A bob of mass M is suspended by an inextensible string. A particle of mass m, moving with velocity v0 strike the bob as shown in the figure. The  particle comes to rest just after collision. If θ=30∘, vo=4 m/s and mM=110 then calculate the velocity of the bob (in cm/s) just after collision.

Here tension in the string will be impulse in nature. Let us make the impulse diagram of the situation.Applying impulse momentum equation for the particlemv0−J1=0 or J1=mv0   ……..(i)As the string is inextensible, the bob cannot move in vertical direction.Hence, Jtension =J1cos⁡θ or Jtension =mvcos⁡θthe horizontal component of impulse J1sin⁡θ will be responsible for motion of the bob in horizontal direction, hence     J1sin⁡θ=MV or V=J1sin⁡θM⇒V=mv0sin⁡θM= 420 m/s=20 cm/sAfter collision the bob starts moving in the circular path.