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AS Daily A Level Physics question

2026-02-04 OCR A Mechanics & Materials (M3) Module 3: Motion and forces — 3.1.1 Kinematics (displacement, velocity, acceleration; graphical and algebraic methods) Module 3: Motion and forces — 3.1.2 Motion with constant acceleration (including free fall with g = 9.81 m s−2)

A small ball is launched straight up from a lab bench and lands back on the bench at the same level. Air resistance is negligible. If the launch speed is doubled, what happens to the total time the ball is in the air? Take g = 9.81 m s−2.

  1. A It doubles, because both the upward and downward durations scale directly with the launch speed when returning to the same height. (correct)
  2. B It increases by about 1.4 times, because the maximum height roughly doubles and time scales with the square root of height.
  3. C It stays the same, because gravity fixes the return time from the top regardless of the launch speed.
  4. D It quadruples, because the kinetic energy is four times larger and the motion lasts in proportion to energy.

Answer

The correct answer is A.

Correct: A — It doubles, because both the upward and downward durations scale directly with the launch speed when returning to the same height. A The time to rise is u/g, so doubling u doubles the rise time; the fall time equals the rise time, so total time doubles (e.g. u=5 m s−1 gives T≈2u/g≈1.02 s; doubling u to 10 m s−1 gives T≈2.04 s). B This assumes height doubles; in fact maximum height scales with u^2 (quadruples), and even using time ∝ √height would predict a factor of 2, not about 1.4. C Gravity does not fix a single return time; the higher peak caused by a larger launch speed increases both ascent and descent times under the same g. D Time under constant acceleration is not proportional to kinetic energy; energy ∝ u^2, whereas the characteristic timescale varies linearly with u.

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