You know the content. Now let's learn how Cambridge examiners test it.
Challenge questions are not about harder facts. They test the same facts you already know — but wrapped in unfamiliar contexts, combined in unexpected ways, or phrased to exploit common misconceptions.
This guide will teach you three things:
1. Where students go wrong — the traps examiners set and how to spot them.
2. How to think through tricky questions — step-by-step reasoning, not guessing.
3. How to tell similar questions apart — because one word can change the answer completely.
Work through each section carefully. By the end, you will not just know the content — you will know how to apply it under pressure.
These are the beliefs that feel true but are not. Examiners love to write wrong answers that match these misconceptions — if you hold the misconception, the wrong answer looks perfect.
Challenge questions often require connecting multiple concepts. Here is how to think through them step by step.
Identify the given values: m1 = 2 kg, v1 = 3 m/s, m2 = 1 kg, v2 = 0 m/s. They “stick together” tells us this is an inelastic collision — after the collision, the two trolleys move as one combined mass.
This is a conservation of momentum problem. Total momentum before = total momentum after. Momentum is always conserved in collisions (no external resultant force).
Before: p = m1v1 + m2v2 = (2)(3) + (1)(0) = 6 kg⋅m/s
After: p = (m1 + m2)v = 3v
Setting equal: 3v = 6, so v = 2 m/s
KE before = ½(2)(3²) = 9 J
KE after = ½(3)(2²) = 6 J
KE is NOT conserved — 3 J has been “lost.”
Put it all together with the correct physics language.
Three sections on the graph: Section 1 (0 to 10 s) — acceleration phase, straight line rising from 0 to 20 m/s. Section 2 (10 s to 30 s) — constant speed phase, horizontal line at 20 m/s. Section 3 (30 s to 35 s) — deceleration phase, straight line falling from 20 to 0 m/s.
Acceleration = gradient of speed-time graph = change in speed ÷ time.
a = (20 − 0) / 10 = 2 m/s²
Distance = area under the speed-time graph.
Section 1 (triangle): ½ × 10 × 20 = 100 m
Section 2 (rectangle): 20 × 20 = 400 m
Section 3 (triangle): ½ × 5 × 20 = 50 m
Total = 100 + 400 + 50 = 550 m
Section 1: Resultant force is forward (driving force > friction), causing acceleration.
Section 2: Resultant force is zero (driving force = friction), constant velocity.
Section 3: Resultant force is backward (friction/braking force > driving force), causing deceleration.
Depth h = 3 m, density ρ = 1000 kg/m³, g = 10 N/kg, atmospheric pressure = 100,000 Pa. The question asks for total pressure.
Total pressure = atmospheric pressure + pressure due to the liquid column. Many students forget atmospheric pressure! The water is open to the atmosphere, so the air above pushes down on the surface, and this pressure is transmitted through the water.
Pliquid = ρgh = 1000 × 10 × 3 = 30,000 Pa
Ptotal = Patm + Pliquid = 100,000 + 30,000 = 130,000 Pa
Mass m = 500,000 kg, initial velocity u = 0 m/s, final velocity v = 100 m/s, time t = 10 s.
KE = ½mv² = 0.5 × 500,000 × (100)²
= 0.5 × 500,000 × 10,000 = 2,500,000,000 J
= 2.5 × 10&sup9; J (2.5 GJ)
Power = Energy / time = 2.5 × 10&sup9; / 10 = 2.5 × 10&sup8; W = 250 MW
The calculation only accounts for kinetic energy gained. But the rocket also:
(1) Does work against gravity — gaining gravitational potential energy as it rises.
(2) Does work against air resistance — energy transferred as thermal energy to the atmosphere.
(3) Has hot exhaust gases carrying away thermal and kinetic energy.
So the actual power output must be significantly greater than 250 MW.
Dimensions: 10 cm × 5 cm × 4 cm. Mass = 160 g. Water density = 1.0 g/cm³.
Volume = 10 × 5 × 4 = 200 cm³
Density = mass / volume = 160 / 200 = 0.8 g/cm³
Since the density of the block (0.8 g/cm³) is less than the density of water (1.0 g/cm³), the block will float. Rule: objects less dense than the fluid float; objects more dense sink.
For a floating object: weight of block = weight of water displaced.
So: mass of block = mass of water displaced.
160 g = 1.0 × Vsubmerged, so Vsubmerged = 160 cm³.
If floating on the 10 × 5 cm face: depth = 160 / (10 × 5) = 160 / 50 = 3.2 cm
These question pairs look almost identical but have different answers. Learning to spot the difference is a Challenge-level skill.
A horizontal line means “the y-axis quantity is not changing.” On a d-t graph, constant distance = not moving. On a v-t graph, constant speed = moving but not accelerating. Same line shape, completely different physical situations. Always check: what is on the y-axis?
Momentum conservation is a universal law (always true in collisions with no external force). Kinetic energy conservation depends on the type of collision. Never confuse the two — the word “conserved” applies differently to each quantity. One word in the question (“momentum” vs “kinetic energy”) changes the answer entirely.
Mass is constant everywhere — same on Earth, Moon, or in deep space. Weight depends on the gravitational field strength (g) and changes with location. Always check whether the question asks for mass (kg) or weight (N). One asks for a property of the object; the other asks for a force.
Constant velocity = zero resultant force (forces balanced). Acceleration = non-zero resultant force (forces unbalanced). Motion does not require force — CHANGE in motion requires force. The word “constant velocity” vs “accelerates” completely changes the force analysis.
Pressure depends on depth (P = ρgh), not on container shape or size. But force depends on both pressure AND area (F = PA). Same pressure but different area means different force. Students who confuse pressure and force get both questions wrong. Watch out for the exact word: “pressure” or “force”?
Every subtopic in Topic 1 links to the others. Understanding these connections is how you handle questions that combine multiple concepts.
Speed, velocity, and acceleration are described by distance-time and speed-time graphs. Gradient and area calculations link graphs directly to the equations of motion.
Resultant force causes acceleration (F = ma). No resultant force means constant velocity (1st Law). Equal and opposite forces act in every interaction (3rd Law).
Mass (kg) is intrinsic to an object. Weight = mg changes with location. Density = m/V determines whether objects float or sink in a fluid.
p = mv is conserved in all collisions. Links force and time through F = Δp/Δt. Whether KE is conserved distinguishes elastic from inelastic collisions.
Work = Fd transfers energy between stores. KE = ½mv², GPE = mgh. Power = E/t measures rate of energy transfer. Efficiency = useful output / total input.
P = F/A for solids on surfaces. P = ρgh for liquids at depth. Connects density, force, and depth in one equation. Acts in all directions in fluids.
These are real student answers that sound reasonable but are wrong. Can you spot the flaw before revealing it?
“The cricket ball lands first because it is heavier, so gravity pulls it harder.”
The student confuses greater gravitational force with greater acceleration. While gravity does exert a greater force on the cricket ball (because W = mg and it has more mass), it also has more inertia (resistance to acceleration). By F = ma, the acceleration a = F/m = mg/m = g for both objects. The mass cancels out. In a vacuum, they land at exactly the same time. In air, the cricket ball may land marginally first because air resistance has less relative effect on it (its weight-to-drag ratio is higher), but the student’s reasoning is fundamentally wrong.
In a vacuum, both land at the same time because gravitational acceleration (g) is independent of mass. In air, the cricket ball lands slightly earlier because air resistance is a smaller fraction of its weight, but both experience the same gravitational acceleration g. The key physics: greater force on the heavier object is exactly compensated by its greater inertia.
“When the auto brakes, a force pushes the passengers forward.”
There is no forward force on the passengers. The student has invented a force that does not exist. The passengers continue moving forward due to inertia (Newton’s First Law). Their bodies were moving at the speed of the auto-rickshaw. When the auto decelerates (a backward force acts on the auto via brakes), the passengers’ bodies tend to continue at the original speed because no backward force has been applied to them yet (until the seatbelt or dashboard stops them).
The passengers continue to move forward at the original speed due to inertia (Newton’s First Law). The auto-rickshaw decelerates underneath them, but their bodies resist the change in motion. They appear to jerk forward relative to the auto, but they are actually just continuing their original motion while the auto slows down. No forward force acts on them — they simply maintain their velocity while the vehicle around them decelerates.
“The ball stops at the top because the force of the throw runs out.”
The “force of the throw” does NOT travel with the ball. Once the ball leaves the hand, the only force acting on it is gravity (and air resistance). There is no upward force from the throw — the hand gave the ball initial velocity, not a lasting force. This is the “impetus misconception” that has been disproven since Newton. The ball decelerates because gravity acts downward while the ball moves upward. At the top, velocity = 0 momentarily, but acceleration is still g = 9.8 m/s² downward.
Once released, the only force on the ball is gravity (downward). Gravity decelerates the ball as it rises (acceleration is 9.8 m/s² downward throughout the entire journey). At the highest point, velocity momentarily equals zero, but the acceleration due to gravity still acts — the ball does not “pause.” It immediately begins falling. The ball stops because gravity has been decelerating it, not because any force “ran out.”
“The diver experiences more pressure in the ocean because there is more water around them.”
Pressure at a given depth depends only on depth (h), density (ρ), and gravitational field strength (g). The formula P = ρgh has no term for the volume or width of the body of water. A 10 m depth of seawater produces the same pressure whether the diver is in an ocean, a deep pool, or a narrow vertical tube (assuming the same water density). The student is confusing total volume of water with the pressure at a specific depth.
The pressure is the same in both cases (assuming the same water density). P = ρgh depends only on depth, not on the total volume of water. At 10 m depth: P = 1000 × 10 × 10 = 100,000 Pa of water pressure (plus atmospheric pressure) in both locations. The amount of water surrounding the diver is irrelevant to the pressure at that depth.
“Yes, the efficiency is 120% because the output energy is greater than the input energy. The motor must be very well designed.”
Efficiency can never exceed 100%. This would violate the law of conservation of energy — you cannot get more energy out of a machine than you put in. If a calculation gives efficiency > 100%, there must be a measurement error, a calculation mistake, or the student has confused input and output values. No machine, no matter how well designed, can create energy from nothing.
No, 120% efficiency is impossible. The law of conservation of energy states that energy cannot be created. The student has likely made a measurement or calculation error — perhaps swapping the input and output values, or using incorrect units. Maximum possible efficiency is 100% (all input energy converted to useful output), but in practice, some energy is always wasted as thermal energy, sound, etc., so real efficiencies are always less than 100%.
These are Challenge-level questions. Read every option carefully before selecting. Each wrong option is designed to match a specific misconception.
This tests whether students can calculate the area under a speed-time graph using the correct geometric shapes. The acceleration phase creates a triangle, not a rectangle. Students who assume the full distance is speed × total time forget that the car was not at top speed for the entire journey.
Lift questions test understanding of apparent weight. The resultant force equation: Scale reading − mg = ma, so Scale reading = m(g + a) when accelerating upward. Option A is the deliberate trap for students who subtract instead of add — a sign error that gives a plausible-looking answer.
The key trap is the bouncing back. When a ball reverses direction, its velocity becomes negative. Students who forget this get the momentum equation wrong: 0.5(4) = 0.5(2) + 1.5v gives v = 0.67, which is not even an option. The correct equation uses −2 for the bounced ball: 0.5(4) = 0.5(−2) + 1.5v.
This is straightforward once you know the rule: objects with density less than the fluid float; objects denser sink. But students sometimes overthink it — “steel ships float!” True, but a steel ship has an overall density less than water because of the air inside. A solid steel bolt does not. Always compare the object’s actual density to the fluid density.
Terminal velocity means constant speed, which means zero resultant force. Option D is the main trap — it describes the situation before terminal velocity is reached. Option A tests whether students confuse “no acceleration” with “no air resistance.” Understanding the sequence (accelerating → balanced forces → constant speed) is essential.
Conservation of energy converts GPE at the top entirely to KE at the bottom (starting from rest, no friction). The mass cancels: mgh = ½mv² gives v = √(2gh), independent of mass! Option C is the classic trap — students calculate v² correctly but forget the square root. Always ask yourself: “did I solve for v or v²?”
Hydraulic systems transmit pressure equally. The key relationship is P1 = P2, so F1/A1 = F2/A2. Students make three typical errors: (1) confusing pressure with force (Option D), (2) inverting the ratio (Option A), (3) arithmetic mistakes (Option B). Always calculate pressure first, then use it to find the unknown force.
Power = work done / time. Work done against gravity = mgh (force × distance = weight × height). The three-step calculation (weight, then work, then power) catches students who skip steps. Option A forgets to convert mass to weight (multiply by g). Option C confuses energy with power. Always check your units: Watts = Joules per second.
Braking distance is proportional to v² (since braking force is constant, work done = Fd, and KE = ½mv²). Double the speed → 4× the KE → 4× the braking distance. Option A is the most common wrong answer — students who think distance is directly proportional to speed. This is a safety-critical concept that appears frequently in exams.
This question combines multiple misconceptions: (1) weight in space — gravity does not disappear in orbit (Option A), (2) speed vs velocity — constant speed with changing direction means acceleration (Option C), (3) Newton’s First Law misapplication — constant speed in a straight line would mean no force, but circular motion requires centripetal force (Option D). Option B requires understanding that “weightlessness” in orbit is free fall, not zero gravity.