IGCSE Physics Paper 4 - Topic 1: Motion, Forces and Energy

(a) [2]

(i) Define speed.
(ii) Write the equation that links speed, distance and time.

Model Answer -- 1(a)

Speed is the distance travelled per unit time [1]

speed = distance / time (v = d / t)

Correct equation stated with correct symbols or words [1]

Mark 1 -- correct definition

Mark 2 -- correct equation

(b) [2]

A car travels a distance of 150 m in 6.0 s. Calculate the speed of the car.

Model Answer -- 1(b)

v = d / t = 150 / 6.0

Correct substitution into the equation [1]

v = 25 m/s

Correct answer with unit: 25 m/s [1]

Mark 1 -- correct substitution

Mark 2 -- correct answer with unit

The car then decelerates uniformly from 25 m/s to 0 m/s in 5.0 s.
Calculate the deceleration of the car.

Model Answer -- 1(c)

a = (v - u) / t

Correct formula stated [1]

a = (0 - 25) / 5.0 = -25 / 5.0

Correct substitution [1]

a = -5.0 m/s² (deceleration = 5.0 m/s²)

Correct answer: deceleration = 5.0 m/s² (or a = -5.0 m/s²) [1]

Mark 1 -- correct formula

Mark 2 -- correct substitution

Mark 3 -- correct answer with unit

(d) [3]

(i) On the axes below, sketch a speed-time graph for the car's journey described in parts (b) and (c). Label the axes and mark the key values.

(ii) Calculate the total distance travelled by the car during the whole journey.

Model Answer -- 1(d)

(i) Graph: Horizontal line at 25 m/s from t = 0 to t = 6 s (constant speed phase), then a straight line sloping down from 25 m/s at t = 6 s to 0 m/s at t = 11 s (deceleration phase) [1]

(ii) Total distance:

Phase 1: d = v x t = 25 x 6.0 = 150 m (rectangle)

Phase 2: d = 1/2 x 25 x 5.0 = 62.5 m (triangle)

Total = 150 + 62.5 = 212.5 m

Correct method (area under graph or separate calculations) [1]

Correct answer: 212.5 m (or 212 m or 213 m) [1]

Mark 1 -- correct shape of graph with key values

Mark 2 -- correct method for distance

Mark 3 -- correct total distance

(e) [2]

Describe what happens to the velocity of a skydiver from the moment they jump out of an aircraft until they reach the ground (before opening the parachute). Explain why they reach terminal velocity.

Model Answer -- 1(e)

Initially the skydiver accelerates because weight is greater than air resistance [1]

As speed increases, air resistance increases until air resistance equals weight -- the resultant force is zero so acceleration is zero -- the skydiver moves at a constant (terminal) velocity [1]

Mark 1 -- explains initial acceleration (weight > air resistance)

Mark 2 -- explains terminal velocity (forces balanced, no resultant force)

(a) [2]

State two differences between mass and weight.

Model Answer -- 2(a)

Mass is a measure of the amount of matter in an object / Weight is the gravitational force acting on an object [1]

Mass is measured in kg / Weight is measured in N OR Mass is constant everywhere / Weight varies with gravitational field strength OR Mass is a scalar / Weight is a vector [1]

Mark 1 -- first correct difference

Mark 2 -- second correct difference

(b) [3]

An astronaut has a mass of 80 kg.
(i) Calculate the astronaut's weight on Earth, where g = 9.8 N/kg.
(ii) Calculate the astronaut's weight on the Moon, where g = 1.6 N/kg.

Model Answer -- 2(b)

W = m x g

Correct formula stated or implied [1]

(i) W = 80 x 9.8 = 784 N

Weight on Earth: 784 N [1]

(ii) W = 80 x 1.6 = 128 N

Weight on Moon: 128 N [1]

Mark 1 -- correct formula (W = mg)

Mark 2 -- correct weight on Earth (784 N)

Mark 3 -- correct weight on Moon (128 N)

A metal cube has sides of length 2.0 cm and a mass of 64 g.
(i) Calculate the density of the metal. Give your answer in g/cm³.
(ii) The density of water is 1.0 g/cm³. State whether the cube would sink or float in water. Explain your answer.

Model Answer -- 2(c)

Volume = 2.0 x 2.0 x 2.0 = 8.0 cm³

Correct volume calculated [1]

Density = mass / volume = 64 / 8.0 = 8.0 g/cm³

Correct density: 8.0 g/cm³ [1]

The cube will sink because its density (8.0 g/cm³) is greater than the density of water (1.0 g/cm³) [1]

Mark 1 -- correct volume (8.0 cm³)

Mark 2 -- correct density (8.0 g/cm³)

Mark 3 -- sinks with correct explanation

(d) [2]

Describe how you would determine the density of an irregularly shaped stone.

Model Answer -- 2(d)

Measure the mass of the stone using a balance [1]

Find the volume by displacement: fill a measuring cylinder with water, record the initial volume, lower the stone in, record the final volume. Volume of stone = final reading - initial reading. Then calculate density = mass / volume [1]

Mark 1 -- measure mass using a balance

Mark 2 -- volume by displacement method described

(a) [3]

A spring has a natural (unstretched) length of 10 cm. When a load of 3.0 N is added, the spring stretches to a length of 16 cm.
Calculate the spring constant of the spring. State the unit.

Model Answer -- 3(a)

Extension = 16 - 10 = 6 cm = 0.06 m

Correct extension calculated (6 cm or 0.06 m) [1]

F = k x e so k = F / e = 3.0 / 0.06

Correct substitution into Hooke's law [1]

k = 50 N/m

Correct answer: 50 N/m [1]

Mark 1 -- correct extension (6 cm or 0.06 m)

Mark 2 -- correct substitution

Mark 3 -- correct answer with unit (50 N/m)

(b) [3]

(i) Sketch a load-extension graph for a spring. Label the limit of proportionality on your graph.
(ii) Explain what happens to the spring when it is stretched beyond the limit of proportionality.

Model Answer -- 3(b)

(i) Straight line through the origin showing force is proportional to extension [1]

Limit of proportionality correctly labelled at the point where the line starts to curve [1]

(ii) Beyond the limit of proportionality, the spring no longer obeys Hooke's law -- the extension is no longer proportional to the load. The spring may be permanently deformed (does not return to original length when the load is removed) [1]

Mark 1 -- straight line through origin

Mark 2 -- limit of proportionality labelled correctly

Mark 3 -- correct explanation of behaviour beyond limit

State Newton's First Law of Motion.

Model Answer -- 3(c)

An object remains at rest or continues to move in a straight line at constant speed [1]

unless acted upon by a resultant (unbalanced / net) force [1]

Mark 1 -- stays at rest or constant velocity in straight line

Mark 2 -- unless resultant/net force acts

(d) [4]

A car of mass 1200 kg accelerates at 2.5 m/s².
(i) Calculate the resultant force acting on the car.
(ii) A friction force of 400 N opposes the motion. Calculate the driving force produced by the engine.

Model Answer -- 3(d)

F = m x a

(i) Correct formula [1]

F = 1200 x 2.5 = 3000 N

Resultant force: 3000 N [1]

Engine force = Resultant force + Friction

Engine force = 3000 + 400 = 3400 N

(ii) Understanding that engine force = resultant + friction [1]

Correct answer: 3400 N [1]

Mark 1 -- correct formula (F = ma)

Mark 2 -- resultant force = 3000 N

Mark 3 -- engine = resultant + friction concept

Mark 4 -- engine force = 3400 N

(a) [2]

(i) Define the moment of a force.
(ii) Write the equation for calculating the moment of a force.

Model Answer -- 4(a)

The moment of a force is the turning effect of a force about a pivot [1]

Moment = force x perpendicular distance from the pivot

Correct equation (M = F x d) with "perpendicular" distance mentioned [1]

Mark 1 -- correct definition (turning effect)

Mark 2 -- correct equation

(b) [3]

A uniform beam of length 3.0 m is pivoted at its centre. A weight of 40 N is hung 1.2 m from the pivot on the left-hand side. Calculate the force needed on the right-hand side at a distance of 0.8 m from the pivot to balance the beam.

Model Answer -- 4(b)

Apply the principle of moments: clockwise moment = anticlockwise moment [1]

40 x 1.2 = F x 0.8

Correct substitution [1]

48 = 0.8F so F = 48 / 0.8 = 60 N

Correct answer: F = 60 N [1]

Mark 1 -- principle of moments stated or used

Mark 2 -- correct substitution

Mark 3 -- correct answer (60 N)

Describe a method to find the centre of gravity of an L-shaped piece of card.

Model Answer -- 4(c)

Make a small hole near one edge of the card and suspend it freely from a pin/nail so it can swing [1]

Hang a plumb line (string with weight) from the same pin and draw a vertical line on the card along the string [1]

Repeat from a different hole. The centre of gravity is where the two lines cross/intersect [1]

Mark 1 -- suspend freely from a hole/pin

Mark 2 -- use plumb line and draw vertical line

Mark 3 -- repeat and find intersection

(d) [2]

Explain why a racing car is designed with a low centre of gravity and a wide wheelbase.

Model Answer -- 4(d)

A low centre of gravity means the car has to be tilted through a larger angle before the line of action of the weight falls outside the base -- making it harder to topple over [1]

A wide wheelbase (wide base) provides a larger area of support, so the car is more stable and less likely to overturn when cornering at high speed [1]

Mark 1 -- low CoG linked to larger tilt angle / harder to topple

Mark 2 -- wide base linked to greater stability

(a) [2]

(i) Define momentum.
(ii) State the SI unit of momentum.

Model Answer -- 5(a)

Momentum is the product of mass and velocity (p = m x v) [1]

Unit: kg m/s (or N s) [1]

Mark 1 -- correct definition (mass x velocity)

Mark 2 -- correct unit (kg m/s or N s)

(b) [4]

A ball of mass 0.50 kg is moving at 8.0 m/s. It collides with a stationary ball of mass 1.5 kg. After the collision, the two balls stick together and move off as one object.

Calculate the velocity of the combined balls after the collision.

Model Answer -- 5(b)

State the principle: total momentum before = total momentum after [1]

Before: p = (0.50 x 8.0) + (1.5 x 0) = 4.0 kg m/s

Correct momentum before collision [1]

After: p = (0.50 + 1.5) x v = 2.0v

Correct expression for momentum after [1]

4.0 = 2.0v so v = 4.0 / 2.0 = 2.0 m/s

Correct answer: v = 2.0 m/s (in the original direction of motion) [1]

Mark 1 -- conservation of momentum principle stated

Mark 2 -- correct total momentum before (4.0 kg m/s)

Mark 3 -- correct expression for combined mass after

Mark 4 -- correct final velocity (2.0 m/s)

Calculate the impulse (change in momentum) experienced by the first ball (0.50 kg) during the collision.

Model Answer -- 5(c)

Impulse = change in momentum = m(v - u)

Impulse = 0.50 x (2.0 - 8.0) = 0.50 x (-6.0)

Correct substitution (using the first ball's initial and final velocities) [1]

Impulse = -3.0 kg m/s (or N s)

Correct answer: -3.0 N s (or 3.0 N s decrease / magnitude 3.0 N s) [1]

Mark 1 -- correct method and substitution

Mark 2 -- correct answer (-3.0 N s or 3.0 N s)

(d) [4]

In a car crash, a driver with a momentum of 1200 kg m/s is brought to rest.

(i) A car with crumple zones takes 0.30 s to stop. Calculate the average force on the driver.
(ii) A car without crumple zones takes 0.050 s to stop. Calculate the average force on the driver.
(iii) Using your answers, explain why crumple zones reduce the risk of injury.

Model Answer -- 5(d)

F = change in momentum / time = delta-p / delta-t

Correct use of F = delta-p / delta-t [1]

(i) F = 1200 / 0.30 = 4000 N

With crumple zones: F = 4000 N [1]

(ii) F = 1200 / 0.050 = 24 000 N

Without crumple zones: F = 24 000 N [1]

(iii) Crumple zones increase the time over which the momentum changes. Since F = delta-p / delta-t, a longer time means a smaller force acts on the driver, reducing the risk of injury [1]

Mark 1 -- correct formula (F = delta-p / delta-t)

Mark 2 -- force with crumple zones (4000 N)

Mark 3 -- force without crumple zones (24 000 N)

Mark 4 -- explanation linking longer time to smaller force

(a) [2]

State the principle of conservation of energy.

Model Answer -- 6(a)

Energy cannot be created or destroyed [1]

It can only be transferred from one form to another (the total energy in a closed system remains constant) [1]

Mark 1 -- energy cannot be created or destroyed

Mark 2 -- only transferred/converted

(b) [4]

A student of mass 60 kg runs up a flight of stairs of vertical height 4.0 m in 3.0 s.
(i) Calculate the gain in gravitational potential energy (GPE) of the student.
(ii) Calculate the power output of the student.

Model Answer -- 6(b)

GPE = m x g x h

(i) Correct formula [1]

GPE = 60 x 9.8 x 4.0 = 2352 J

Correct answer: GPE = 2352 J (accept 2350 J) [1]

Power = energy transferred / time = 2352 / 3.0

(ii) Correct formula and substitution [1]

Power = 784 W

Correct answer: Power = 784 W (accept 783 W or 780 W) [1]

Mark 1 -- correct GPE formula

Mark 2 -- GPE = 2352 J

Mark 3 -- correct power formula and substitution

Mark 4 -- Power = 784 W

A roller coaster car of mass 500 kg starts from rest at the top of a hill of height 30 m. Assuming no friction, calculate the speed of the car at the bottom of the hill.

Model Answer -- 6(c)

GPE at top = KE at bottom

mgh = 1/2 mv²

Equating GPE to KE (mass cancels) [1]

gh = 1/2 v² so v² = 2gh = 2 x 9.8 x 30 = 588

Correct substitution [1]

v = sqrt(588) = 24.2 m/s

Correct answer: v = 24.2 m/s (accept 24 m/s) [1]

Mark 1 -- GPE = KE or mgh = 1/2 mv²

Mark 2 -- correct substitution

Mark 3 -- correct answer (24.2 m/s)

(d) [3]

In practice, friction is present and the car reaches a speed of only 22 m/s at the bottom.
Calculate the efficiency of the energy transfer.

Model Answer -- 6(d)

Useful KE at bottom = 1/2 x 500 x 22² = 1/2 x 500 x 484 = 121 000 J

Correct KE at actual speed [1]

Total GPE at top = 500 x 9.8 x 30 = 147 000 J

Correct GPE input [1]

Efficiency = (useful output / total input) x 100%

Efficiency = (121 000 / 147 000) x 100% = 82.3%

Correct answer: 82.3% (accept 82%) [1]

Mark 1 -- correct actual KE (121 000 J)

Mark 2 -- correct GPE (147 000 J)

Mark 3 -- correct efficiency (82.3% or 82%)

(a) [2]

(i) Define pressure.
(ii) State the SI unit of pressure.

Model Answer -- 7(a)

Pressure is the force per unit area (or force acting per unit area) [1]

p = F / A

SI unit: pascal (Pa) or N/m² [1]

Mark 1 -- correct definition (force per unit area)

Mark 2 -- correct SI unit (Pa or N/m²)

(b) [3]

A woman weighing 600 N stands on one heel of area 1.0 cm².
Calculate the pressure exerted on the floor. Give your answer in Pa.

Model Answer -- 7(b)

Area = 1.0 cm² = 1.0 x 10^-4 m² = 0.0001 m²

Correct conversion of area to m² [1]

p = F / A = 600 / 0.0001

Correct substitution [1]

p = 6 000 000 Pa = 6.0 x 10⁶ Pa

Correct answer: 6 000 000 Pa (or 6.0 x 10⁶ Pa) [1]

Mark 1 -- correct unit conversion (1 cm² = 10^-4 m²)

Mark 2 -- correct substitution into p = F/A

Mark 3 -- correct answer (6 000 000 Pa or 6.0 x 10⁶ Pa)

Calculate the pressure due to the water at the bottom of a swimming pool that is 2.5 m deep.
(Density of water = 1000 kg/m³, g = 9.8 N/kg)

Model Answer -- 7(c)

p = rho x g x h

Correct formula stated [1]

p = 1000 x 9.8 x 2.5

Correct substitution [1]

p = 24 500 Pa

Correct answer: 24 500 Pa (or 2.45 x 10⁴ Pa) [1]

Mark 1 -- correct formula (p = rho g h)

Mark 2 -- correct substitution

Mark 3 -- correct answer (24 500 Pa)

(d) [4]

Compare solar cells and nuclear power as energy resources. Discuss each of the following:
- renewability
- reliability
- environmental impact

Model Answer -- 7(d)

Renewability: Solar energy is renewable -- the Sun will not run out (on a human timescale). Nuclear fuel (uranium) is non-renewable -- supplies are limited and will eventually run out [1]

Reliability: Solar cells are unreliable as they depend on sunlight -- they do not work at night or in cloudy conditions. Nuclear power is reliable -- it can produce a constant (baseload) supply of electricity 24/7 regardless of weather [1]

Environmental (Solar): Solar cells produce no greenhouse gases or pollution during operation. Manufacturing panels requires energy and materials. No waste products during use. Visual impact on landscape [1]

Environmental (Nuclear): Nuclear power produces no greenhouse gases during operation but produces radioactive waste that is hazardous and difficult to store safely for thousands of years. Risk of nuclear accidents (though very rare). Thermal pollution of waterways used for cooling [1]

Mark 1 -- correct comparison of renewability

Mark 2 -- correct comparison of reliability

Mark 3 -- environmental impact of solar discussed

Mark 4 -- environmental impact of nuclear discussed

IGCSE Physics Paper 4 (Theory / Extended)

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