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AS Daily A Level Physics question

2026-03-11 OCR A Forces and motion (Module 3.1.1): Linear motion with g = 9.81 m s^-2 OCR-A Module 3.1.1 Motion — kinematics, uniformly accelerated motion (suvat), free fall with g = 9.81 m s^-2

In a safe lab drop test, a rubber ball is released from rest from 1.25 m above a soft mat. It takes about 0.50 s to reach the mat and its impact speed is about 4.9 m s^-1. Neglect air resistance and take g = 9.81 m s^-2. If the ball is instead dropped from 5.00 m (same conditions), which option best estimates the new fall time and impact speed?

  1. A About 1.0 s; about 9.8 m s^-1 (both roughly double). (correct)
  2. B About 1.0 s; about 4.9 m s^-1 (time doubles; speed similar).
  3. C About 2.0 s; about 9.8 m s^-1 (time quadruples; speed doubles).
  4. D About 0.71 s; about 6.9 m s^-1 (both increase by factor √2).

Answer

The correct answer is A.

Correct: A — About 1.0 s; about 9.8 m s^-1 (both roughly double). From rest under constant g, both fall time and impact speed scale with the square root of height, so quadrupling height doubles each; numerically 0.50 s → ~1.0 s and 4.9 m s^-1 → ~9.8 m s^-1. B is wrong because if time doubles under the same g, the impact speed (v ≈ g t from rest) must also double, not stay similar. C is wrong because time does not scale linearly with height; quadrupling height doubles time, not quadruples it. D is wrong because the factor √2 applies to doubling the height, not quadrupling it; the given numbers correspond to a twofold, not fourfold, increase in height.