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AS Daily A Level Physics question

2026-01-27 OCR A Kinematics: linear motion under gravity (g = 9.81 m s^-2) OCR-A Module 3.1 Motion — Kinematics: displacement, velocity, acceleration OCR-A Module 3.1 Motion — Equations of motion (constant acceleration, free fall with g ≈ 9.81 m s^-2)

In a lab test of a floor sensor, a technician drops identical steel bearings from rest from heights 0.80 m and 3.2 m above the sensor. Air resistance is negligible and g = 9.81 m s^-2. Compared with the 0.80 m drop, what happens to the impact speed at 3.2 m, and why?

  1. A Four times larger, because speed is proportional to the height fallen.
  2. B Unchanged, because the ball’s weight is the same in both drops.
  3. C Twice as large, because the ball accelerates for four times as long.
  4. D About twice as large, because from rest the impact speed is proportional to the square root of height. (correct)

Answer

The correct answer is D.

Correct: D — About twice as large, because from rest the impact speed is proportional to the square root of height. A speed ∝ height is a common mistake; from v^2 ∝ h, quadrupling h only doubles v (e.g., 0.80 m gives ~4.0 m s^-1; 3.2 m gives ~8.0 m s^-1). A is wrong because v is not directly proportional to h. B is wrong because having the same weight does not fix the impact speed; a greater drop height gives a greater speed. C is wrong because the time to fall scales with the square root of height, not fourfold; the correct reason for doubling v is v^2 ∝ h. D is correct: v ∝ √h, so increasing h by a factor of 4 increases v by √4 = 2.

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