Question 1Multiple choice · 1 markA* challenge
A car accelerates uniformly from 72 km/h to 108 km/h in 6.0 s.
What is its acceleration?
- A 1.7 m/s²
- B 5.0 m/s²
- C 6.0 m/s²
- D 8.3 m/s²
Show answer
Answer: A
10 free IGCSE Physics Motion, Forces & Energy questions written in the style of Cambridge Physics papers: 5 multiple-choice and 5 structured, each with its answer and mark scheme. 3 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Physical quantities & measurement, Motion, Mass & weight, Density, Effects of forces, Turning effect & centre of gravity, Momentum, Energy, work & power, Energy resources, Pressure, Pendulums & timing. Work through the questions below, then generate a full Motion, Forces & Energy paper with a fresh set every time.
Generate a full Motion, Forces & Energy paper →A car accelerates uniformly from 72 km/h to 108 km/h in 6.0 s.
What is its acceleration?
Answer: A
A tall, empty lorry is parked on a slope.
1 A wider wheelbase makes the lorry less likely to topple.
2 Loading heavy boxes on the floor of the lorry makes it more stable.
3 The lorry topples if its centre of gravity is directly above the base.
Which statements are correct?
Answer: B
Why does a student time 20 oscillations of a pendulum rather than one?
Answer: A
What is the unit of the spring constant?
Answer: A
A box of weight 80 N rests on an area of 0.05 m². What pressure does it exert on the floor?
Answer: D
A uniform plank of weight 200 N and length 4.0 m rests on a pivot 1.0 m from one end. A box rests on the short end to keep it horizontal.
State where the weight of the uniform plank acts.
Calculate the weight of the box.
Calculate the upward force of the pivot on the plank.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | at its centre at the centre of the plank, 2.0 m from either end B1 | 1 |
| (b) | 200 N 200 × 1.0 = W × 1.0 C1 200 N A1 | 2 |
| (c) | 400 N 400 N (sum of downward forces) B1 | 1 |
A trolley A of mass 0.50 kg moves at 3.0 m/s along a horizontal track towards a stationary trolley B of mass 0.30 kg. The trolleys collide and stick together.
State the principle of conservation of momentum.
Calculate the speed of the trolleys immediately after the collision.
Show that kinetic energy is not conserved in the collision, and state what happens to the energy lost.
Explain why a spring buffer between the trolleys would reduce the force on each trolley.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | total momentum before = total momentum after; no external force the total momentum of a system remains constant B1 provided no external (resultant) force acts B1 | 2 |
| (b) | 1.88 m/s momentum before = 0.50 × 3.0 = 1.50 kg m/s C1 (0.50 + 0.30) v = 1.50 C1 v = 1.88 m/s A1 | 3 |
| (c) | before 2.25 J; after 1.41 J; transferred to thermal / sound KE before = 2.25 J C1 KE after = 1.41 J (less) C1 energy transferred to the thermal store of the trolleys / surroundings (and sound) B1 | 3 |
| (d) | longer collision time; same change of momentum so smaller force the collision lasts longer B1 the change of momentum is the same, so the force (= change of momentum ÷ time) is smaller B1 | 2 |
A car of mass 1500 kg travels along a straight, horizontal road at 25 m/s. The driver sees a hazard. After a delay of 0.7 s the brakes are applied and the car decelerates uniformly at its maximum deceleration of 8 m/s² until it stops.
Sketch the speed–time graph for the car from the moment the driver sees the hazard until it stops. Label the values on both axes.
Determine the total distance travelled from the moment the driver sees the hazard until the car stops.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | horizontal line at 25 m/s to 0.7 s; straight line down to 0 at 3.83 s horizontal line at 25 m/s for 0.7 s B1 straight line sloping down to zero B1 reaching zero at 3.83 s B1 | 3 |
| (b) | 56.6 m distance at constant speed = 25 × 0.7 = 17.5 m C1 distance while decelerating = area = ½ × 25 × 3.13 = 39.1 m C1 56.6 m A1 | 3 |
A car A of mass 1400 kg moving at 16.0 m/s collides with a stationary car B of mass 1900 kg. The cars stick together.
Calculate the momentum of car A before the collision.
Calculate the speed of the cars immediately after the collision.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 22400 kg m/s p = mv C1 22400 A1 | 2 |
| (b) | 6.79 m/s Total momentum conserved C1 6.79 A1 | 2 |
A student wants to measure the thickness of one sheet of paper with a ruler.
Explain why measuring one sheet with the ruler is not accurate.
Describe how she can find an accurate value.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | too thin compared with ruler precision the sheet is much thinner than the smallest division on the ruler (1 mm) B1 | 1 |
| (b) | measure a stack and divide measure the thickness of a large stack (e.g. 100 sheets) B1 divide by the number of sheets B1 | 2 |
In the 0625 syllabus, Motion, Forces & Energy covers Physical quantities & measurement, Motion, Mass & weight, Density, Effects of forces, Turning effect & centre of gravity, Momentum, Energy, work & power, Energy resources, Pressure, Pendulums & timing.
This page has 10 Motion, Forces & Energy questions with answers. For more, the free generator on mentorriyaz.in builds a new Motion, Forces & Energy paper with a full mark scheme each time you click.
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Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in