A steel ball of mass `0.5 kg` is fastened to a cord `20 cm` long and fixed at the far end and is released when the cord is horizontal. At the bottom of its path the ball strikes a `2.5 kg` steel block initially at rest on a frictionless surface. The collision is elastic. The speed of the block just after the collision will be.

Explanation

Let the ball strike the block with a speed `u_(1)`. Since the initial speed (speed before collision) of the block `= u_(2)=0` for the perfectly elastic collision, the speed of the block is given as
`v_(2)=(2m_(1))/(m_(1)+m_(2))u_(1)=(2(m_(1))/(m_(2)))/((m_(1))/(m_(2))+1)u_(1)`
where `u_(1)` can be found by conserving energy of the ball between the position A and B before collision of `m_(1)`.
`rArr (1)/(2)m_(1)u_(1)^(2)-m_(1)gl=0`
`rArr u_(1)=sqrt(2gl)=sqrt(2xx10xx20)=20` m/sec.
`(m_(1))/(m_(2))=(0.5 kg)/(2.5 kg) =(1)/(5)`
By putting `u_(1)=20` m/sec and `(m_(1))/(m_(2))=(1)/(5)` in (a) we obtain
`v_(2)={(2)/((1)/(5)+1)}{(1)/(5)}(20)=(20)/(3)m//s`