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A2 Daily A Level Physics question

2026-03-01 OCR A Exam practice set A — Misconceptions clinic: Newton’s 3rd OCR-A Module 3.5 Newton’s laws of motion and momentum (Newton’s third law; impulse; conservation of momentum in isolated systems)

In a low-friction lab demo, a 60 kg student on a wheeled board and a 120 kg trolley are initially at rest. The student pushes the trolley for a short time; horizontal external forces are negligible. Immediately after they lose contact, the trolley moves at 0.60 m/s to the right. Which statement must be true about the student's motion just after separation?

  1. A A The student moves left at 0.60 m/s, because they experienced the same force as the trolley.
  2. B B The student moves left at 0.30 m/s, because their mass is smaller.
  3. C C The student moves left at 1.2 m/s, because both received impulses of equal magnitude in opposite directions. (correct)
  4. D D The student remains at rest, because only the trolley was pushed forward.

Answer

The correct answer is C.

Correct: C — C The student moves left at 1.2 m/s, because both received impulses of equal magnitude in opposite directions. With negligible external horizontal forces, the equal-and-opposite forces act for the same time, so impulses are equal in magnitude; hence m_student Δv_student = m_trolley Δv_trolley, giving Δv_student = (120/60) × 0.60 = 1.2 m/s in the opposite direction. A The forces are equal but the masses are not, so equal force does not imply equal speeds; the lighter student must have the larger speed. B This reverses the mass–speed relationship: for equal and opposite impulses, the smaller mass gains the larger speed, not the smaller one. C Uses equal impulse and momentum conservation correctly to find the speed and direction. D Both objects are pushed by internal forces of the pair; starting from rest they must recoil in opposite directions, so the student cannot remain at rest.