JEE Main 2024PhysicsCenter of Mass Momentum and CollisionHardMCQ

JEE Main 2024 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2024 (01 Feb Shift 1)

Question

A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass 10-2 kg moving with a speed of 2×102 m s-1. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use g=10 m s-2)

Choose an option

Show full solutionCorrect option: B
Correct answer
B0.20 m

Step-by-step explanation

Linear momentum will be conserved, therefore

mu=M+mV

⇒10−2×2×102≅1×V

⇒V≅2 m s-1

Again from energy conservation, we get height achieved, h=V22g=0.2m.

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Center of Mass Momentum and Collision chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2024, 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.