Particle A of mass m 1 moving with velocity 3 i ^ + j ^ ms − 1 collides with another particle B of…

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 :
  1. 15°
  2. 60°
  3. −45°
  4. 105°

Solution

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°

   

Asked in: JEE Main 2020 (06 Sep Shift 2)

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