Two identical balls A and B, each of mass 2 kg and radius R, are suspended vertically from inextensible strings as shown in figure. The third ball C of mass 1 kg and radius r=(2−1)R falls and hits A and B symmetrically with 10 m/s. Speed of both A and B just after the collision is 3 m/s.Speed of C just after collision isImpulse provided by each string during collision isThe value of coefficient of restitution is

First calculate, θ=45∘.  As the balls A and B are constrained to move horizontally (immediately after collision), if ‘I’ be the impulse imparted by ball 'C' to each of A and B, the impulse received by ball C from them would be 2Isin⁡θ. Now, each of ball B and C received impulse 'I' as shown in figure, but moves horizontally as its vertical comp. gets balanced by impulse imparted to ball B and C by the respective strings and hence,Icos⁡θ=MAVA=MBVB⇒ I=MAVAcos⁡θ (I= magnitude of impulse )Now, for ball C, if its final velocity is Vc′ downwards, we haveMcVc′=McVc−2Isin⁡θ⇒ Vc′=Vc−2MAMCVA ∵θ=45∘=−2m/s (-ve sign indicates that it is directed upwards) II method: Icosθ=mvf-0=2×3=6Ns---(1) 2Isinθ=v0-(-10)-------(2) divide (2) ÷(1) 2IsinθIcosθ=v0+106 v0=2m/sImpulse provided by each string I1=Isin⁡θ=62 sin45=6Nse= Velocity of separation  Velocity of approach ⇒ e=3cos⁡θ+VC′sin⁡θ10sin⁡θ⇒e=3+210=12