Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

77 of 130

Question 77 of 130

A ball of mass 0.15kg0.15kg hits the wall with its initial speed of 12ms−112m{s}^{-1} and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100N100N. calculate the time duration of the contact of ball with the wall.

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Answer: (B) 0.036s0.036s

Given: m=1.5kg&u=12m{s}^{-1}
Change in the momentum of the ball after the collision=2mv=2mv
=2×1.5×12=2\times 1.5\times 12
=36Ns=36Ns
As the force applied during collision is equal to 100N100N and if tt is the duration of collision, so
100×t=Δp100\times t=\Delta p
⇒t=36100s\Rightarrow t=\dfrac{36}{100}s
t=36×10−2s=0.036st=36\times {10}^{-2}s=0.036s

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