AQA AS Level Physics Paper 2, June 2024: Question 22

1 mark · Medium difficulty · Multiple Choice

Calculate the average force exerted on a cricket ball of mass 0.16 kg when it changes velocity from 35 m s^-1 towards a bat to 30 m s^-1 in the opposite direction during a contact time of 52 ms.

Practise this question

Question

Multiple choice question 22. A cricket ball of mass 0.16 kg travels at a speed of 35 m s^-1 towards a bat, is in contact for 52 ms, and travels in the opposite direction with a speed of 30 m s^-1. Four options are given: A 0.015 N, B 0.20 N, C 15 N, D 200 N.
Question text

22 A cricket ball of mass 0.16 kg travels at a speed of 35 m s−1 towards a bat.

When the ball is hit, it is in contact with the bat for 52 ms.

As a result, the ball travels in the opposite direction with a speed of 30 m s−1.

What is the average force on the ball from the bat?

[1 mark]

A 0.015 N

B 0.20 N

C 15 N

D 200 N

Mark scheme

Show the mark scheme Mark scheme for question 22 showing the correct answer is D, corresponding to 200 N.

22 D 200 N

How to answer it

Calculating Average Force from Momentum Change

What this question tests

This question assesses your understanding of Newton's Second Law of Motion expressed in terms of rate of change of momentum, vector directions (signs), and handling metric prefixes like milliseconds (ms). You must calculate impulse and deduce the correct average force during a collision.

Question Part 2.2 [1 Mark]

Multiple Choice Question: Average Force on a Ball

✅ Correct Answer

D: 200 N

Mark Scheme: Option D is awarded the single available mark.

💡 Key Knowledge

  • Force is equal to the rate of change of momentum: F = Δp / Δt = m(v - u) / Δt
  • Momentum is a vector quantity. Reversing direction means a change in sign (+ / -).
  • Prefix conversion: 52 ms = 52 × 10⁻³ s .

🧠 Exam Technique

  • Define a positive direction before starting (e.g., motion towards the bat is positive, away from the bat is negative, or vice versa).
  • Always check for hidden traps like time prefixes (milliseconds) and vector subtractions.

❌ Common Errors

  • Substraction error: Doing 35 - 30 instead of accounting for the direction change, leading to 35 - (-30) = 65 . Failing to do this gives option C (15 N).
  • Prefix omission: Forgetting to convert 52 ms into seconds, dividing by 52 instead of 0.052 .

📐 Step-by-Step Calculation

  1. Establish sign convention: Let motion towards the bat be positive ( u = +35 m s⁻¹ ). Therefore, motion away from the bat after being hit is negative ( v = -30 m s⁻¹ ).
  2. Calculate change in velocity (Δv):
    Δv = v - u = (-30) - (+35) = -65 m s⁻¹ (magnitude is 65).
  3. Substitute into the force equation:
    F = m × (Δv / Δt)
    F = 0.16 kg × (-65 m s⁻¹) / (52 × 10⁻³ s)
  4. Evaluate final value:
    F = -10.4 / 0.052 = -200 N (The magnitude of the average force is 200 N).

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.