AQA AS Level Physics Paper 2, November 2021: Question 25
1 mark · Medium difficulty · Multiple Choice
Calculate the speed of a ball of mass 0.044 kg immediately after being hit, using a force-time graph with a peak force of 1400 N and a total duration of 5 ms.
Practise this questionQuestion
Question text
25 A stationary ball is free to move. The ball is hit with a bat.
The graph shows how the force of the bat on the ball changes with time.
The ball has a mass of 0.044 kg.
What is the speed of the ball immediately after being hit?
[1 mark]
A 13 m s−1
B 60 m s−1
C 80 m s−1
D 160 m s−1
Mark scheme
Show the mark scheme
25 C 80 m s−1
How to answer it
Determining Speed from a Force-Time Graph
What this question tests
This question assesses your ability to link graphical analysis with momentum and Newton's laws of motion. Specifically, you must recognize that the area under a force-time graph represents impulse (change in momentum), convert metric prefixes correctly, and apply the relationship between impulse, mass, and velocity to calculate final speed.
Finding the Speed of the Ball Immediately After Impact
✅ Correct Answer
C: 80 m s⁻¹
💡 Key Knowledge
- Impulse Definition: Impulse is defined as change in momentum, given by Δ(mv) = F × Δt.
- Graphical Representation: The area under a force-time graph equals the impulse delivered to the object.
- Triangle Area Formula: Area = ½ × base × height.
🧠 Exam Technique
- Watch out for time axes given in milliseconds ( ms ). Always convert to seconds by multiplying by 10⁻³.
- Since the ball starts from rest, its initial velocity is zero, meaning the impulse is simply equal to its final momentum ( m × v ).
❌ Common Errors
- Forgetting the prefix: Forgetting to convert milliseconds to seconds leads to an answer that is out by a factor of 1,000.
- Geometry mistakes: Using base × height instead of ½ × base × height for the triangular peak.
📐 Step-by-Step Calculation
- Calculate the Impulse (Area under the graph):
Base = 5 ms = 5 × 10⁻³ s
Height = 1400 N
Area = ½ × (5 × 10⁻³ s) × (1400 N) = 3.5 N s (or kg m s⁻¹) - Relate Impulse to Change in Momentum:
Impulse = m × (v - u)
Since initial velocity u = 0 , Impulse = m × v - Rearrange for Velocity ( v ):
v = Impulse / mass
v = 3.5 / 0.044 = 79.545... m s⁻¹ - Round to Appropriate Significant Figures:
Rounding to 2 significant figures (matching the data given in the question like 0.044 kg) yields 80 m s⁻¹ (Option C).
Topics
Physics · 3.4 Mechanics and materials
Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.