JEE Main 2015PhysicsCenter of Mass Momentum and CollisionMediumMCQ

JEE Main 2015 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2015 (04 Apr)

Question

A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

Choose an option

Show full solutionCorrect option: D
Correct answer
D56%

Step-by-step explanation

The initial momentum of system is Pi→=m2Vi^+2mV j^

According to question as



On perfectly inelastic collision the particles stick to each other.

Pf→=3m Vf→

By conservation of linear momentum

Pf→=Pi→ ⇒3m Vf→=m2V i^+2mVj^

⇒Vf→=2V3i^+j^⇒ | V f | =223V

∴ loss in KE. of system =Kinitial-Kfinal

=12m2V2+122mV2-123m 2 2 V⁡ 3 2

=2mV2+mV2-43mV2=3mV2-4mV23=53mV2

% Loss in KE =100×∆KKi=53mv23mV2=59×100=56%

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About this question

This is a previous-year question from JEE Main 2015, 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.