Given: m=1.5\mathrm{kg}&u=12m{s}^{-1}
Change in the momentum of the ball after the collision=2mv
=2×1.5×12
=36Ns
As the force applied during collision is equal to 100N and if t is the duration of collision, so
100×t=Δp
⇒t=10036s
t=36×10−2s=0.036s
JEE Main 2022 — Physics Mechanics
A ball of mass 0.15kg hits the wall with its initial speed of 12ms−1 and bounces back without changing its initial speed. If the force applied by the wall on the ball during the contact is 100N. calculate the time duration of the contact of ball with the wall.
Held on 26 Jul 2022 · Verified 6 Jul 2026.
0.018s
0.036s
0.009s
0.072s
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