JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Conservation of Momentum

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100 balls each of mass m moving with speed v simultaneously strike a wall normally and reflected back with same speed, in time

t

s. The total force exerted by the balls on the wall is

(A) $\frac{100 m v}{t}$
(B) $\frac{200 m v}{t}$ ✓ Correct
(C)200 mvt
(D) $\frac{m v}{100 t}$

Answer: (B)

Solution

Each ball reverses its velocity on reflection, so its momentum changes in magnitude by

Δ p = m v - (- m v) = 2 m v

. All

100

balls strike within the time

t

, so the total momentum change delivered is

100 \times 2 m v = 200 m v

. The average force is this momentum change divided by the time over which it acts, and by Newton's third law the balls push on the wall with equal magnitude,

F = \frac{200 m v}{t}

JEE Main · 2023 · 31 Jan 2023 · Shift 1 · Single correct option

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