In Fig. there are `n` identical suspended with wires of equal length. The spheres are almost in contact with each other. Sphere `1` is pulled aside and released. If sphere `1` strikes sphere `2` with velocity `u`. find an expression for velocity `v_(n)` of the `nth` sphere immediately after being struck by the one adjacent to it. The coefficient of restitution for all the impacts is e.

Explanation

Let velocities of sphere `1` and `2` after collision be `v_(1)` and `v_(2)`, respectively, then
`(v_(2)-v_(1))/u=e` and `mv_(2)+mv_(1)="mu"`
From above equations `v_(2)=((1+e)u)/2`
Now sphere `2` will collide with sphere `3` and affter collision velocity of `3` can be formed as
`v_(3)=((1+e)/2)^(2)u`
Hence, similarly velocity of `nth` sphere can be formed as
`v_(n)=((1+e)/2)^(n-1)u`