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.

Options

A. `10/3ms^(-1)`
B. `20/3ms^(-1)`
C. `5ms^(-1)`
D. `5/3ms^(-1)`

Answer

B

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
`m_(1)u_(1)=m_(1)v_(1)+m_(2)v_(2)`

`e=1=v_(1)/(v_(2)-v_(1))impliesv_(2)-v_(1)=u_(1)`
`0.5u_(1)=0.5v_(1)2.5v_(2)`
`u_(1)=-v_(1)+v_(2)impliesv_(2)=u_(1)/3=(sqrt(2gl))/3=20/3 ms^(-1)`