 × (−50) = −8.0 kg m/s

Step 3: Find the force

F = Δ(mv) ÷ Δt = −8.0 ÷ 0.02 = −400 N

Step 4: Interpret

The negative sign means the force acts in the opposite direction to the ball's original motion (the bat pushes the ball back). The magnitude is 400 N.

Answer: (a) Impulse = 8.0 kg m/s (in reverse direction). (b) Force = 400 N on the ball.

Worked Example A Namma Metro train applies a braking force of 50,000 N for 12 seconds. What is the impulse?

Step 1: Write what you know

F = 50,000 N, Δt = 12 s

Step 2: Use impulse formula

Impulse = F × Δt = 50,000 × 12 = 600,000 N s

Answer: The impulse is 600,000 N s (or 600,000 kg m/s). This is the change in the train's momentum during braking.

Conservation of Momentum

This is one of the most powerful ideas in all of physics! Here is the rule:

In any collision (or explosion), the total momentum before = total momentum after, as long as no external forces act on the system.

Think about playing carrom. When the striker hits a coin, the striker slows down (loses momentum) and the coin speeds up (gains momentum). The total momentum of striker + coin stays the same! Momentum is simply transferred from one object to the other.

🧠 Memory Trick

"Momentum is like money — it cannot be created or destroyed, only transferred!" In a collision, one object gives momentum to the other, just like paying money from one pocket to another. The total stays the same.

Types of Collisions

Property	Elastic Collision	Inelastic Collision
Momentum conserved?	Yes	Yes
Kinetic energy conserved?	Yes	No (some converted to heat/sound)
Objects after collision	Bounce apart	May stick together
Example	Carrom striker hitting coin	Two cars crashing and sticking together

Worked Example A carrom striker (mass 0.015 kg) moves at 2 m/s and hits a stationary coin (mass 0.005 kg). After the collision, the striker moves at 1 m/s in the same direction. What is the velocity of the coin?

Step 1: Total momentum before

p(before) = m₁ × u₁ + m₂ × u₂ = (0.015 × 2) + (0.005 × 0) = 0.030 kg m/s

Step 2: Total momentum after (must equal before)

p(after) = m₁ × v₁ + m₂ × v₂ = (0.015 × 1) + (0.005 × v₂) = 0.015 + 0.005v₂

Step 3: Set equal and solve

0.030 = 0.015 + 0.005v₂
0.005v₂ = 0.015
v₂ = 0.015 ÷ 0.005 = 3 m/s

Answer: The coin moves at 3 m/s in the same direction. Notice — the lighter coin moves faster than the heavier striker, which makes perfect sense when you play carrom!

Worked Example An auto-rickshaw (mass 400 kg) travelling at 10 m/s crashes into a stationary car (mass 1,200 kg) and they stick together. What is their velocity after the collision?

Step 1: Momentum before

p(before) = (400 × 10) + (1,200 × 0) = 4,000 kg m/s

Step 2: After collision they stick — combined mass

Combined mass = 400 + 1,200 = 1,600 kg
p(after) = 1,600 × v

Step 3: Conservation of momentum

4,000 = 1,600 × v
v = 4,000 ÷ 1,600 = 2.5 m/s

Answer: They move together at 2.5 m/s in the original direction. The combined vehicle is much slower because the momentum is shared across a larger mass.

Worked Example A 0.5 kg Diwali rocket is initially at rest. It expels 0.05 kg of gas at 100 m/s downwards. What is the velocity of the rocket?

Step 1: Total momentum before (at rest)

p(before) = 0 (everything was stationary)

Step 2: After explosion

Let upwards be positive. Gas goes down: p(gas) = 0.05 × (−100) = −5 kg m/s
Remaining rocket mass = 0.5 − 0.05 = 0.45 kg
p(rocket) = 0.45 × v

Step 3: Conservation of momentum

0 = 0.45v + (−5)
0.45v = 5
v = 5 ÷ 0.45 = 11.1 m/s upwards

Answer: The rocket moves at 11.1 m/s upwards. The gas pushes down, the rocket goes up — conservation of momentum in action!

Force and Rate of Change of Momentum

Newton's Second Law can also be written using momentum. The resultant force on an object equals the rate of change of its momentum:

F = Δp ÷ Δt

F = resultant force in newtons (N) Δp = change in momentum in kg m/s Δt = time taken in seconds (s)

This is actually the original way Newton wrote his second law! F = ma is a special case of this when mass stays constant.

Real-world application: This is why cars have crumple zones and airbags. In a crash, Δp (change in momentum) is fixed — the car goes from moving to stopped. By making the crumple zone collapse slowly (increasing Δt), the force F on passengers is reduced. Same idea as catching a cricket ball with soft hands!

⚠ Exam Tip

The formula F = Δp/Δt and the impulse formula FΔt = Δp are the SAME equation, just rearranged. In exams, choose whichever version is easier for the question you are given.

Worked Example A 70 kg passenger in an auto-rickshaw is travelling at 12 m/s. The auto crashes and stops in 0.5 s (without seatbelt). What force acts on the passenger? If the auto had a crumple zone that extended the stopping time to 2 s, what would the force be?

Step 1: Change in momentum

Δp = mv − mu = (70 × 0) − (70 × 12) = −840 kg m/s

Step 2: Force without crumple zone

F = Δp ÷ Δt = 840 ÷ 0.5 = 1,680 N

Step 3: Force with crumple zone

F = Δp ÷ Δt = 840 ÷ 2 = 420 N

Answer: Without crumple zone: 1,680 N. With crumple zone: 420 N. The crumple zone reduces the force by 4 times! That is why modern cars are designed to crumple — they sacrifice the car to save the passengers.

⚠ Exam Tip

Common exam question: "Explain why airbags/crumple zones/helmets reduce injury." Answer: They increase the time over which momentum changes (Δt increases), so the force on the person decreases (F = Δp/Δt). Always mention: same change in momentum, longer time, smaller force.

🔍 Apply It: Real-World Physics

Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.

Indian Railways is testing a new anti-collision system. A 50,000 kg freight train is moving at 20 m/s and collides with a stationary goods wagon of mass 30,000 kg. They couple together after the collision.

What is their combined speed after the collision?

▼

Identify the Physics

This is a perfectly inelastic collision — the two objects stick together after colliding. We use the conservation of momentum: total momentum before = total momentum after.

Work It Out

Step 1: Calculate total momentum before the collision.
Momentum of train = m × v = 50,000 × 20 = 1,000,000 kg m/s
Momentum of wagon = 30,000 × 0 = 0 kg m/s (stationary)
Total momentum before = 1,000,000 + 0 = 1,000,000 kg m/s

Step 2: After the collision, they move together.
Combined mass = 50,000 + 30,000 = 80,000 kg
Total momentum after = combined mass × v
1,000,000 = 80,000 × v

Step 3: Solve for v.
v = 1,000,000 ÷ 80,000 = 12.5 m/s

💡 The Aha! Moment

When two objects stick together, the combined speed is always less than the original speed of the moving object. This makes sense — the same momentum is now shared across a bigger mass. Remember: momentum is ALWAYS conserved in collisions (as long as no external forces act), but kinetic energy is NOT conserved in inelastic collisions — some is converted to heat and sound.

In a cricket match, Virat Kohli hits a 0.16 kg ball travelling towards him at 40 m/s. The ball leaves his bat at 50 m/s in the opposite direction.

What is the change in momentum of the ball? If the bat was in contact with the ball for 0.005 seconds, what was the average force on the ball?

▼

Identify the Physics

This is about change in momentum (impulse) and the relationship F = Δp ÷ Δt. The ball changes direction, so we must be very careful with signs.

Work It Out

Step 1: Define a positive direction.
Let’s say “away from the bat” (the direction the ball leaves) is positive.
So: initial velocity u = −40 m/s (towards the bat = negative)
Final velocity v = +50 m/s (away from the bat = positive)

Step 2: Calculate the change in momentum.
Δp = m × v − m × u = m(v − u)
Δp = 0.16 × (50 − (−40))
Δp = 0.16 × 90 = 14.4 kg m/s

Step 3: Calculate the average force.
F = Δp ÷ Δt = 14.4 ÷ 0.005 = 2880 N

💡 The Aha! Moment

The force is nearly 2900 N from a 160 g ball! The key is the direction change — you must add the speeds when the ball reverses direction (40 + 50 = 90), not subtract them. This is the most common mistake in momentum questions. If the ball bounces back, the change in momentum is bigger than if it just stops.

A Diwali rocket (mass 0.2 kg) sits on the ground. It expels 0.05 kg of hot gas downward at 80 m/s.

What is the velocity of the remaining rocket immediately after? Explain why the rocket continues to accelerate even after the initial explosion.

▼

Identify the Physics

This is about conservation of momentum in an explosion (recoil). Before the explosion, everything is stationary, so the total momentum is zero. After the explosion, the total momentum must still be zero — the gas goes one way and the rocket goes the other.

Work It Out

Step 1: Total momentum before = 0 (everything is stationary).

Step 2: Total momentum after must also = 0.
Mass of remaining rocket = 0.2 − 0.05 = 0.15 kg
Momentum of gas (downward) = 0.05 × 80 = 4 kg m/s (downward)
Momentum of rocket (upward) = 0.15 × v (upward)

Step 3: Set total momentum = 0.
0.15v (up) − 4 (down) = 0
0.15v = 4
v = 4 ÷ 0.15 = 26.7 m/s upward

Why does it keep accelerating?
The rocket continues to burn fuel and expel gas. Each burst of gas provides another “push” (impulse). As fuel burns away, the rocket gets lighter, so the same force produces an even greater acceleration (F = ma — smaller m means bigger a).

💡 The Aha! Moment

In an explosion, total momentum before = total momentum after = 0. The gas and rocket carry equal and opposite momenta. Don’t forget to subtract the mass of the expelled gas from the rocket’s original mass! The remaining rocket is only 0.15 kg, not 0.2 kg.

Modern cars are designed with crumple zones — sections of the car body that deliberately collapse in a crash. A 1200 kg car travelling at 15 m/s hits a wall and stops completely.

The change in momentum is the same whether the car stops in 0.05 s (rigid car) or 0.3 s (car with crumple zones). Calculate the force on passengers in each case. Why are crumple zones safer?

▼

Identify the Physics

This is about impulse: the same change in momentum can happen over a short time (huge force) or a long time (smaller force). The equation is F = Δp ÷ Δt.

Work It Out

Step 1: Calculate the change in momentum (same in both cases).
Δp = m × v − m × u = 1200 × 0 − 1200 × 15 = −18,000 kg m/s
The magnitude of the change = 18,000 kg m/s

Step 2: Rigid car (stops in 0.05 s).
F = Δp ÷ Δt = 18,000 ÷ 0.05 = 360,000 N

Step 3: Car with crumple zones (stops in 0.3 s).
F = Δp ÷ Δt = 18,000 ÷ 0.3 = 60,000 N

The crumple zone reduces the force by a factor of 6!
360,000 ÷ 60,000 = 6 times less force on the passengers.

💡 The Aha! Moment

Crumple zones, seat belts, and airbags all work on the same principle: they increase the time over which the momentum changes, which reduces the force on the person. The momentum change is fixed (you go from moving to stopped), but the time is what you can control. This is a favourite exam question — always explain BOTH parts: longer time AND smaller force.

Two auto-rickshaws are at a traffic signal on MG Road, Bangalore. Auto A (mass 400 kg, speed 10 m/s) rear-ends stationary Auto B (mass 350 kg). After the collision, Auto A moves forward at 3 m/s.

What is the speed of Auto B after the collision? Is kinetic energy conserved?

▼

Identify the Physics

This is about conservation of momentum in a collision where the objects separate afterwards. We also need to check if kinetic energy is conserved to determine whether the collision is elastic or inelastic.

Work It Out

Step 1: Total momentum before.
p(before) = (400 × 10) + (350 × 0) = 4000 + 0 = 4000 kg m/s

Step 2: Total momentum after (must equal 4000 kg m/s).
p(after) = (400 × 3) + (350 × v)
4000 = 1200 + 350v
350v = 2800
v = 2800 ÷ 350 = 8 m/s

Step 3: Check kinetic energy.
KE before = ½ × 400 × 10² = ½ × 400 × 100 = 20,000 J
KE after = (½ × 400 × 3²) + (½ × 350 × 8²)
KE after = (½ × 400 × 9) + (½ × 350 × 64)
KE after = 1800 + 11,200 = 13,000 J

KE before (20,000 J) ≠ KE after (13,000 J)
Kinetic energy is NOT conserved — 7000 J was converted to heat, sound, and deformation. This is an inelastic collision.

💡 The Aha! Moment

Momentum is ALWAYS conserved in collisions (no exceptions!), but kinetic energy is only conserved in perfectly elastic collisions (which barely exist in real life). The “missing” 7000 J wasn’t destroyed — it was converted into the crunch sound, the heat of the metal bending, and the permanent dents in both autos. Always check both momentum AND energy when asked whether a collision is elastic.

Practice Questions: 1.6 Momentum

20 multiple choice questions — tap an option to check your answer

Your Score 0 / 20

Question 1

What is the momentum of a 50 kg girl running at 4 m/s?

A 12.5 kg m/s

B 200 kg m/s

C 54 kg m/s

D 46 kg m/s

p = m × v = 50 × 4 = 200 kg m/s. Simply multiply mass by velocity.

Question 2

Momentum is a vector quantity. This means it has:

A Only magnitude

B Both magnitude and direction

C Only direction

D Neither magnitude nor direction

Momentum is a vector because it depends on velocity (which is a vector). It has both magnitude (how much) and direction (which way the object moves).

Question 3

The unit of momentum is:

A kg m/s²

B kg m/s

C N m

D kg/m

p = m × v. Mass is in kg, velocity is in m/s, so momentum is in kg m/s. This is also equal to N s (newton seconds).

Question 4

A 0.4 kg ball moving at 5 m/s is stopped by a goalkeeper in 0.1 s. What is the force on the ball?

A 2 N

B 20 N

C 200 N

D 0.2 N

Δp = 0.4 × (0 − 5) = −2 kg m/s. F = Δp/Δt = 2/0.1 = 20 N.

Question 5

Two trolleys collide and stick together. This is an example of:

A An elastic collision

B An inelastic collision

C An explosion

D Conservation of energy only

When objects stick together after a collision, it is inelastic. Momentum is conserved, but kinetic energy is not — some is converted to heat and sound.

Question 6

A 3 kg object moving at 4 m/s collides with a 1 kg stationary object. They stick together. What is their combined velocity?

A 3 m/s

B 4 m/s

C 12 m/s

D 1 m/s

Before: p = 3 × 4 + 1 × 0 = 12 kg m/s. After: p = (3+1) × v = 12, so v = 12/4 = 3 m/s.

Question 7

Why does a gun recoil (kick back) when a bullet is fired?

A The bullet pushes air backwards

B The explosion creates a loud noise

C Conservation of momentum — the bullet gains forward momentum, so the gun gains equal backward momentum

D The gun is lighter than the bullet

Before firing, total momentum = 0 (both at rest). After: bullet has forward momentum, so the gun must have equal backward momentum. Total stays zero — conservation of momentum.

Question 8

A 2 kg toy rocket expels 0.1 kg of gas at 50 m/s. What is the velocity of the remaining rocket?

A 5 m/s

B 50 m/s

C 2.63 m/s

D 25 m/s

Initial momentum = 0. After: 0.1 × (−50) + 1.9 × v = 0. So 1.9v = 5, v = 5/1.9 = 2.63 m/s (opposite direction to gas).

Question 9

What is the impulse when a 0.5 kg ball changes velocity from 8 m/s to 2 m/s?

A 5 N s

B 3 N s

C 4 N s

D 1 N s

Impulse = Δ(mv) = m(v − u) = 0.5 × (2 − 8) = 0.5 × (−6) = −3 N s. Magnitude = 3 N s.

Question 10

Airbags in cars reduce injury because they:

A Reduce the change in momentum

B Increase the change in momentum

C Increase the time over which momentum changes, reducing the force

D Decrease the mass of the passenger

Airbags increase Δt (the stopping time). Since F = Δp/Δt, a bigger Δt means a smaller force. The change in momentum stays the same — the airbag just spreads it over a longer time.

Question 11

A force of 500 N acts on an object for 0.2 s. The impulse is:

A 2500 N s

B 100 N s

C 500.2 N s

D 499.8 N s

Impulse = F × Δt = 500 × 0.2 = 100 N s.

Question 12

A 1,000 kg car moving at 20 m/s crashes into a wall and stops in 0.5 s. What is the average force on the car?

A 10,000 N

B 40,000 N

C 20,000 N

D 500 N

Δp = 1,000 × (0 − 20) = −20,000 kg m/s. F = Δp/Δt = 20,000/0.5 = 40,000 N.

Question 13

Object A (2 kg, 6 m/s right) collides head-on with Object B (2 kg, 4 m/s left). They stick together. What is their velocity after collision?

A 1 m/s to the right

B 5 m/s to the right

C 10 m/s to the left

D 0 m/s (they stop)

Take right as positive. p(before) = (2 × 6) + (2 × −4) = 12 − 8 = 4 kg m/s. After: (2+2) × v = 4, so v = 4/4 = 1 m/s to the right.

Question 14

In an elastic collision, which quantities are conserved?

A Only momentum

B Only kinetic energy

C Both momentum and kinetic energy

D Neither momentum nor kinetic energy

In an elastic collision, BOTH momentum and kinetic energy are conserved. In inelastic collisions, only momentum is conserved — some kinetic energy is converted to heat/sound.

Question 15

A stationary 60 kg person jumps off a 200 kg boat. The person moves at 3 m/s towards the shore. What is the boat's velocity?

A 3 m/s away from shore

B 0.9 m/s away from shore

C 0.9 m/s towards shore

D 1.5 m/s away from shore

Initial p = 0. After: 60 × 3 + 200 × v = 0. 200v = −180, v = −0.9 m/s. The negative means 0.9 m/s away from shore (opposite direction to person).

Question 16

Which has greater momentum: a 0.01 kg bullet at 800 m/s or a 1,500 kg car at 2 m/s?

A The bullet

B The car

C They have equal momentum

D Cannot be determined

Bullet: p = 0.01 × 800 = 8 kg m/s. Car: p = 1,500 × 2 = 3,000 kg m/s. The car has far more momentum despite being much slower, because its mass is enormous.

Question 17

A cricket ball (0.16 kg) moving at 25 m/s is caught by a fielder who moves his hands back over 0.4 s. What is the average force on the fielder's hands?

A 10 N

B 100 N

C 4 N

D 40 N

Δp = 0.16 × 25 = 4 kg m/s. F = Δp/Δt = 4/0.4 = 10 N. Moving hands back increases Δt and reduces the force — that is why fielders "give" with the ball!

Question 18

Which statement about conservation of momentum is correct?

A It only applies to elastic collisions

B It only applies when objects stick together

C It applies to all collisions and explosions when no external force acts

D It only applies in space where there is no gravity

Conservation of momentum applies to ALL collisions (elastic and inelastic) AND explosions, as long as no external resultant force acts on the system.

Question 19

A 5 kg trolley moving at 3 m/s collides with a 3 kg stationary trolley. After collision, the 5 kg trolley moves at 1 m/s. What is the velocity of the 3 kg trolley?

A 5 m/s

B 2 m/s

C 3.33 m/s

D 4 m/s

Before: p = 5 × 3 = 15 kg m/s. After: 5 × 1 + 3 × v = 15. So 3v = 10, v = 10/3 = 3.33 m/s.

Question 20

A 0.5 kg Diwali rocket at rest expels gas. The rocket moves up at 10 m/s. If 0.05 kg of gas was expelled, what was the speed of the gas?

A 10 m/s

B 50 m/s

C 90 m/s

D 100 m/s

Initial p = 0. Rocket mass after = 0.5 − 0.05 = 0.45 kg. After: 0.45 × 10 + 0.05 × (−v) = 0. So 0.05v = 4.5, v = 4.5/0.05 = 90 m/s downwards.

1.7 Energy, Work and Power ▼

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Energy, Work and Power - IGCSE Physics

1.7.1 Energy Stores

Energy is the ability to do work. It cannot be seen directly, but we can see its effects — things moving, heating up, glowing, or growing. Energy comes in many forms, which physicists call energy stores. Think of energy stores like different kinds of bank accounts — the money (energy) is stored differently but it's all still money!

⚠ Exam Tip

The IGCSE syllabus lists exactly 8 energy stores. Learn them all! Questions often ask you to identify which energy store is involved in a situation. Remember: energy is always stored somewhere and transferred from one store to another.

The 8 Energy Stores

Energy Store	What It Means	Indian Examples	Symbol / Formula
Kinetic	Energy stored in moving objects — anything that moves has kinetic energy. The faster it moves or the more mass it has, the more kinetic energy it stores.	Namma Metro train moving through Bangalore; a cricket ball bowled by Jasprit Bumrah; a BMTC bus on MG Road; you running to catch an auto-rickshaw	E_k = ½mv²
Gravitational Potential	Energy stored in an object because of its height above the ground. The higher up it is, the more it has. It's "potential" because it has the potential to fall.	Water stored behind the Tehri Dam in Uttarakhand; a coconut at the top of a palm tree in Kerala; a book on a shelf; water in an overhead tank on a rooftop in Koramangala	ΔE_p = mgΔh
Chemical	Energy stored in chemical bonds between atoms. Released during chemical reactions (burning, digestion, battery discharge).	LPG cylinder used for cooking; food like dosas and idlis (which give you energy to study!); a car battery; coal from Jharkhand; petrol in a two-wheeler; a torch battery	—
Elastic (Strain)	Energy stored in a stretched or compressed object. When you stretch or squash something elastic, it stores energy that it releases when it returns to its original shape.	A stretched rubber catapult (gulel) used by children; the compressed spring inside a ballpoint pen; a stretched rubber band; the bent bow in archery at the Olympics	—
Nuclear	Energy stored in the nucleus (centre) of atoms. Released during nuclear fission (splitting heavy atoms) or nuclear fusion (joining light atoms). This is an enormous amount of energy from a tiny amount of matter.	Uranium fuel rods at the Kudankulam Nuclear Power Plant in Tamil Nadu; the Sun (which uses nuclear fusion); nuclear fuel at Tarapur Atomic Power Station	E = mc²
Thermal	Energy stored in the random movement of particles inside a substance. The hotter something is, the more thermal energy it stores. Also called internal energy or heat energy.	A hot cup of chai; a hot tawa (griddle) after making rotis; the hot engine of a motorcycle; hot water in a geyser; the surface of a road on a hot Bangalore afternoon	—
Electrostatic	Energy stored in separated electric charges. When positive and negative charges are separated, they have the potential to come together, releasing energy.	Electric charges stored in storm clouds before a lightning strike over the Western Ghats; a charged capacitor in a circuit; static electricity when you rub a balloon on your hair	—
Magnetic	Energy stored in a magnetic field. Two magnets that repel or attract each other store magnetic energy in the space between them.	A fridge magnet holding a child's drawing; an electromagnet in the Namma Metro's electric motor; the magnetic field in a loudspeaker; MRI machines in hospitals like Manipal Hospital	—

💡 Memory Trick

Conservation of Energy

This is one of the most important principles in all of physics. Here it is:

Energy cannot be created or destroyed — it can only be transferred from one store to another.

This is called the Principle of Conservation of Energy The total amount of energy in the universe always stays the same Energy that seems to "disappear" has just moved to a less useful store (usually thermal)

Let's see this with some Indian examples of energy transfers:

Situation	Energy Transfer	Energy "Lost" To
Burning wood in a chulha (wood stove)	Chemical → Thermal + Light	Thermal energy in surroundings (smoke, heated air)
Namma Metro accelerating from a station	Electrical → Kinetic + Thermal (friction)	Thermal energy in rails and brake pads
Coconut falling from a palm tree	Gravitational potential → Kinetic	Thermal + Sound when it hits the ground
Charging your phone	Electrical → Chemical (stored in battery)	Thermal energy (phone gets warm while charging)
Water flowing from Tehri Dam to turbines	Gravitational potential → Kinetic → Electrical	Thermal (friction in turbines and generators)
Lighting a diya (oil lamp) during Diwali	Chemical → Thermal + Light	Thermal energy in surrounding air

⚠ Exam Tip

In exams, when you write an energy transfer, always go from the source store to the destination store using an arrow (→). For example: "Chemical energy in petrol → Kinetic energy of car + Thermal energy (wasted)". Always mention wasted thermal energy — it shows the examiner you understand conservation of energy!

Supplement (Extended)

Kinetic Energy Formula

We can calculate exactly how much kinetic energy a moving object has using this formula:

E_k = ½mv²

E_k = kinetic energy in joules (J) m = mass in kilograms (kg) v = speed in metres per second (m/s) Note: v is squared — so doubling your speed quadruples your kinetic energy!

💡 Memory Trick

Think of it as: "Half a Mass of Velocity squared". The ½ is because kinetic energy is derived from the work-energy theorem using calculus — but for your exam, just remember the formula! Notice that speed (v) is squared, which means speed has a BIG effect on kinetic energy.

Worked Example A BMTC bus has a mass of 8000 kg and is travelling at 15 m/s along Outer Ring Road in Bangalore. Calculate its kinetic energy.

Step 1: Write down what you know

Mass of bus: m = 8000 kg
Speed of bus: v = 15 m/s
Find: Kinetic energy E_k = ?

Step 2: Write the formula

E_k = ½mv²

Step 3: Substitute and calculate

E_k = ½ × 8000 × (15)²
E_k = ½ × 8000 × 225
E_k = 4000 × 225
E_k = 900,000 J

Answer: E_k = 900,000 J = 900 kJ. That's a LOT of kinetic energy — which is why buses are so dangerous in accidents!

Worked Example A cricket ball of mass 0.16 kg is bowled at 40 m/s by a fast bowler. Calculate its kinetic energy. How does it compare to the same ball thrown at 20 m/s?

Step 1: Write down what you know

Mass of ball: m = 0.16 kg
Speed 1: v₁ = 40 m/s
Speed 2: v₂ = 20 m/s
Find: E_k at each speed

Step 2: Calculate at 40 m/s

E_k = ½mv² = ½ × 0.16 × (40)²
E_k = ½ × 0.16 × 1600
E_k = 0.08 × 1600 = 128 J

Step 3: Calculate at 20 m/s

E_k = ½ × 0.16 × (20)²
E_k = ½ × 0.16 × 400
E_k = 0.08 × 400 = 32 J

Step 4: Compare

128 J ÷ 32 J = 4 times more kinetic energy at 40 m/s
Even though speed only doubled (×2), the kinetic energy quadrupled (×4). This is because v is squared in the formula!

Answer: E_k at 40 m/s = 128 J; E_k at 20 m/s = 32 J. Doubling the speed gives 4 times the kinetic energy.

Worked Example Tara has a mass of 45 kg and is running to catch her school bus at 6 m/s. Calculate her kinetic energy. If she slows down to 3 m/s, what is her new kinetic energy?

Step 1: Identify known values

m = 45 kg, v₁ = 6 m/s, v₂ = 3 m/s

Step 2: Calculate at 6 m/s

E_k = ½ × 45 × (6)² = ½ × 45 × 36 = 22.5 × 36 = 810 J

Step 3: Calculate at 3 m/s

E_k = ½ × 45 × (3)² = ½ × 45 × 9 = 22.5 × 9 = 202.5 J

Answer: At 6 m/s, E_k = 810 J. At 3 m/s, E_k = 202.5 J. Halving speed gives one-quarter the kinetic energy!

Gravitational Potential Energy Formula

When an object is raised to a height, it gains gravitational potential energy (GPE). This is the energy stored due to its position in the Earth's gravitational field. If it falls, this energy converts to kinetic energy.

ΔE_p = mgΔh

ΔE_p = change in gravitational potential energy in joules (J) m = mass in kilograms (kg) g = gravitational field strength = 10 N/kg (on Earth's surface) Δh = change in height in metres (m) Δ (delta) means "change in" — so Δh means "change in height"

⚠ Exam Tip

In IGCSE Physics, always use g = 10 N/kg (or 10 m/s²) unless the question tells you otherwise. Some questions might give g = 9.8 N/kg — use whatever the question provides. Also remember: Δh is the vertical height gained, not the distance along a slope!

Worked Example A trekker of mass 60 kg climbs from the base of Nandi Hills to the top, gaining a vertical height of 600 m. Calculate the gain in gravitational potential energy. (g = 10 N/kg)

Step 1: Write down what you know

Mass: m = 60 kg
g = 10 N/kg
Height gained: Δh = 600 m
Find: ΔE_p = ?

Step 2: Write the formula

ΔE_p = mgΔh

Step 3: Substitute and calculate

ΔE_p = 60 × 10 × 600
ΔE_p = 60 × 6000
ΔE_p = 360,000 J

Answer: ΔE_p = 360,000 J = 360 kJ. All this energy came from the chemical energy stored in the trekker's food and muscles!

Worked Example A woman in a village lifts a bucket of water of mass 12 kg from a well. The water is raised a vertical height of 8 m to the surface. Calculate the gain in gravitational potential energy. (g = 10 N/kg)

Step 1: Identify known values

m = 12 kg, g = 10 N/kg, Δh = 8 m

Step 2: Apply the formula

ΔE_p = mgΔh = 12 × 10 × 8 = 960 J

Answer: ΔE_p = 960 J. This energy was transferred from the woman's chemical energy (muscles) to the gravitational potential energy of the water.

Worked Example A coconut of mass 1.5 kg falls from the top of a palm tree at a height of 12 m. Assuming all its gravitational potential energy converts to kinetic energy, what is its speed just before hitting the ground? (g = 10 N/kg)

Step 1: Write down what you know

m = 1.5 kg, g = 10 N/kg, Δh = 12 m
By conservation of energy: E_p lost = E_k gained

Step 2: Calculate the GPE lost

ΔE_p = mgΔh = 1.5 × 10 × 12 = 180 J
So the coconut gains 180 J of kinetic energy.

Step 3: Use E_k = ½mv² to find v

180 = ½ × 1.5 × v²
180 = 0.75 × v²
v² = 180 ÷ 0.75 = 240
v = √240 ≈ 15.5 m/s

Answer: The coconut hits the ground at approximately 15.5 m/s. This beautifully shows conservation of energy — GPE converts to KE!

1.7.2 Work Done

In everyday life, "work" means any effort you make. But in physics, work has a very specific meaning: work is done only when a force causes an object to move in the direction of that force. If you push against a wall all day and it doesn't move — you've done zero work in physics terms (even though you're exhausted!).

Work done = Energy transferred

When a force does work on an object, it transfers energy to that object Both work done and energy are measured in joules (J) 1 joule = 1 newton × 1 metre (J = N·m)

The Work Formula

W = F × d

W = work done in joules (J) F = force applied in newtons (N) d = distance moved in the direction of the force in metres (m)

⚠ Exam Tip

The distance must be in the direction of the force! If you carry a heavy bag horizontally, the vertical weight force does no work (because the bag moves horizontally, not vertically). If you carry it upstairs, the vertical component of displacement matters. This trips up many students — always check the direction!

Worked Example A vegetable vendor pushes a cart with a horizontal force of 200 N along Chickpet Market for a distance of 150 m. Calculate the work done by the vendor.

Step 1: Write down what you know

Force: F = 200 N (horizontal)
Distance: d = 150 m (horizontal — same direction as force)
Find: W = ?

Step 2: Write the formula

W = F × d

Step 3: Substitute and calculate

W = 200 × 150
W = 30,000 J

Answer: Work done = 30,000 J = 30 kJ. This energy came from the vendor's chemical energy (food) and was transferred to the cart (kinetic energy) and to the surroundings as thermal energy (friction).

Worked Example A student carries a stack of physics textbooks, applying an upward force of 50 N. She walks up two floors in her school building, climbing a vertical height of 6 m. Calculate the work done against gravity.

Step 1: Identify the relevant force and distance

The force working against gravity: F = 50 N (upward)
The distance in the direction of this force: d = 6 m (vertical height only)
Note: the horizontal distance walked along the corridor does NOT count!

Step 2: Apply the formula

W = F × d = 50 × 6 = 300 J

Answer: Work done = 300 J. This equals the gain in gravitational potential energy of the books (check: ΔE_p = mgh. If g=10, then m = F/g = 50/10 = 5 kg, so ΔE_p = 5×10×6 = 300 J ✓ — they match!)

Worked Example At Bengaluru airport, a traveller pulls a suitcase with a force of 80 N over a distance of 400 m along the terminal. Calculate the work done. If this work is done in 5 minutes, calculate the power.

Step 1: Calculate work done

F = 80 N, d = 400 m
W = F × d = 80 × 400 = 32,000 J

Step 2: Convert time to seconds

t = 5 minutes = 5 × 60 = 300 s

Step 3: Calculate power

P = W ÷ t = 32,000 ÷ 300 ≈ 106.7 W

Answer: Work done = 32,000 J = 32 kJ. Power ≈ 107 W (about the same as a light bulb — humans aren't that powerful!)

1.7.3 Energy Resources

The world needs energy for everything — electricity, transport, heating, cooking, industry. We get this energy from various energy resources. These fall into two main categories: renewable (won't run out, naturally replenished) and non-renewable (limited supply, will eventually run out).

The Sun: The Ultimate Source

Here's something remarkable: almost all of Earth's energy resources ultimately come from the Sun! Think about it:

Energy Resource	Connection to the Sun
Fossil fuels (coal, oil, gas)	Ancient plants and animals used sunlight for photosynthesis millions of years ago. Their remains became fossil fuels — so fossil fuels are really stored ancient sunlight!
Biofuels (wood, biogas, ethanol)	Plants grow using sunlight (photosynthesis). Burning them releases that solar energy.
Wind energy	The Sun heats the Earth unevenly, causing air to move (wind). So wind energy = solar energy!
Hydroelectric (water)	The Sun evaporates water from oceans; it falls as rain on mountains; rivers flow downhill to reservoirs. The Sun drives the water cycle!
Solar (photovoltaic/thermal)	Direct use of sunlight.
Tidal	Caused mainly by the Moon's gravity (not the Sun!) — one exception to the Sun-rule.
Geothermal	Heat from inside the Earth (radioactive decay in the Earth's core) — another exception.
Nuclear fission	Uranium was formed in ancient star explosions — not from our Sun, but from stars! Another exception.

⚠ Exam Tip

The three energy resources NOT from the Sun are: Tidal (Moon's gravity), Geothermal (Earth's internal heat), and Nuclear (energy from atomic nuclei). All others — fossil fuels, wind, hydro, solar, biofuels — trace back to the Sun. This is a very common exam question!

Renewable vs Non-Renewable Energy Resources

Non-renewable resources took millions of years to form and are being used faster than they are replaced. Once gone, they're gone forever. Renewable resources are naturally replenished and will not run out on a human timescale.

Energy Resource	Type	How it works	Indian Example	Advantages	Disadvantages
Coal / Oil / Gas (Fossil Fuels)	Non-renewable	Burnt to boil water → steam drives turbines → electricity generated	Coal from Jharkhand's Jharia coalfields; NTPC Ramagundam power plant in Telangana	Reliable, available 24/7; high energy density; cheap and existing infrastructure	Produces CO₂, main cause of climate change; limited supply; causes air pollution; oil spills damage ecosystems
Biofuels (Wood, Biogas)	Renewable	Organic matter burnt or decomposed (biogas) → heat or electricity	Biogas plants in rural Gujarat; chulha (wood stoves) in villages; biogas from cow dung (gobar gas)	Carbon-neutral (CO₂ released = CO₂ absorbed during growth); uses waste materials; helps rural communities	Can contribute to deforestation; burning releases particulates causing health problems; needs large land area
Hydroelectric	Renewable	Water stored at height falls through turbines → electricity	Tehri Dam (Uttarakhand); Sharavathi Hydroelectric Project (Karnataka); Bhakra-Nangal Dam (Punjab/Himachal)	No greenhouse gas emissions during operation; very reliable; can store energy (pump water uphill); long lifespan	Destroys ecosystems; displaces communities (Tehri dam displaced 100,000 people); depends on rainfall; expensive to build
Tidal	Renewable	Tidal movement of sea water drives turbines	Potential in Gulf of Kutch (Gujarat) and Sundarbans; India exploring tidal energy in coastal regions	Predictable and reliable; no fuel cost; no greenhouse gas emissions	Very expensive to build; limited locations (need large tidal range); affects coastal ecosystems; no large operational plants in India yet
Geothermal	Renewable	Heat from Earth's interior heats water → steam drives turbines	Puga Valley (Ladakh) has geothermal potential; hot springs at Manikaran (Himachal Pradesh)	Available 24/7 regardless of weather; low operating costs; small land footprint; no greenhouse gas emissions	Limited to geologically active areas; can release sulfur gases; expensive drilling; potential for local earthquakes
Nuclear Fission	Non-renewable (uranium is limited)	Uranium atoms split in a controlled chain reaction → huge heat → steam → turbines → electricity	Kudankulam Nuclear Power Plant (Tamil Nadu) — India's largest nuclear plant; Tarapur Atomic Power Station (Maharashtra)	No CO₂ during operation; very high energy density (small amount of fuel = huge energy); reliable 24/7; low fuel costs	Radioactive waste that remains dangerous for thousands of years; risk of accidents (Chernobyl, Fukushima); very expensive to build; uranium mining is hazardous; nuclear weapons proliferation risk
Solar	Renewable	Photovoltaic (PV) cells convert sunlight directly to electricity; solar thermal uses sunlight to heat water	Bhadla Solar Park (Rajasthan) — one of the world's largest; rooftop solar panels across Bangalore; PM KUSUM scheme for farmers	No greenhouse gas emissions; free fuel (sunlight); low maintenance; scalable from a phone charger to a power station; rapidly falling costs	Only works in daylight; much less effective in cloudy/rainy weather; batteries needed for storage (expensive); large land area for utility-scale power
Wind	Renewable	Moving air rotates wind turbine blades → generator → electricity	Muppandal Wind Farm (Tamil Nadu) — one of Asia's largest; Jaisalmer Wind Farm (Rajasthan); Karnataka has growing wind capacity	No greenhouse gas emissions; no fuel cost; can share land with agriculture; rapidly becoming cheapest electricity source	Intermittent (only when wind blows); visual impact; noise; affects birds and bats; needs backup power for calm days; best wind often in remote areas (transmission costs)

💡 Memory Trick

For non-renewable: "Coal Needs No Renewal" — Coal, oil, Natural gas, Nuclear are non-renewable.
Everything else — Solar, Wind, Hydro, Tidal, Geothermal, Biofuel — is renewable!

1.7.4 Efficiency

Nothing is perfect. Whenever energy is transferred from one store to another, some energy is always wasted — usually as thermal energy (heat). This wasted energy is transferred to the surroundings, where it becomes spread out and is no longer useful. Efficiency tells us how good a device is at converting energy into the form we actually want.

For example, an ordinary incandescent light bulb converts only about 5% of the electrical energy into light — the other 95% becomes heat! That's why LED bulbs (which are about 80% efficient) have largely replaced them.

Efficiency = (Useful energy output ÷ Total energy input) × 100%

Efficiency is expressed as a percentage (%) Useful energy output = the energy that does what we actually want (in joules) Total energy input = all the energy put into the device (in joules) Efficiency is always between 0% and 100% — it can NEVER exceed 100%!

⚠ Exam Tip

Efficiency can be expressed as a decimal (0 to 1) or a percentage (0% to 100%). If the question asks for percentage, multiply by 100. If it asks for the ratio/decimal, don't multiply. Also remember: you can NEVER have efficiency greater than 100% — that would violate conservation of energy, which is impossible!

Sankey Diagrams

A Sankey diagram is a visual way to show energy transfers in a device. The key features are:

A thick arrow enters from the left showing the total energy input
The arrow splits: a thick arrow continues straight ahead showing useful energy output
Thinner arrows bend downward showing wasted energy (usually thermal)
The width of each arrow is proportional to the amount of energy it represents
The total widths of all output arrows must equal the width of the input arrow (conservation of energy!)

Example — Ceiling Fan: A ceiling fan in an Indian home receives 75 J of electrical energy. 60 J is transferred usefully as kinetic energy (moving air). 15 J is wasted as thermal energy (friction in the motor and bearings) and some sound. In a Sankey diagram: a wide arrow (75 J) enters, a moderately wide arrow (60 J) goes straight (kinetic energy of moving air), and a thin arrow (15 J) curves down (thermal + sound).

Worked Example A ceiling fan in a Bengaluru home uses 80 J of electrical energy. It transfers 60 J to the air as kinetic energy (wind) and wastes the rest as thermal energy and sound. Calculate the efficiency of the fan.

Step 1: Write down what you know

Total energy input: 80 J (electrical)
Useful energy output: 60 J (kinetic energy of air)
Wasted energy: 80 - 60 = 20 J (thermal + sound)
Find: efficiency = ?

Step 2: Apply the efficiency formula

Efficiency = (Useful energy output ÷ Total energy input) × 100%
Efficiency = (60 ÷ 80) × 100%

Step 3: Calculate

Efficiency = 0.75 × 100% = 75%

Answer: The ceiling fan is 75% efficient. 25% of the input energy is wasted as thermal energy and sound.

Worked Example A solar water heater on a rooftop in Pune receives 5000 J of solar energy. It heats water with 3500 J of useful thermal energy. Calculate the efficiency. How much energy is wasted?

Step 1: Identify values

Total energy input: 5000 J (solar)
Useful energy output: 3500 J (heat in water)

Step 2: Calculate efficiency

Efficiency = (3500 ÷ 5000) × 100%
Efficiency = 0.70 × 100% = 70%

Step 3: Find wasted energy

Wasted energy = Total input − Useful output
Wasted energy = 5000 − 3500 = 1500 J
This 1500 J is lost to the surroundings as thermal radiation, conduction, and convection from the heater's surface.

Answer: Efficiency = 70%. Energy wasted = 1500 J.

Worked Example An LED bulb receives 10 J of electrical energy and produces 8 J of light. An old incandescent bulb of the same brightness receives 100 J of electrical energy and produces 5 J of light. Calculate and compare their efficiencies.

Step 1: Calculate LED efficiency

Efficiency_LED = (8 ÷ 10) × 100% = 80%

Step 2: Calculate incandescent efficiency

Efficiency_incandescent = (5 ÷ 100) × 100% = 5%

Step 3: Compare

The LED is 80% efficient vs 5% for the incandescent bulb.
The incandescent bulb wastes 95 J as thermal energy (that's why it gets so hot!)
The LED wastes only 2 J — it stays cool and uses far less electricity for the same brightness.

Answer: LED = 80% efficient, incandescent = 5% efficient. LEDs are 16 times more efficient — which is why India's UJALA scheme distributed 370 million LED bulbs to save electricity!

Supplement (Extended)

Efficiency Using Power

Efficiency can also be calculated using power instead of energy. Since power = energy ÷ time, and both input and output are measured over the same time, the time cancels out:

Efficiency = (Useful power output ÷ Total power input) × 100%

Useful power output = the rate at which useful work is done (in watts, W) Total power input = the total rate at which energy is supplied (in watts, W) This formula is equivalent to the energy formula — use whichever the question gives you data for

Worked Example An electric motor in a Namma Metro train has a power input of 50,000 W. It delivers 42,000 W of mechanical power to drive the wheels. Calculate the efficiency of the motor. What happens to the remaining power?

Step 1: Identify values

Total power input: 50,000 W
Useful power output: 42,000 W (mechanical)

Step 2: Apply the formula

Efficiency = (42,000 ÷ 50,000) × 100%
Efficiency = 0.84 × 100% = 84%

Step 3: Find wasted power

Wasted power = 50,000 − 42,000 = 8,000 W
This 8,000 W of power is dissipated as thermal energy in the motor windings and friction — which is why metro trains have cooling systems.

Answer: Efficiency = 84%. The remaining 16% (8,000 W) is wasted as thermal energy in the motor.

Worked Example A pump at a Sharavathi hydroelectric plant has an electrical power input of 200 kW. It raises water with a useful mechanical power output of 150 kW. Calculate its efficiency.

Step 1: Use consistent units

Total power input: 200 kW = 200,000 W
Useful power output: 150 kW = 150,000 W
(Or you can keep both in kW — the units cancel in the ratio anyway!)

Step 2: Calculate efficiency

Efficiency = (150 ÷ 200) × 100% = 0.75 × 100% = 75%

Answer: The pump is 75% efficient. The remaining 25% is wasted as thermal energy due to friction in the pump mechanism.

1.7.5 Power

We've talked about energy and work, but they don't tell us how fast the work is done. That's what power measures. A strong person might do the same work as a weak person, but a powerful person does it much faster.

Power is defined as work done per unit time, or equivalently, the rate of energy transfer.

P = W / t

P = power in watts (W) W = work done (or energy transferred) in joules (J) t = time taken in seconds (s) 1 watt = 1 joule per second (1 W = 1 J/s)

💡 Memory Trick

Power is how fast energy is transferred. A 100 W bulb transfers 100 joules every second. A 1000 W electric kettle transfers 1000 joules every second — that's why it boils water faster! The unit Watt (W) is named after Scottish inventor James Watt, who improved the steam engine.

⚠ Exam Tip

Always convert time to seconds before using P = W/t! If time is given in minutes, multiply by 60. If in hours, multiply by 3600. This is one of the most common mistakes — using minutes instead of seconds.

Worked Example A student climbs the stairs in her school, doing 2400 J of work in 8 seconds. Calculate her power output.

Step 1: Write down what you know

Work done: W = 2400 J
Time taken: t = 8 s
Find: P = ?

Step 2: Apply the formula

P = W / t

Step 3: Calculate

P = 2400 / 8 = 300 W

Answer: P = 300 W. For comparison, a resting human produces about 80 W, so climbing stairs at 300 W is a real workout!

Worked Example A water pump at a village well in Rajasthan lifts water by doing 180,000 J of work. It runs for 3 minutes. Calculate its power output in watts.

Step 1: Write down what you know

Work done: W = 180,000 J
Time taken: t = 3 minutes = 3 × 60 = 180 s (convert to seconds!)

Step 2: Calculate power

P = W / t = 180,000 / 180 = 1000 W

Answer: P = 1000 W = 1 kW. This is about the same power as an electric kettle — a reasonably powerful pump!

Worked Example A Kudankulam nuclear reactor generates 1,000,000,000 J of electrical energy in 10 seconds during peak production. Calculate its power output in megawatts (MW).

Step 1: Identify values

W = 1,000,000,000 J = 1 × 10³ MJ
t = 10 s

Step 2: Calculate power in watts

P = W / t = 1,000,000,000 / 10 = 100,000,000 W

Step 3: Convert to megawatts

P = 100,000,000 W ÷ 1,000,000 = 100 MW
(Kudankulam Unit 1 actually produces about 1000 MW — this example is simplified!)

Answer: P = 100,000,000 W = 100 MW. Nuclear power plants are incredibly powerful — they produce electricity for millions of homes!

Supplement (Extended)

Power = Force × Velocity (P = Fv)

There's another very useful formula for power when an object moves at constant velocity with a constant force applied to it. This is derived from combining P = W/t with W = Fd:

P = F × v

P = power in watts (W) F = force in newtons (N) v = velocity (speed) in metres per second (m/s) Derivation: P = W/t = (F×d)/t = F×(d/t) = F×v

⚠ Exam Tip

Use P = Fv when the question gives you force and speed but NOT time or distance directly. This formula is especially useful for vehicles! If a car moves at constant speed, the driving force equals the drag force (they're in equilibrium), so F is the driving force = drag force.

Worked Example A BMTC bus moves at a constant speed of 12 m/s along Hosur Road. The engine exerts a forward driving force of 8000 N to overcome friction and air resistance. Calculate the power output of the engine.

Step 1: Write down what you know

Force: F = 8000 N
Speed: v = 12 m/s (constant speed means force = drag)
Find: P = ?

Step 2: Apply P = Fv

P = F × v = 8000 × 12 = 96,000 W

Answer: P = 96,000 W = 96 kW. That's about 129 horsepower — a typical bus engine rating!

Worked Example A cyclist in Bangalore rides at a constant speed of 5 m/s. She exerts a forward force of 60 N on the pedals (which overcomes friction and air resistance). Calculate her power output. Then find how far she could travel in 10 minutes using P = W/t.

Step 1: Calculate power using P = Fv

P = F × v = 60 × 5 = 300 W

Step 2: Find work done in 10 minutes using P = W/t

t = 10 × 60 = 600 s
W = P × t = 300 × 600 = 180,000 J

Step 3: Find distance using W = Fd

d = W / F = 180,000 / 60 = 3000 m = 3 km
(Check: distance = speed × time = 5 × 600 = 3000 m ✓)

Answer: Power = 300 W. She travels 3 km in 10 minutes. Her body converts chemical energy (from food) at a rate of 300 W — but since humans are only about 25% efficient, her body is actually using about 1200 W of chemical energy!

🔍 Apply It: Real-World Physics

Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.

The Pavagada Solar Park in Karnataka is one of the world’s largest solar farms. A single solar panel receives 1000 W of solar energy per square metre. The panel has an area of 2 m² and is 20% efficient.

How much electrical power does one panel produce? If a typical Bangalore home uses 5 kWh of electricity per day, how many panels would be needed?

▼

Identify the Physics

This is about efficiency (useful output ÷ total input) and power (rate of energy transfer). We need to find the useful electrical power from a solar panel, then figure out how many panels can supply a home’s daily energy needs.

Work It Out

Step 1: Find the total solar power hitting the panel.
Solar power per m² = 1000 W
Panel area = 2 m²
Total solar power input = 1000 × 2 = 2000 W

Step 2: Apply efficiency to find electrical power output.
Efficiency = 20% = 0.20
Useful electrical power = 0.20 × 2000 = 400 W

Step 3: Find energy produced per panel per day.
Assume about 6 hours of good sunshine per day (realistic for Karnataka).
Energy per panel per day = 400 W × 6 h = 2400 Wh = 2.4 kWh

Step 4: Find the number of panels needed.
Home uses 5 kWh per day.
Number of panels = 5 ÷ 2.4 ≈ 2.1 panels
Since you can’t have a fraction of a panel, you need 3 panels (round up to be safe).

💡 The Aha! Moment

80% of the solar energy is “wasted” — it reflects off the panel or heats it up instead of becoming electricity. But even at 20% efficiency, just 3 panels on your rooftop in Bangalore can power your home! In exams, always use efficiency = useful output ÷ total input, and remember that “useful” depends on the purpose of the device.

Tara climbs the 100 steps of the Nandi Hills viewpoint. Each step is 0.2 m high, so the total height is 20 m. Tara’s mass is 50 kg. She takes 2 minutes to climb.

Calculate the work done against gravity and her useful power output. If her body is only 25% efficient, how much chemical energy (from her breakfast idli!) did she actually use?

▼

Identify the Physics

This is about work done against gravity (W = mgh), power (P = W ÷ t), and efficiency of the human body as a machine that converts chemical energy from food into gravitational potential energy.

Work It Out

Step 1: Calculate work done against gravity.
W = mgh = 50 × 10 × 20 = 10,000 J (10 kJ)

Step 2: Convert time to seconds.
t = 2 minutes = 2 × 60 = 120 s

Step 3: Calculate useful power output.
P = W ÷ t = 10,000 ÷ 120 = 83.3 W

Step 4: Find total chemical energy used.
Body efficiency = 25% = 0.25
Efficiency = useful energy output ÷ total energy input
So total energy input = useful energy output ÷ efficiency
Total chemical energy = 10,000 ÷ 0.25 = 40,000 J (40 kJ)

That means 30,000 J was “wasted” as thermal energy (which is why Tara feels hot and sweaty at the top!).

💡 The Aha! Moment

Your body is only about 25% efficient — for every 4 J of energy from your idli, only 1 J actually lifts you up. The rest becomes thermal energy (heat). That’s why you get warm when you exercise! In exams, “efficiency of the human body” is a favourite twist — always check whether they want the useful output or the total input.

A roller coaster at Wonderla amusement park near Bangalore starts at the top of a 30 m hill. A car (total mass with passengers = 500 kg) starts from rest at the top.

Assuming no friction, what speed does it reach at the bottom? In reality, it reaches only 20 m/s. What is the efficiency of the ride? Where did the “lost” energy go?

▼

Identify the Physics

This is about conservation of energy: gravitational potential energy (GPE) at the top converts to kinetic energy (KE) at the bottom. In a perfect (frictionless) world, all GPE becomes KE. In reality, some energy is “lost” (dissipated), so the ride has an efficiency less than 100%.

Work It Out

Step 1: Calculate GPE at the top.
GPE = mgh = 500 × 10 × 30 = 150,000 J

Step 2: Find the theoretical speed at the bottom (no friction).
All GPE converts to KE: mgh = ½mv²
The mass cancels! gh = ½v²
v² = 2gh = 2 × 10 × 30 = 600
v = √600 ≈ 24.5 m/s

Step 3: Calculate actual KE at the bottom.
Actual speed = 20 m/s
Actual KE = ½ × 500 × 20² = ½ × 500 × 400 = 100,000 J

Step 4: Calculate efficiency.
Efficiency = useful energy output ÷ total energy input × 100%
Efficiency = 100,000 ÷ 150,000 × 100% = 66.7%

Step 5: Where did the lost energy go?
Energy “lost” = 150,000 − 100,000 = 50,000 J
This was dissipated as thermal energy (friction between wheels and track, air resistance) and sound energy (the screaming doesn’t count — that’s the passengers!).

💡 The Aha! Moment

Notice that mass cancelled out in the frictionless calculation — a heavy car and a light car would reach the same speed! This is just like Galileo’s famous discovery. But in reality, friction means some energy always gets “lost” to the surroundings. In physics, energy is never destroyed — it just becomes less useful (thermal energy scattered into the environment).

A construction crane at a building site in Hebbal, Bangalore, lifts a 2000 kg concrete block 15 m high in 30 seconds. The crane motor is rated at 15 kW.

Calculate the useful work done and the useful power of the crane. What is the crane’s efficiency? Draw a simple Sankey diagram showing the energy flow.

▼

Identify the Physics

This is about work done (W = Fd = mgh for lifting), power (P = W ÷ t), efficiency, and Sankey diagrams that show how energy flows through a system from input to useful and wasted outputs.

Work It Out

Step 1: Calculate useful work done.
W = mgh = 2000 × 10 × 15 = 300,000 J (300 kJ)

Step 2: Calculate useful power.
P = W ÷ t = 300,000 ÷ 30 = 10,000 W (10 kW)

Step 3: Calculate efficiency.
Motor rated power (total input power) = 15 kW = 15,000 W
Total energy input in 30 s = 15,000 × 30 = 450,000 J (450 kJ)
Efficiency = useful output ÷ total input × 100%
Efficiency = 300,000 ÷ 450,000 × 100% = 66.7%

Step 4: Sankey diagram.
Think of it as a flow diagram with arrows whose width shows the amount of energy:
• Input arrow (left): 450 kJ of electrical energy (full width)
• Useful output arrow (right, going straight): 300 kJ of gravitational PE (2/3 width)
• Wasted arrow (going down): 150 kJ of thermal energy + sound (1/3 width)
The input arrow width = the sum of the output arrows. Energy is conserved!

💡 The Aha! Moment

A Sankey diagram is just a visual way of saying “energy in = energy out.” The key rule: the width of each arrow is proportional to the energy it carries. Examiners love asking you to draw these — always label every arrow with the type of energy AND the amount in joules. The wasted arrow usually goes downward.

A cyclist on Bangalore’s new cycling track at Ulsoor Lake is pedaling at a constant speed of 6 m/s against a total friction force of 40 N.

What power is she generating? If she switches to an e-bike that has a 250 W motor and 80% efficiency, what useful power reaches the wheel? At the same friction force, what maximum speed can the e-bike maintain?

▼

Identify the Physics

This is about the relationship P = Fv (power = force × velocity). At constant speed, the driving force exactly balances friction, so we can use the friction force as the driving force. We also need efficiency to find the useful power from the e-bike motor.

Work It Out

Step 1: Calculate the cyclist’s power.
At constant speed, driving force = friction force = 40 N
P = Fv = 40 × 6 = 240 W

Step 2: Find useful power from the e-bike motor.
Motor total power = 250 W
Efficiency = 80% = 0.80
Useful power = 0.80 × 250 = 200 W

Step 3: Find maximum speed of the e-bike.
At maximum constant speed, all useful power is used to overcome friction:
P = Fv, so v = P ÷ F
v = 200 ÷ 40 = 5 m/s
(That’s 18 km/h — a comfortable cruising speed!)

💡 The Aha! Moment

The formula P = Fv is incredibly useful — it connects power, force, and speed in one neat equation. Notice that the cyclist (240 W) actually produces MORE useful power than the e-bike motor (200 W)! But the e-bike doesn’t get tired. At constant speed, the driving force MUST equal the friction force — if it didn’t, the object would accelerate or decelerate.

Practice Questions: 1.7 Energy, Work and Power

20 multiple choice questions — tap an option to check your answer

Your Score 0 / 20

Question 1

A BMTC bus is moving along MG Road in Bangalore. Which energy store does the bus have because it is moving?

A Gravitational potential energy

B Kinetic energy

C Elastic energy

D Chemical energy

Any moving object has kinetic energy because of its motion. The bus has kinetic energy due to its mass and speed. Gravitational potential energy requires height, elastic requires stretching/compressing, and chemical is stored in fuel (which is used up as the bus moves).

Question 2

Water is stored behind the Tehri Dam in Uttarakhand at a height of 260 m above the turbines. What is the main energy store of this water?

A Kinetic energy

B Thermal energy

C Gravitational potential energy

D Electrostatic energy

An object at height stores gravitational potential energy (GPE = mgh). The water is stationary (not kinetic), not hot (not thermal), and not charged (not electrostatic). This GPE converts to kinetic energy as the water falls to drive the turbines.

Question 3

A stretched catapult (gulel) is about to be released. What energy store does it have?

A Chemical energy

B Magnetic energy

C Elastic (strain) energy

D Nuclear energy

Elastic (strain) energy is stored in objects that are stretched, compressed, or bent. When the rubber band of the catapult is stretched, it stores elastic energy. When released, this converts to kinetic energy of the stone.

Question 4

Which of the following is an example of THERMAL energy store?

A An LPG cylinder in a kitchen

B Uranium fuel rods at Kudankulam

C A charged storm cloud before lightning

D A hot cup of chai

Thermal energy is stored in the random movement of particles — hotter objects have more thermal energy. A hot cup of chai has thermal energy. LPG is chemical energy; uranium rods are nuclear energy; a charged cloud has electrostatic energy.

Question 5

A diya (oil lamp) is lit during Diwali. Which statement correctly describes the energy transfer?

A Kinetic energy → Chemical energy + Light

B Chemical energy → Thermal energy + Light energy

C Thermal energy → Chemical energy + Kinetic energy

D Light energy → Chemical energy + Thermal energy

Burning releases energy stored in chemical bonds in the oil. This is transferred as thermal energy (heat) and light energy. The chemical energy in the fuel is the input; thermal and light are the outputs. Note: the total output equals the input — conservation of energy!

Question 6

Which statement about the conservation of energy is correct?

A Energy can be created by burning fuel and destroyed by friction

B The total energy in a closed system increases over time

C Energy cannot be created or destroyed — it can only be transferred from one store to another

D Energy can be destroyed when a machine wastes it as heat

The principle of conservation of energy states that energy cannot be created or destroyed — only transferred between stores. When a machine "wastes" energy as heat, that thermal energy still exists in the surroundings. The total remains constant. No process in the universe can create or destroy energy.

Question 7 ★ Supplement

A cricket ball has a mass of 0.16 kg and is travelling at 30 m/s. What is its kinetic energy?

A 4.8 J

B 48 J

C 72 J

D 144 J

E_k = ½mv² = ½ × 0.16 × (30)² = ½ × 0.16 × 900 = 0.08 × 900 = 72 J. Remember to square the velocity first — many students forget this and get 4.8 J (which would be ½ × 0.16 × 30 without squaring).

Question 8 ★ Supplement

A car's speed doubles from 10 m/s to 20 m/s. What happens to its kinetic energy?

A It doubles

B It triples

C It quadruples (becomes 4 times larger)

D It remains the same

E_k = ½mv². Since speed appears squared, doubling speed multiplies E_k by 2² = 4. At 10 m/s: E_k = ½m(100). At 20 m/s: E_k = ½m(400) = 4 × ½m(100). This is why speeding is so dangerous — twice the speed means four times the kinetic energy to absorb in a crash!

Question 9 ★ Supplement

A 50 kg student climbs from the ground floor to the second floor of her school, gaining a height of 8 m. Calculate the increase in her gravitational potential energy. (g = 10 N/kg)

A 400 J

B 500 J

C 4000 J

D 4800 J

ΔE_p = mgΔh = 50 × 10 × 8 = 4000 J. Common mistake: forgetting to multiply by g (giving 50 × 8 = 400 J) or using the wrong value for g. Always write out all three values before calculating!

Question 10 ★ Supplement

A 2 kg coconut falls from a height of 5 m. Assuming all its gravitational potential energy converts to kinetic energy, what is its speed just before hitting the ground? (g = 10 N/kg)

A 5 m/s

B 7.5 m/s

C 10 m/s

D 100 m/s

GPE lost = KE gained: mgΔh = ½mv². The mass cancels: gΔh = ½v². So v² = 2gΔh = 2 × 10 × 5 = 100. v = √100 = 10 m/s. Notice the mass cancelled — the speed of a falling object doesn't depend on its mass (as Galileo discovered!).

Question 11

A person pushes a heavy almirah (wardrobe) with a force of 300 N but it does not move. How much work is done on the almirah?

A 0 J

B 300 J

C It depends on how long they push

D It depends on the mass of the almirah

W = F × d. Since the almirah does not move, the distance d = 0. Therefore W = 300 × 0 = 0 J. No matter how hard you push, if there's no movement, no work is done (in the physics sense). The person does feel tired, but that energy is wasted as thermal energy in their muscles, not transferred to the almirah!

Question 12

A force of 500 N is used to push a cart 20 m along a flat market lane. Calculate the work done.

A 25 J

B 500 J

C 10,000 J

D 100,000 J

W = F × d = 500 × 20 = 10,000 J = 10 kJ. The force and displacement are in the same direction (both horizontal), so we use the full force value. Work done = energy transferred to the cart.

Question 13

What is the unit of work done?

A Newton (N)

B Watt (W)

C Joule (J)

D Newton per metre (N/m)

Work done (and energy) are measured in joules (J). One joule = one newton × one metre (J = N·m). Newtons measure force; watts measure power; N/m is actually the unit for spring constant. Remember: work done = energy transferred, so they share the same unit — joules!

Question 14

Which energy resource is correctly classified as RENEWABLE?

A Coal from Jharkhand

B Uranium fuel rods at Kudankulam

C Natural gas from the Krishna-Godavari basin

D Wind from the Muppandal wind farm in Tamil Nadu

Wind energy is renewable — the wind keeps blowing, naturally replenished by the Sun heating Earth unevenly. Coal, natural gas, and nuclear fuel (uranium) are all non-renewable — they are finite and will eventually run out. India is rapidly expanding wind capacity, especially in Tamil Nadu and Karnataka.

Question 15

Which energy resource does NOT ultimately derive its energy from the Sun?

A Wind energy

B Hydroelectric energy

C Biofuel energy

D Geothermal energy

Geothermal energy comes from heat deep inside the Earth (radioactive decay of elements in the Earth's core and mantle) — not from the Sun. Wind is driven by the Sun heating air unevenly; hydroelectric relies on the Sun-driven water cycle; biofuels come from plants that use sunlight. Tidal and nuclear are also exceptions to the "from the Sun" rule.

Question 16

What is a disadvantage of solar energy?

A It produces large amounts of carbon dioxide

B It is a non-renewable resource

C It does not produce electricity at night or when it is very cloudy

D It produces dangerous radioactive waste

A major disadvantage of solar energy is that it is intermittent — it only generates electricity during daylight hours and is much less effective on cloudy days. Solar panels produce no CO₂, solar is renewable (not non-renewable), and radioactive waste is associated with nuclear energy, not solar. India's Bhadla Solar Park uses battery storage to extend availability.

Question 17

An electric motor receives 500 J of electrical energy and produces 350 J of useful mechanical energy. What is the efficiency of the motor?

A 35%

B 57%

C 70%

D 143%

Efficiency = (Useful output ÷ Total input) × 100% = (350 ÷ 500) × 100% = 0.70 × 100% = 70%. The remaining 30% (150 J) is wasted as thermal energy in the motor windings. Option D (143%) is impossible — efficiency can never exceed 100%, as that would mean creating energy from nothing!

Question 18

A ceiling fan uses 75 J of electrical energy. 15 J is wasted as thermal energy and sound. What is the efficiency of the fan?

A 15%

B 20%

C 25%

D 80%

Useful energy output = 75 − 15 = 60 J. Efficiency = (60 ÷ 75) × 100% = 80%. First find the useful output by subtracting the wasted energy from the total input. The useful output of a fan is the kinetic energy transferred to the air (which cools you down).

Question 19

A student does 6000 J of work in climbing stairs in 30 seconds. What is her power output?

A 20 W

B 180,000 W

C 200 W

D 2000 W

P = W ÷ t = 6000 ÷ 30 = 200 W. This is a realistic value for a person climbing stairs — about twice the resting metabolic rate. Watch out for option B (180,000 W = P × t instead of P = W/t!) and option A (6000 ÷ 300, wrong time conversion).

Question 20 ★ Supplement

A BMTC bus moves at a constant speed of 10 m/s. The driving force from the engine is 6000 N (which balances the drag and friction). What is the power output of the engine?

A 600 W

B 6000 W

C 60,000 W

D 600,000 W

Using P = Fv = 6000 × 10 = 60,000 W = 60 kW. This is the Supplement formula for power: P = Fv. It applies when a constant force acts on an object moving at constant speed. 60 kW is about 80 horsepower — a typical bus engine output at that speed. At constant speed, the engine force exactly equals the resistive forces (drag + friction).

1.8 Pressure ▼

Watch on YouTube

Pressure - IGCSE Physics

What is Pressure?

Imagine you are lying on a bed. Now imagine someone pokes you with one finger versus laying their whole palm flat on your shoulder. Which hurts more? The single finger, right! Even though the person pushes with the same force, the finger concentrates all that force onto a tiny area. That concentration of force is exactly what pressure is!

Here is the formal definition that you need to know for your exam:

Pressure is defined as the force acting per unit area. The force must act perpendicular (at right angles) to the surface.

p = F / A

p = pressure in pascals (Pa) F = force in newtons (N) A = area in m²

The unit of pressure is the pascal (Pa). One pascal equals one newton per square metre:

1 Pa = 1 N/m²

⚠ Exam Tip

The most common mistake is leaving the area in cm² instead of converting to m². Always convert before using the formula!

1 cm² = 0.0001 m² = 1 × 10⁻⁴ m²
(Because 1 cm = 0.01 m, so 1 cm² = 0.01 × 0.01 = 0.0001 m²)

💡 Memory Trick

Think of it like sharing a pizza! If 3 friends share a pizza, each gets a decent slice (low pressure on the pizza box). If 1 person takes the whole pizza and puts all their weight on a tiny spot — that spot feels a LOT of pressure. Same force, smaller area = more pressure.

Rearranging the Formula

You need to be able to find any one of the three quantities if you know the other two. Here are all three forms:

p = F / A | F = p × A | A = F / p

Use the first to find pressure, the second to find force, the third to find area

Worked Examples: p = F / A

Worked Example 1 An elephant weighs 40,000 N and stands on all four feet. Each foot has an area of 0.2 m². Calculate the pressure the elephant exerts on the ground.

Step 1: List what you know

Total weight (force) = 40,000 N
Number of feet = 4
Area of each foot = 0.2 m²
Total contact area = 4 × 0.2 = 0.8 m²

Step 2: Write the formula

p = F / A

Step 3: Substitute and calculate

p = 40,000 N ÷ 0.8 m²
p = 50,000 Pa

✓ Answer: The elephant exerts a pressure of 50,000 Pa (50 kPa) on the ground.

💡 Interesting fact!

This is actually quite low because the elephant's weight is spread across such a large total area. A woman in high heels (see next example) can exert FAR more pressure!

Worked Example 2 A woman of weight 600 N is wearing high heels. Each heel has a contact area of 1 cm² (0.0001 m²). She also has flat sandals where each sole covers 150 cm² (0.015 m²). She stands on one foot at a time. Calculate the pressure for each type of footwear and compare with the elephant above.

Step 1: High heels — identify values

Force = 600 N (standing on one heel)
Area = 0.0001 m²

Step 2: Calculate pressure (high heels)

p = F / A
p = 600 ÷ 0.0001
p = 6,000,000 Pa = 6 MPa

Step 3: Flat sandals — identify values

Force = 600 N
Area = 0.015 m²

Step 4: Calculate pressure (flat sandals)

p = F / A
p = 600 ÷ 0.015
p = 40,000 Pa = 40 kPa

✓ Answer: High heels: 6,000,000 Pa | Flat sandals: 40,000 Pa

💡 Comparison

The high-heeled woman exerts 120 times more pressure than the elephant! This is why high heels damage soft floors. The flat sandal gives almost the same pressure as the elephant — just shows how important area is!

Worked Example 3 Tara carries a school bag that weighs 80 N on her shoulder. Her thin school bag strap has a width of 2 cm and contacts a length of 3 cm on her shoulder (area = 6 cm²). Her mother suggests a wide padded strap with a contact area of 60 cm². Calculate the pressure in each case and explain why the wider strap is more comfortable.

Step 1: Convert areas to m²

Thin strap: 6 cm² = 6 × 0.0001 = 0.0006 m²
Wide strap: 60 cm² = 60 × 0.0001 = 0.006 m²

Step 2: Calculate pressure — thin strap

p = F / A = 80 / 0.0006 = 133,333 Pa ≈ 133 kPa

Step 3: Calculate pressure — wide strap

p = F / A = 80 / 0.006 = 13,333 Pa ≈ 13.3 kPa

✓ Answer: Thin strap: ~133,000 Pa | Wide strap: ~13,300 Pa

💡 Explanation

The wider strap reduces the pressure on Tara's shoulder by 10 times because it spreads the same force (the weight of the bag) over a larger area. That is why padded, wide school bag straps are much more comfortable!

Pressure in Everyday Life

Once you understand pressure, you will start noticing it everywhere! Here are some brilliant examples from daily life in India and around the world:

Situation	Area	Pressure	Effect / Why?
Sharp knife blade	Tiny (very thin edge)	Very HIGH	Cuts through food easily — same hand force, tiny area
Drawing pin (thumb tack)	Tiny point	Very HIGH	Pushes into a wall or notice board easily
Elephant feet	Very large (wide, flat feet)	LOW	Weight spread over large area — doesn't sink into soft ground
High heels	Tiny heel tip	Extremely HIGH	More pressure than an elephant! Damages soft floors
Snowshoes / Skis	Very large	LOW	Spreads body weight across snow — you don't sink!
Wide tractor tyres	Large	LOW	Spreads tractor's weight on soft farm soil in Karnataka
Camel's flat wide feet	Large and flat	LOW	Doesn't sink into desert sand — nature's snowshoe!
Bed of nails (fakir)	Hundreds of nail tips together = large area	LOW per nail	Weight spread across 500+ nails means pressure per nail is tiny
Wide building foundation	Very large base	LOW	Heavy building spreads weight over large area — doesn't sink into ground

⚠ Exam Tip

In "explain" questions about pressure, always mention TWO things:
1. What happens to the area (increases or decreases)
2. The effect on pressure (so pressure increases/decreases)

Example: "Wide tractor tyres increase the contact area, which reduces the pressure on soft soil, so the tractor does not sink."

How Pressure Varies with Force and Area

From the formula p = F / A, we can see two clear relationships:

Force increases → Pressure increases: Push a drawing pin harder → more pressure → goes deeper into the wall. Same area, more force = more pressure.
Area decreases → Pressure increases: A sharp knife cuts better than a blunt one. Same force but smaller area = much more pressure at the cutting edge.

More examples from India:

Pressure cooker: The sealed lid traps steam. As heat builds up, the pressure inside rises — the high pressure raises the boiling point of water above 100°C, so rice and dal cook faster. (The cooker whistles when the safety valve releases excess pressure!)
Drinking through a straw: You reduce the air pressure inside the straw by sucking. The higher air pressure outside then pushes the drink up into your mouth.
Why surgeons use thin, sharp scalpels: The tiny blade area concentrates force to give enormous pressure — the blade cuts through skin easily with minimal force applied.
Why it hurts to kneel on a hard floor: Your body weight concentrated on the small area of your knees creates high pressure on the hard floor — and the floor pushes back with equal pressure on your knees.

Supplement (Extended)

Pressure in Liquids: p = ρgh

Have you ever dived to the bottom of a swimming pool and felt your ears hurt? Or noticed that the walls of a dam are much thicker at the bottom than at the top? Both of these observations are about how pressure in a liquid increases with depth.

Think about it this way: if you are at a depth of 2 metres underwater, you have a column of water 2 metres tall sitting above you, pushing down on you. The deeper you go, the more water sits above you, and the greater the pressure.

p = ρgh

p = pressure difference in pascals (Pa) ρ = density of the liquid in kg/m³ (rho) g = gravitational field strength = 10 N/kg (on Earth) h = depth below the surface in metres (m)

⚠ Exam Tip

This formula gives the pressure due to the liquid column only (pressure difference). To find the total absolute pressure at depth h, you would also add atmospheric pressure (about 100,000 Pa). However, in most IGCSE questions, you are asked to find the pressure difference caused by the liquid, so p = ρgh is what you use.

💡 Memory Trick for ρgh

Remember "Roh-gah" — ρ (rho) × g × h. Or think: Dense Gravity Height — density, gravity, how high (deep) the column is. The denser the liquid, the stronger gravity, the deeper you are — all three increase pressure.

Key facts about liquid pressure:

Pressure depends only on depth, not on the shape or total volume of the container
At the same depth, a wide tank and a narrow tube have the same pressure
Pressure increases with depth (more liquid above = more weight pushing down)
Pressure increases with density (denser liquids push harder — mercury is 13.6 times denser than water, so it creates 13.6 times the pressure at the same depth)

Density of common liquids to know:

Liquid	Density (kg/m³)
Fresh water	1000
Sea water	1025
Mercury	13,600
Mango juice (approximately)	~1050

Worked Examples: p = ρgh

Worked Example 4 A swimming pool in Bangalore is 2 m deep. Calculate the pressure due to the water at the bottom of the pool. (Density of water = 1000 kg/m³, g = 10 N/kg)

Step 1: List known values

ρ = 1000 kg/m³
g = 10 N/kg
h = 2 m (depth at bottom)

Step 2: Write the formula

p = ρgh

Step 3: Substitute and calculate

p = 1000 × 10 × 2
p = 20,000 Pa = 20 kPa

✓ Answer: The water pressure at the bottom of the pool is 20,000 Pa.

💡 Why do your ears hurt?

At 2 m depth, the water exerts 20,000 Pa extra pressure on your eardrums. Added to atmospheric pressure (~100,000 Pa), the total pressure is 120,000 Pa — your ears feel the extra 20,000 Pa and it can be uncomfortable!

Worked Example 5 The Tehri Dam in Uttarakhand is one of India's tallest dams. The reservoir has a maximum depth of 260 m. Calculate the water pressure at the base of the dam wall. (ρ = 1000 kg/m³, g = 10 N/kg) Explain why the dam wall is much thicker at the bottom.

Step 1: List known values

ρ = 1000 kg/m³
g = 10 N/kg
h = 260 m

Step 2: Apply formula

p = ρgh
p = 1000 × 10 × 260
p = 2,600,000 Pa = 2.6 MPa

✓ Answer: The pressure at the base is 2,600,000 Pa (2.6 megapascals — enormous!)

💡 Why is the dam thicker at the bottom?

At the surface (h = 0), pressure due to water = 0 Pa. Near the bottom (h = 260 m), pressure = 2,600,000 Pa. The pressure increases with depth, so the dam experiences far greater force near the base. Engineers make it thicker at the bottom to withstand this huge pressure — if it were the same thickness all the way down, the bottom would crack!

Worked Example 6 A glass of fresh mango juice is 15 cm tall and is completely full. The density of the mango juice is 1050 kg/m³. Calculate the pressure difference between the surface and the bottom of the glass. (g = 10 N/kg)

Step 1: Convert depth to metres

h = 15 cm = 15 / 100 = 0.15 m

Step 2: List all values

ρ = 1050 kg/m³
g = 10 N/kg
h = 0.15 m

Step 3: Calculate

p = ρgh
p = 1050 × 10 × 0.15
p = 1575 Pa

✓ Answer: The pressure at the bottom of the glass is 1575 Pa higher than at the surface.

Pressure in a Fluid Acts Equally in All Directions

Here is something fascinating: at any single point inside a fluid (a liquid or gas), the pressure acts equally in all directions — up, down, sideways, diagonally — all the same!

At point P inside the fluid, pressure pushes equally in every direction

Why does this matter? Here are brilliant examples:

Round balloons: When you blow air into a balloon, the air pressure inside acts equally outward in all directions — that is why the balloon becomes roughly spherical (round). If pressure were stronger in one direction, the balloon would be lopsided!
Toothpaste tube: Poke a hole anywhere in the tube and toothpaste squirts out. The pressure inside the paste is equal in every direction, so it escapes from wherever there is an opening.
A diver underwater: Pressure acts on the diver from all sides equally — not just from above. This is why deep-sea pressure can crush submarine hulls — it pushes in from every direction at once.
Hydraulic systems (Pascal's Principle): Because pressure in a fluid acts equally in all directions, when you apply pressure at one point, it transmits equally throughout the fluid. This is how car brakes, JCB excavators, and hydraulic lifts in car service centres work!
A drinking straw: When you suck on a straw, you reduce pressure at the top. The atmospheric pressure acting equally in all directions on the surface of the liquid pushes the drink up the straw.

⚠ Exam Tip

If asked to explain why pressure acts in all directions, say: "In a fluid, the molecules are free to move. They collide with any surface, regardless of orientation. So the fluid exerts a force on surfaces facing in any direction — the pressure is the same at a given depth in all directions."

Quick Summary

Formula	What it finds	Syllabus level
p = F / A	Pressure (Pa) from force (N) and area (m²)	Core
F = p × A	Force (N) from pressure (Pa) and area (m²)	Core
A = F / p	Area (m²) from force (N) and pressure (Pa)	Core
p = ρgh	Pressure difference (Pa) due to liquid depth	Supplement

🔍 Apply It: Real-World Physics

Cambridge examiners LOVE testing familiar concepts in unfamiliar situations. Can you spot the physics hiding in these real-world scenarios? Tap each one to reveal the answer.

A hydraulic car jack at a Bangalore auto repair shop has a small piston with area 5 cm² and a large piston with area 200 cm². A mechanic pushes down on the small piston with a force of 100 N.

What force is generated at the large piston? Can this lift a 1500 kg Maruti car? What is the trade-off (hint: think about distance moved)?

▼

Identify the Physics

This is about hydraulic pressure transmission. In a hydraulic system, pressure is transmitted equally through the liquid. Since pressure = force ÷ area, a small force on a small piston creates the same pressure as a large force on a large piston. This gives mechanical advantage.

Work It Out

Step 1: Find the pressure created by the small piston.
Convert area to m²: 5 cm² = 5 × 10−⁴ m² = 0.0005 m²
Pressure = F ÷ A = 100 ÷ 0.0005 = 200,000 Pa

Step 2: This same pressure acts on the large piston.
Convert area: 200 cm² = 200 × 10−⁴ m² = 0.02 m²
Force on large piston = p × A = 200,000 × 0.02 = 4000 N

Step 3: Can it lift the Maruti car?
Weight of car = mg = 1500 × 10 = 15,000 N
The jack produces only 4000 N, so no — one push of 100 N is not enough. The mechanic would need to push with at least 375 N, or pump multiple times (which is exactly how real hydraulic jacks work!).

Step 4: The trade-off.
The force is multiplied by 40 (the ratio of the areas: 200 ÷ 5 = 40). But the large piston moves 40 times less distance than the small piston. Energy is conserved: small force × large distance = large force × small distance.

💡 The Aha! Moment

Hydraulics don’t give you free energy — they trade distance for force. The force multiplier equals the area ratio. That’s why car jacks need many pumps to lift a car just a few centimetres. In exams, always remember: pressure is the same throughout the liquid, so p₁ = p₂, which means F₁/A₁ = F₂/A₂.

Tara is trekking to the summit of Mullayanagiri, Karnataka’s highest peak (1930 m altitude). At sea level, atmospheric pressure is 101,325 Pa. At the summit, it drops to about 80,000 Pa.

Explain why water boils at a lower temperature at the summit (about 93°C instead of 100°C). Why does a sealed bag of chips from Bangalore puff up as Tara climbs higher?

▼

Identify the Physics

This is about atmospheric pressure variation with altitude. The higher you go, the less air is above you, so the atmospheric pressure decreases. This affects boiling points and the behaviour of sealed containers.

Work It Out

Why water boils at a lower temperature:
Water boils when its vapour pressure equals the atmospheric pressure pushing down on its surface. At sea level, atmospheric pressure is high (101,325 Pa), so water needs to reach 100°C before its vapour pressure is strong enough to overcome that.

At the summit (80,000 Pa), atmospheric pressure is lower, so water’s vapour pressure can match it at only about 93°C. The water molecules don’t need as much energy to escape into the gas phase.

Why the chip bag puffs up:
The chip bag was sealed at Bangalore (around 920 m, approximately 91,000 Pa). The air inside the bag was at that pressure. As Tara climbs to 1930 m, the outside atmospheric pressure drops to 80,000 Pa.

The air inside the bag is still at the original higher pressure. Since the inside pressure is now greater than the outside pressure, the air inside pushes outward, causing the bag to puff up and expand.

💡 The Aha! Moment

Atmospheric pressure is caused by the weight of the air column above you. Go higher, less air above you, less pressure. This has real consequences — cooking rice takes longer at high altitudes because water boils at a lower temperature! In exams, remember that boiling point is NOT a fixed number — it depends on the external pressure.

An Indian Navy submarine is operating at a depth of 200 m in the Arabian Sea. The density of seawater is 1025 kg/m³. The submarine hull has a circular hatch with diameter 0.8 m.

Calculate the pressure due to the water at this depth. Add atmospheric pressure (101,325 Pa) to find the total pressure. What total force does the water exert on the hatch?

▼

Identify the Physics

This is about liquid pressure using the formula p = ρgh, and then finding the force from pressure using F = pA. Remember that total pressure at depth includes both the water pressure AND atmospheric pressure pushing down on the surface.

Work It Out

Step 1: Calculate the pressure due to the water.
p = ρgh = 1025 × 10 × 200 = 2,050,000 Pa (2050 kPa)

Step 2: Find the total pressure.
Total pressure = water pressure + atmospheric pressure
Total pressure = 2,050,000 + 101,325 = 2,151,325 Pa ≈ 2151 kPa

That’s about 21 times atmospheric pressure!

Step 3: Calculate the area of the circular hatch.
Diameter = 0.8 m, so radius = 0.4 m
Area = πr² = π × 0.4² = π × 0.16 ≈ 0.503 m²

Step 4: Calculate the force on the hatch.
F = p × A = 2,151,325 × 0.503 ≈ 1,082,000 N

That’s over 1 million newtons — equivalent to the weight of about 108 tonnes pressing on one hatch!

💡 The Aha! Moment

The pressure at 200 m depth is enormous — over a million newtons on a single hatch! That’s why submarines need incredibly strong hulls. In exams, don’t forget to add atmospheric pressure when asked for “total pressure” at a depth. The formula p = ρgh gives only the extra pressure due to the liquid, not the total.

An elephant at Bannerghatta National Park near Bangalore has a mass of 5000 kg. Each of its four feet has an area of about 0.08 m². Tara’s mother wears stiletto heels to a wedding — the heel tip has an area of about 0.5 cm² (0.00005 m²) and her mass is 55 kg.

Calculate the pressure each exerts on the ground. Who exerts more pressure?

▼

Identify the Physics

This is about pressure = force ÷ area. A heavy object does not necessarily exert more pressure than a light one — it depends on the area over which the force is spread. This is one of the most surprising and exam-favourite applications of pressure!

Work It Out

Step 1: Calculate the elephant’s pressure.
Weight = mg = 5000 × 10 = 50,000 N
Total foot area = 4 × 0.08 = 0.32 m²
Pressure = F ÷ A = 50,000 ÷ 0.32 = 156,250 Pa ≈ 156 kPa

Step 2: Calculate the stiletto heel pressure.
Weight = mg = 55 × 10 = 550 N
Assume she stands on one heel (worst case): area = 0.00005 m²
Pressure = F ÷ A = 550 ÷ 0.00005 = 11,000,000 Pa = 11,000 kPa

Step 3: Compare!
Stiletto: 11,000 kPa vs Elephant: 156 kPa
The stiletto exerts about 70 times more pressure than the elephant!

💡 The Aha! Moment

A 55 kg woman in stilettos exerts 70 times more pressure than a 5-tonne elephant! That’s why stiletto heels damage wooden floors but elephants don’t — it’s all about the area. Pressure is force per unit area, so a tiny area means enormous pressure, even from a small force. This is also why knives are sharp (small area), drawing pins have points, and tractors have wide tyres.

A dam on the Cauvery River holds back water to a depth of 40 m. An engineer needs to design the dam wall. The density of water is 1000 kg/m³.

Why is the dam wall thicker at the bottom than at the top? Calculate the pressure at the bottom of the dam due to the water. If a crack 0.1 m wide and 0.01 m tall appears at the very bottom, what force would the water exert through this crack?

▼

Identify the Physics

This is about liquid pressure increasing with depth (p = ρgh). The deeper you go, the greater the weight of water above, and the greater the pressure. This has direct consequences for how dams, swimming pools, and submarine hulls are designed.

Work It Out

Step 1: Why is the dam thicker at the bottom?
Pressure increases with depth (p = ρgh). At the bottom of the dam, the water pressure is at its maximum because there are 40 m of water above. At the top, there is almost no water above, so the pressure is very low. The dam wall must be thicker at the bottom to withstand this greater pressure without cracking.

Step 2: Calculate the pressure at the bottom.
p = ρgh = 1000 × 10 × 40 = 400,000 Pa (400 kPa)

That’s about 4 times atmospheric pressure.

Step 3: Calculate the force through the crack.
Crack area = width × height = 0.1 × 0.01 = 0.001 m²
Force = p × A = 400,000 × 0.001 = 400 N

That’s like the weight of a 40 kg child pushing through a tiny crack — enough to widen it rapidly and potentially cause a catastrophic failure!

💡 The Aha! Moment

Even a tiny crack at the bottom of a dam experiences huge force because the pressure is so high at that depth. This is why dam engineers constantly monitor for cracks — water pressure at the bottom would force water through any opening and erode it wider. In exams, remember that liquid pressure depends only on depth, density, and g — NOT on the total volume of water behind the dam!

Practice Questions: 1.8 Pressure

20 multiple choice questions — tap an option to check your answer

Your Score 0 / 20

Question 1

Which of the following is the correct definition of pressure?

A Force multiplied by area

B Force divided by area

C Area divided by force

D Force added to area

Pressure is defined as force per unit area: p = F / A. Think of it as how much force is concentrated on each square metre.

Question 2

What is the SI unit of pressure?

A Newton (N)

B Joule (J)

C Pascal (Pa)

D Watt (W)

The SI unit of pressure is the pascal (Pa). 1 Pa = 1 N/m². It is named after the French scientist Blaise Pascal.

Question 3

1 pascal is equivalent to which of the following?

A 1 N cm²

B 1 N/m²

C 1 N m²

D 1 kg/m²

1 pascal = 1 newton per square metre (1 N/m²). This follows directly from the definition p = F / A, where F is in newtons and A is in m².

Question 4

A force of 200 N acts on an area of 0.5 m². What is the pressure?

A 100 Pa

B 100,000 Pa

C 400 Pa

D 0.0025 Pa

p = F / A = 200 / 0.5 = 400 Pa. Divide the force by the area to get pressure.

Question 5

A pressure of 5000 Pa acts over an area of 0.02 m². What force is exerted?

A 250,000 N

B 0.000004 N

C 100 N

D 4996 N

F = p × A = 5000 × 0.02 = 100 N. Rearranging p = F/A gives F = p × A.

Question 6

A rectangular block exerts a force of 360 N on the ground. The block is 30 cm × 40 cm. What pressure does it exert? (Convert area carefully!)

A 3 Pa

B 30 Pa

C 3000 Pa

D 30,000 Pa

Area = 0.30 m × 0.40 m = 0.12 m². p = 360 / 0.12 = 3000 Pa. Remember to convert cm to m FIRST: 30 cm = 0.30 m, 40 cm = 0.40 m.

Question 7

A pressure of 80,000 Pa is applied to a surface. If the force applied is 400 N, what is the area of the surface?

A 0.005 m²

B 0.5 m²

C 200 m²

D 32,000,000 m²

A = F / p = 400 / 80,000 = 0.005 m². Rearranging p = F/A gives A = F/p.

Question 8

Why are knives made with thin, sharp blades?

A To increase the area and reduce pressure

B To increase the force applied by the knife

C To decrease the area so the same force creates a larger pressure

D To reduce friction between the knife and food

A thin, sharp blade has a very small contact area. Since p = F/A, a smaller area gives a larger pressure for the same cutting force. This is why a sharp knife cuts through a tomato much more easily than a blunt one.

Question 9

A man lying on a bed of nails does not get hurt. Which statement best explains why?

A The force of his weight is reduced by the nails

B The nails push upward with extra force to help

C His weight is spread across hundreds of nail tips, so the pressure per nail is very small

D The nails are not sharp enough to cause pressure

With hundreds of nail tips, the total contact area is large. Even though each tip is sharp, the total force (body weight) is shared across all the tips, so p = F/A becomes small per nail. This is a classic example of pressure depending on area!

Question 10

Camels have large, flat, wide feet. How does this help them walk on desert sand?

A It increases the pressure on the sand, helping them grip

B It increases the contact area, reducing the pressure on the sand so they don't sink

C It reduces the force the camel exerts on the ground

D It reduces friction between the camel's feet and the sand

Large, flat feet increase the contact area. More area means less pressure (p = F/A with larger A). So the same body weight creates less pressure on the sand, and the camel does not sink. Nature's own pressure engineering!

Question 11

A drawing pin is pushed into a notice board with a force of 5 N. If the same force is applied by pressing with a thumb (area = 5 cm² = 0.0005 m²) instead of the sharp point (area = 0.5 mm² = 5 × 10⁻⁷ m²), what happens to the pressure?

A Pressure is the same in both cases

B The thumb exerts more pressure than the pin point

C The pin point exerts far more pressure than the thumb

D No pressure is exerted because the force is the same

Pin point: p = 5 / (5 × 10⁻⁷) = 10,000,000 Pa. Thumb: p = 5 / 0.0005 = 10,000 Pa. The pin point exerts 1000 times more pressure because its area is 1000 times smaller. Same force, tiny area = huge pressure!

Question 12

Two identical bricks are placed on soft clay. Brick A stands upright on its smallest face (area = 0.01 m²). Brick B lies flat on its largest face (area = 0.06 m²). Both bricks weigh 30 N. Which brick sinks deeper into the clay?

A Brick A, because it exerts greater pressure

B Brick B, because it has a larger area

C Both sink equally because they have the same weight

D Neither sinks because both are resting on the ground

Brick A: p = 30/0.01 = 3000 Pa. Brick B: p = 30/0.06 = 500 Pa. Brick A exerts 6 times more pressure on the clay, so it sinks deeper. This is why you should lay a brick on its flat side if you don't want it to sink.

Question 13 Supplement

What is the pressure due to a column of water of depth 5 m? (Density of water = 1000 kg/m³, g = 10 N/kg)

A 500 Pa

B 5000 Pa

C 50,000 Pa

D 500,000 Pa

p = ρgh = 1000 × 10 × 5 = 50,000 Pa. Remember the formula: density × gravitational field strength × depth.

Question 14 Supplement

A diver is at a depth of 20 m in the sea (density of seawater = 1025 kg/m³, g = 10 N/kg). What is the pressure due to the seawater at this depth?

A 2050 Pa

B 20,500 Pa

C 205,000 Pa

D 2,050,000 Pa

p = ρgh = 1025 × 10 × 20 = 205,000 Pa. Note this is the pressure due to the water only. Added to atmospheric pressure (~100,000 Pa), the total pressure on the diver would be about 305,000 Pa!

Question 15 Supplement

A liquid exerts a pressure of 6000 Pa at a depth of 60 cm. What is the density of the liquid? (g = 10 N/kg)

A 100 kg/m³

B 360 kg/m³

C 1000 kg/m³

D 10,000 kg/m³

First convert: h = 60 cm = 0.60 m. Rearrange p = ρgh to get ρ = p / (gh) = 6000 / (10 × 0.60) = 6000 / 6 = 1000 kg/m³. This is the density of fresh water!

Question 16 Supplement

Two tanks, A and B, contain fresh water (ρ = 1000 kg/m³). Tank A is wide with a small depth of 1 m. Tank B is narrow with a greater depth of 4 m. Compare the pressure at the bottom of each tank.

A Tank A has more pressure because it is wider

B The pressure is the same in both tanks

C Tank B has more pressure because it is deeper

D Tank B has more pressure because it is narrower and has more volume of water above

Pressure in a liquid depends ONLY on depth (h), not on width or shape of the container. Tank A: p = 1000 × 10 × 1 = 10,000 Pa. Tank B: p = 1000 × 10 × 4 = 40,000 Pa. Tank B is deeper, so greater pressure — width is irrelevant!

Question 17 Supplement

Why is the base of a dam wall made much thicker than the top?

A Because the base is heavier and needs more support from the ground

B To reduce the pressure at the base by increasing the wall's area

C Because water pressure increases with depth, so the dam experiences greater outward force at its base

D Because the water is colder and denser at the bottom

Since p = ρgh, pressure increases with depth. At the surface, water pressure on the dam is nearly zero. At the base, it is enormous. The dam must be thicker at the base to withstand this greater outward force without cracking or collapsing.

Question 18 Supplement

A balloon, when inflated, becomes approximately round (spherical). Which statement best explains this?

A The rubber of the balloon is thicker at the sides, so it expands equally

B The air pressure inside acts equally in all directions, pushing the balloon walls outward equally

C Gravity pulls the air inside downward, making the balloon expand at the bottom first

D The balloon is round because the rubber is elastic

In a fluid (including air inside a balloon), pressure at any point acts equally in all directions. The air pressure pushes outward on the balloon walls with the same force per unit area in every direction — so the balloon expands equally in all directions and becomes roughly spherical.

Question 19 Supplement

A submarine is at a depth of 300 m in the ocean. A physicist says the submarine's hull experiences the same pressure from all directions. Is this correct, and why?

A No — pressure only acts downward because gravity pulls the water down

B No — pressure is greater on the sides than on the top

C Yes — at any point in a fluid, pressure acts equally in all directions

D Yes — but only because the submarine is completely submerged

At a given depth in a fluid, pressure acts equally in every direction. The water molecules at 300 m depth collide with the submarine hull from all angles — top, bottom, sides — with equal force per unit area. This is why submarines need to be extremely strong in all directions, not just on top.

Question 20 Challenge

A hydraulic press has a small piston of area 0.002 m² and a large piston of area 0.1 m². A force of 50 N is applied to the small piston. Using the principle that pressure in a fluid acts equally in all directions, what force does the large piston exert? (Hint: find pressure first, then use F = p × A)

A 1 N

B 50 N

C 1000 N

D 2500 N

Step 1 — Find pressure: p = F/A = 50/0.002 = 25,000 Pa. Step 2 — Because pressure in a fluid acts equally in all directions, this same pressure of 25,000 Pa acts on the large piston. Step 3 — Force on large piston: F = p × A = 25,000 × 0.1 = 2500 N. A small input force of 50 N produces a large output force of 2500 N — this is how hydraulic car lifts, JCB diggers, and braking systems work!