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Question: Answered & Verified by Expert
Particle A of mass m1 moving with velocity 3i^+j^ms1 collides with another particle B of mass m2 which is at rest initially. Let v1 and v2 be the velocities of particles A and B after collision respectively. If m1=2m2 and after collision v1i^+3j^ms1, the angle between v1 and v2 is :
PhysicsCenter of Mass Momentum and CollisionJEE MainJEE Main 2020 (06 Sep Shift 2)
Options:
  • A 15°
  • B 60°
  • C 45°
  • D 105°
Solution:
2646 Upvotes Verified Answer
The correct answer is: 105°

m1u1+m2u2=m1v1+m2v2

2m2(3i^+j^)+m2×0=2m2(i^+3j^)+m2×v2

23i^+2j^=2i^+23j^+v2

v2=(3-1)i^-(3-1)j^

v1=i^+3j^

Angle between v1 and v2 is 105°

   

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