JEE Main 2020PhysicsCenter of Mass Momentum and CollisionHardMCQ

JEE Main 2020 — Center of Mass Momentum and Collision Question with Solution

JEE Main 2020 (06 Sep Shift 2)

Question

Particle A of mass m1 moving with velocity 3i^+j^ms−1 collides with another particle B of mass m2 which is at rest initially. Let v→1 and v→2 be the velocities of particles A and B after collision respectively. If m1=2m2 and after collision v→1−i^+3j^ms−1, the angle between v→1 and v→2 is :

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Show full solutionCorrect option: D
Correct answer
D105°

Step-by-step explanation

m1u→1+m2u→2=m1v→1+m2v→2

2m2(3i^+j^)+m2×0=2m2(i^+3j^)+m2×v→2

23i^+2j^=2i^+23j^+v→2

v→2=(3-1)i^-(3-1)j^

v→1=i^+3j^

Angle between v→1 and v→2 is 105°

   

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About this question

This is a previous-year question from JEE Main 2020, 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.