JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Conservation of Momentum

A ball of mass 0.15 kg hits the wall with its initial speed of

12 ms^{- 1}

and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100 N, calculate the time duration of the contact of ball with the wall.

(A)0.018 s
(B)0.036 s✓ Correct
(C)0.009 s
(D)0.072 s

Answer: (B)

Solution

Initial momentum

P_{i} = 0.15 \times 12 = 1.8

kg·m/s (toward wall). Final momentum

P_{f} = 0.15 \times 12 = 1.8

kg·m/s (away from wall). Change in momentum magnitude

∣Δ P ∣ = 1.8 + 1.8 = 3.6

kg·m/s. Using impulse-momentum theorem:

F Δ t = ∣Δ P ∣

, so

100 \times Δ t = 3.6

, giving

Δ t = 0.036

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option

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