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°