OxfordAQA International GCSE Physics: Momentum, collisions and safety in road transport (9203) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover sections 3.1.4 (Momentum) and 3.1.5 (Safety in public transport) of the OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards. The qualification is untiered. Learn the content first from the study guide. Related pages: the OxfordAQA International GCSE Physics hub, the checklist and the free diagnostics.

Questions

1. A golf ball of mass 0.046 kg leaves the club at 55 m/s. (a) Calculate the momentum of the ball. [2] (b) State the unit of momentum. [1]

2. A trolley of mass 1.5 kg moving at 2.0 m/s collides with a stationary trolley of mass 0.50 kg. The trolleys stick together. Calculate their velocity after the collision. [4]

3. Two trolleys of mass 0.80 kg and 0.30 kg are at rest, touching, with a compressed spring between them. The spring is released and the 0.30 kg trolley moves off at 1.6 m/s. (a) State the total momentum of the trolleys before release. [1] (b) Calculate the velocity of the 0.80 kg trolley. [3]

4. At a fairground, a dodgem car and driver of total mass 220 kg moves right at 3.0 m/s. A second car and driver of total mass 180 kg moves left at 2.0 m/s. They collide head on and stick together. Calculate their velocity immediately after the collision, giving its direction. [5]

5. A ball of mass 0.050 kg hits a wall at 24 m/s and rebounds along the same line at 16 m/s. The contact lasts 0.010 s. Calculate the average force on the ball. [4]

6. Explain how an air bag and a seat belt together reduce the risk of serious injury to a driver in a head-on collision. [4]

7. A gymnast of mass 50 kg lands at 5.0 m/s and stops completely. (a) Calculate the force on the gymnast if she stops in 0.050 s on a hard floor. [3] (b) Calculate the force if she stops in 0.40 s on a crash mat. [2] (c) Explain why a cycle helmet protects a cyclist’s head in a similar way. [2]

8. A lorry of mass 8000 kg moves along a straight, level road at a steady speed. The driving force is 6200 N. (a) State the size of the resistive forces. [1] (b) The driver increases the driving force to 7400 N. The resistive forces stay the same. Calculate the initial acceleration. [3]

9. A van of mass 2400 kg travels at 15 m/s. The driver’s reaction time is 0.80 s. The braking force is 7200 N. (a) Calculate the thinking distance. [2] (b) Calculate the kinetic energy of the van. [2] (c) Calculate the braking distance. [2] (d) Calculate the stopping distance, and the deceleration while braking. [3]

10. (a) State two factors that can increase a driver’s reaction time. [2] (b) Explain which part of the stopping distance is affected by a longer reaction time. [2] (c) A car brakes with the same force at 30 m/s and at 15 m/s. Compare the kinetic energies and the braking distances. [3] (d) Explain why a very large deceleration is dangerous. [2]

11. A car of mass 1250 kg travels at 25 m/s on a dry road. The driver’s reaction time is 0.60 s and the braking force is 6250 N. (a) Calculate the stopping distance. [5] (b) The driver has been drinking alcohol and the reaction time becomes 1.4 s. Calculate the new stopping distance. [2] (c) On a wet road the braking force is halved. Calculate the braking distance and explain why it changes. [3] (d) Describe the energy transfers in the brakes as the car stops. [2]

12. A truck of mass 2400 kg moves at 6.0 m/s and runs into the back of a stationary car of mass 1200 kg. The two vehicles stay joined after impact. (a) Find their common velocity. [3] (b) Calculate the change in momentum of the car. [2] (c) The contact lasts 0.30 s. Calculate the average force on the car. [2] (d) Explain how a crumple zone on the car would change this force. [2] (e) Show that the kinetic energy has decreased in the collision. [3]

Answers

1. (a) p = mv = 0.046 × 55 [1] = 2.53, so 2.5 kg m/s (2 significant figures) [1] [2] (b) kg m/s [1] [1] Examiner insight: Show the substitution before the answer. Give the unit with the number as well as stating it separately.

2. Take the original direction as positive. Momentum before = 1.5 × 2.0 + 0.50 × 0 [1] = 3.0 kg m/s [1]. Momentum after = (1.5 + 0.50) × v = 2.0v [1]. Conservation of momentum: v = 3.0 / 2.0 = 1.5 m/s in the original direction [1] [4] Examiner insight: Using the combined mass after the collision is the key step; writing a line with total momentum before and after makes your method clear to the reader.

3. (a) 0 [1] [1] (b) Take the direction of the 0.30 kg trolley as positive. Total after must be zero: 0.30 × 1.6 + 0.80 × v = 0 [1]. 0.48 + 0.80v = 0, so 0.80v = −0.48 [1]. v = −0.60 m/s, so the 0.80 kg trolley moves at 0.60 m/s, opposite to the other trolley [1] [3] Examiner insight: State the direction as well as the speed. A negative sign with no explanation of what it means leaves the answer incomplete.

4. Take right as positive. Before: 220 × 3.0 + 180 × (−2.0) [1] = 660 − 360 [1] = 300 kg m/s [1]. After: (220 + 180) × v = 400v [1]. v = 300 / 400 = 0.75 m/s to the right [1] [5] Examiner insight: The sign of the second car’s momentum is the common slip. Mark the positive direction on a sketch first.

5. Take the direction towards the wall as positive. Initial momentum = 0.050 × 24 = 1.2 kg m/s [1]. Final momentum = 0.050 × (−16) = −0.80 kg m/s [1]. Δp = −0.80 − 1.2 = −2.0 kg m/s, size 2.0 kg m/s [1]. F = Δp / t = 2.0 / 0.010 = 200 N [1] [4] Examiner insight: When an object rebounds, the velocity changes sign, so the change is the sum of the two speeds, not their difference.

6. The driver’s momentum must change to zero in the collision [1]. The seat belt stretches slightly and the air bag inflates and cushions the head and chest, so the driver stops over a longer time [1]. F = Δp / t, so a longer time means a smaller force [1]. The air bag also spreads the force over a larger area [1] [4] Examiner insight: Give the chain: longer time, momentum changes more slowly, smaller force. Saying only that the features “protect” or “absorb the impact” does not explain.

7. (a) Δp = 50 × 5.0 = 250 kg m/s [1]. F = 250 / 0.050 [1] = 5000 N [1] [3] (b) F = 250 / 0.40 [1] = 625 N [1] [2] (c) The foam inside the helmet crushes, so the head takes longer to stop [1]. The change in momentum happens over a longer time, so the force on the head is smaller [1] [2] Examiner insight: Part (c) asks for the same idea in a new context; reusing the time-and-force reasoning, with the helmet foam named, is the point of the question.

8. (a) 6200 N [1] [1] (b) Resultant force = 7400 − 6200 = 1200 N [1]. a = F / m = 1200 / 8000 [1] = 0.15 m/s² [1] [3] Examiner insight: The resultant force is a difference, not the driving force itself. Subtracting the resistive force before using F = ma is the step to show.

9. (a) Thinking distance = speed × reaction time = 15 × 0.80 [1] = 12 m [1] [2] (b) KE = ½mv² = ½ × 2400 × 15² [1] = 270 000 J [1] [2] (c) Work done by the brakes = kinetic energy lost, so 7200 × d = 270 000 [1]. d = 37.5 m [1] [2] (d) Stopping distance = 12 + 37.5 = 49.5 m [1]. a = F / m = 7200 / 2400 [1] = 3.0 m/s² [1] [3] Examiner insight: Using the braking distance alone as the stopping distance is a common loss. Naming the two parts and adding them shows the method clearly.

10. (a) Any two of: tiredness, distractions, drugs, alcohol [1] [1] [2] (b) The thinking distance [1], because the vehicle keeps moving at its original speed during the reaction time [1] [2] (c) KE is proportional to speed squared, so the KE at 30 m/s is four times the KE at 15 m/s [1]. The same braking force does the same amount of work per metre, so the braking distance is four times as long [1] [1] [3] (d) Brakes can overheat and become less effective [1]; the wheels may skid so the driver loses control [1] [2] Examiner insight: Part (c) is about a comparison. Give the ratio (four times), not just “greater”, and state why using kinetic energy.

11. (a) Thinking distance = 25 × 0.60 = 15 m [1]. KE = ½ × 1250 × 25² [1] = 390 625 J [1]. Braking distance = 390 625 / 6250 = 62.5 m [1]. Stopping distance = 15 + 62.5 = 77.5 m [1] [5] (b) Thinking distance = 25 × 1.4 = 35 m [1]. Stopping distance = 35 + 62.5 = 97.5 m [1] [2] (c) Braking distance = 390 625 / 3125 = 125 m [1]. On a wet road there is less friction between the tyres and the road, so the largest braking force available is smaller [1], and the car needs a longer distance to lose its kinetic energy [1] [3] (d) Friction between the brake pads and the wheel does work [1]; kinetic energy is transferred to thermal energy and the temperature of the brakes rises [1] [2] Examiner insight: In part (c), halving the force doubles the braking distance because the energy to remove is the same. Say that the thinking distance is unchanged.

12. (a) Take the truck’s direction as positive. Before: 2400 × 6.0 = 14 400 kg m/s [1]. After: (2400 + 1200) × v = 3600v [1]. v = 4.0 m/s [1] [3] (b) Δp = 1200 × 4.0 [1] = 4800 kg m/s [1] [2] (c) F = Δp / t = 4800 / 0.30 [1] = 16 000 N [1] [2] (d) The crumple zone would collapse, increasing the contact time [1]; the same change in momentum over a longer time gives a smaller force on the car and its passengers [1] [2] (e) KE before = ½ × 2400 × 6.0² = 43 200 J [1]. KE after = ½ × 3600 × 4.0² = 28 800 J [1]. 28 800 < 43 200, so kinetic energy has decreased [1] [3] Examiner insight: Write out both kinetic energy totals and compare them. Momentum is conserved here but kinetic energy is not, and saying so shows understanding.

Where marks are usually lost

• Forgetting to define a positive direction, then adding momenta that point in opposite directions.
• Using only one object’s mass after two objects stick together, or doubling braking distance when speed doubles (it is four times).
• Saying a safety feature “reduces the momentum” or “absorbs the force” rather than increasing the time.

Next steps

• Review the revision notes and study guide.
• Go to the OxfordAQA International GCSE Physics hub and the checklist.
• Find your weak spots with all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International GCSE Physics (9203) specification, for exams May/June 2018 onwards, published by OxfordAQA.