AQA A-Level Physics Paper 1, June 2022: Question 26

Tennis Ball Change in Momentum

Question 26 • Multiple Choice • 1 Mark

What this question tests

This question assesses your ability to combine kinematic equations of motion (or conservation of energy) with the definition of momentum and vector subtraction. Specifically, it tests your competence in handling vector directions during rebounds and managing unit conversions (grams to kilograms).

Question 26

Calculate the change in momentum during the collision with the ground

✅ Correct Answer

D • 0.614 N s

Mark Scheme: Option D is the only correct choice based on proper vector subtraction of velocities before and after impact.

💡 Key Knowledge

Momentum (p): Calculated as p = m × v , measured in N s or kg m s⁻¹ .
Conservation of Energy / Kinematics: Velocity just before impact can be found using v = √(2gh) or v² = u² + 2as .
Vector Nature: Change in momentum is Δp = final momentum - initial momentum . Taking upward as positive means Δp = m(+v_up) - m(-v_down) = m(v_up + v_down) .

🧠 Exam Technique

In multiple-choice calculation questions, common errors correspond to incorrect distractor options. Always calculate intermediate values carefully on your calculator and write down full unrounded figures before your final step to avoid rounding errors.

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❌ Common Errors

Sign convention mistake: Subtracting velocities instead of adding them ( v_final - v_initial where directions are ignored) leads to distractor C ( 0.215 N s ).
Mass unit failure: Forgetting to convert 58 g into 0.058 kg .
Kinematic trap: Squaring heights incorrectly or mixing up initial drop height with rebound height.

📐 Step-by-Step Calculation Guide

Convert units for mass:
m = 58 g = 0.058 kg
Find velocity of the ball just before hitting the ground (downward, v₁):
Using energy ( mgh = ½mv² ) or kinematics ( v² = u² + 2as ):
v₁ = √(2 × 9.81 × 1.8) = √35.316 ≈ 5.9427 m s⁻¹ (taking downwards as negative: -5.9427 m s⁻¹ )
Find velocity of the ball just after bouncing (upward, v₂):
v₂ = √(2 × 9.81 × 1.1) = √21.582 ≈ 4.6456 m s⁻¹ (taking upwards as positive: +4.6456 m s⁻¹ )
Calculate the change in momentum (Δp):
Δp = m(v₂ - v₁)
Δp = 0.058 × (4.6456 - (-5.9427))
Δp = 0.058 × (4.6456 + 5.9427)
Δp = 0.058 × 10.5883 = 0.6141 N s
Round to appropriate significant figures:
Matches 0.614 N s (Option D).