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Question

A particle of mass m moving with velocity 1 m/s collides elastically with another particle of mass 2m (initialy at rest). If the incident particle is deflected by 90. The angle θ made by the heavier mass with the initial direction of motion of mass m will be equal to:
(consider collision to be perfectly elastic)


Your Answer
A
60
Your Answer
B
45
Your Answer
C
15
Correct Answer
D
30

Solution

The correct option is B 30
During collison, Fext=0 on the system of particles.
Applying conservation of momentum along incident direction,
(m×1)+0=0+2mv2cos θ
1=2v2cosθ ....(1)
Conservation of momentum along perpendicular direction, considering upwards as +vedirection.
(m×0)+(2m×0)=mv12mv2sin θ
v1=2v2sinθ ....(2)

From energy conservation during perfectly elastic collision,
12m×(1)2=12mv21+122mv22
v21+2v22=1 ....(3)
Squaring and adding Eq. (1) and Eq. (2),
1+v21=4v22(sin2θ+cos2θ)
1+v21=4v22 ....(4)
From Eq.(3) and Eq.(4),
v22=26=13
v2=13
and v1=431=13
From Eq.(2),
sinθ=v12v2=12
θ=30

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