AQA AS Level Physics Paper 2, June 2024: Question 29

1 mark · Medium difficulty · Multiple Choice

Calculate the speed and direction of movement of glider X immediately after a collision with glider Y using conservation of momentum.

Practise this question

Question

Multiple choice question 29 featuring a diagram of two gliders, X and Y, on a horizontal frictionless track moving towards each other. Glider X has a mass of 400 g and moves right at 1.8 m s^-1, while glider Y has a mass of 300 g and moves left at 0.7 m s^-1. After colliding, Y moves right at 0.9 m s^-1, and four options A through D give potential speeds and directions for X.
Question text

29 Glider X of mass 400 g travels at 1.8 m s−1 to the right on a horizontal, frictionless track.

Glider Y of mass 300 g travels towards X at 0.7 m s−1.

X and Y collide.

Immediately after the collision, Y travels to the right at a speed of 0.9 m s−1.

What are the speed and direction of movement of X immediately after the collision?

[1 mark]

A 0.6 m s−1 to the left

B 0.6 m s−1 to the right

C 1.7 m s−1 to the left

D 1.7 m s−1 to the right

Mark scheme

Show the mark scheme Mark scheme table showing question number 29 with the correct answer option B, corresponding to 0.6 m s^-1 to the right.

29 B 0.6 m s−1 to the right

How to answer it

Glider Collision and Conservation of Momentum

What this question tests

This question assesses your understanding of the Principle of Conservation of Linear Momentum in a closed system, vector sign conventions for velocities in opposite directions, and unit conversions (grams to kilograms).

Question 29 • Multiple Choice • 1 Mark

Exam Question Analysis

✅ Correct Answer

B • 0.6 m s⁻¹ to the right

Awarded for correctly applying momentum conservation with proper vector signs.

💡 Key Knowledge

  • Momentum formula: p = m × v (where p is momentum, m is mass, v is velocity).
  • Vector Direction: Rightward velocities are positive (+); leftward velocities are negative (-).
  • Conservation Law: Total momentum before collision = Total momentum after collision.

🧠 Exam Technique

  • Always define a positive direction before writing out your equations. Here, taking to the right as positive is the natural choice.
  • Double-check whether masses need to be converted to SI units (kg). In ratios/multiplications where all masses are in grams, conversion isn't strictly necessary, but it is safer practice to convert them.

❌ Common Errors

  • Sign errors: Forgetting to make Glider Y's initial velocity negative ( -0.7 m s⁻¹ ) because it travels to the left.
  • Mass mixing: Mismatching the masses and velocities of Glider X and Glider Y during substitution.

📐 Step-by-Step Calculation

  1. State the sign convention: Let velocities to the right be positive (+).
  2. Calculate total momentum before collision (p_before):
    p_before = (m_X × v_X) + (m_Y × v_Y)
    p_before = (0.400 kg × +1.8 m s⁻¹) + (0.300 kg × -0.7 m s⁻¹)
    p_before = +0.72 - 0.21 = +0.51 kg m s⁻¹
  3. Set up momentum after collision (p_after):
    Let v_X' be the final velocity of Glider X.
    p_after = (m_X × v_X') + (m_Y × v_Y')
    p_after = (0.400 × v_X') + (0.300 × +0.9)
  4. Equate momentum before and after (p_before = p_after):
    0.51 = (0.400 × v_X') + 0.27
    0.400 × v_X' = 0.51 - 0.27
    0.400 × v_X' = +0.24
    v_X' = 0.24 / 0.400 = +0.6 m s⁻¹
  5. Interpret the final sign: Since the result is positive, the velocity is 0.6 m s⁻¹ to the right , which matches option B.

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.