JEE Main 2022PhysicsCenter of Mass Momentum and CollisionMediumMCQ

JEE Main 2022 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2022 (26 Jul Shift 2)

Question

A ball of mass 0.15 kg hits the wall with its initial speed of 12 m s-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.

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Show full solutionCorrect option: B
Correct answer
B0.036 s

Step-by-step explanation

Given: m=1.5 kg & u=12 m s-1

Change in the momentum of the ball after the collision=2mv

=2×1.5×12

=36 N s

As the force applied during collision is equal to 100 N and if t is the duration of collision, so

100×t=Δp

⇒t=36100 s

t=36×10-2 s=0.036 s

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About this question

This is a previous-year question from JEE Main 2022, covering the Center of Mass Momentum and Collision chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.