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Q.

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 V1→ and V2→  be the velocities of particles A and B after collision respectively.  If  m1=2m2 and after collision V1→=i^+3j^ms−1 , the angle between  V1→  and V2→  is

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a

15°

b

−45°

c

60°

d

105°

answer is D.

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Detailed Solution

Let  V→2=Vxi^+Vyj^Applying momentum conservation along  x and y direction,m1×3=m1+m2Vx ⇒Vx=23−1  And  m1=3m1+m2Vy ⇒Vy=−23−1So  V→2=23−1i^−23−1j^, which is making angle 45° with x axis as shownand V→1 =i^+3j^ , is making 600 with x axis as shown in figure So angle between  V→1 and V→2 is 105°
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