JEE Main 2023PhysicsCenter of Mass Momentum and CollisionMediumMCQ

JEE Main 2023 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2023 (30 Jan Shift 2)

Question

A machine gun of mass 10 kg fires 20 g bullets at the rate of 180 bullets per minute with a speed of 100 m s–1 each. The recoil velocity of the gun is :

Choose an option

Show full solutionCorrect option: D
Correct answer
D0.6 m s-1

Step-by-step explanation

Number of bullets fired per second 18060=3.

Applying momentum conservation, we get

320×10-3×100=10v

⇒v=0.6 m s-1

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 2023, 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.