A ball of mass 2kg moving with speed 5m/s collides directly with another ball of mass 3kg moving in the same direction with speed 4m/s. The coefficient os restitution is 2/3. Find the velocities of both ball after the collision.

Explanation

Denoting the first ball by A and the second ball by B velocities immediately before and after the impact are shown in the figure.

Applying principle of conservation of momentum, we have
`m_(B)v_(B) + m_(A)v_(A) = m_(A)u_(A) + m_(B)u_(b) rarr 3v_(B) + 2v_(A) = 2 xx 5 + 3 xx 4`
`3v_(B) + 2v_(A) = 22` ...(i)
Applying equation of coefficient of restitution, we have
`v_(B) - v_(A) = e(u_(A) - u_(B)) rarr v_(B)- v_(A) = 2/3(5 - 4)`
`3v_(B) - 3v_(A) = 2` ...(ii)
From equation (i) and (ii), we have `v_(4) = 4m//s` and `v_(B) = 4.67m//s`.