JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Conservation of Momentum

Two billiard balls of mass 0.05 kg each moving in opposite directions with

10 ms^{- 1}

collide and rebound with the same speed. If the time duration of contact is

t = 0.005 s

, then what is the force exerted on the ball due to each other?

(A)100 N
(B)200 N✓ Correct
(C)300 N
(D)400 N

Answer: (B)

Solution

For one ball, the change in momentum is:

∣Δ p ∣ = m ∣ v_{f} - v_{i} ∣ = 0.05 \times ∣10 - (- 10) ∣ = 0.05 \times 20 = 1 kg\cdotm/s

The average force exerted is:

F_{avg} = \frac{∣Δ p ∣}{Δ t} = \frac{1}{0.005} = 200 N

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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