JEE Main 2022PhysicsCenter of Mass Momentum and CollisionHardNumerical

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

JEE Main 2022 (27 Jun Shift 1)

Question

A pendulum of length 2 m consists of a wooden bob of mass 50 g. A bullet of mass 75 g is fired towards the stationary bob with a speed v. The bullet emerges out of the bob with a speed v3 and the bob just completes the vertical circle. The value of v is _____  m s-1

(if g=10 m s-2).

Enter your answer

Show full solutionCorrect answer: 10
Correct answer
10

Step-by-step explanation

For the bob to completer the vertical circle, vb=5gl.

From the conservation of momentum, 

m1v=m2vb+m1v3⇒75v=505gl+75v3

⇒752v3=505×10×2

⇒752v3=50×10

⇒v=500×3150=50050=10 m s-1

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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.