A sphere of mass m moving with velocity v collides head -on on another sphere of same mass which is at rest. The ratio of final velocity of second sphere to the initial velocity of the first sphere is (where, e is coefficient of restitution and collision is inelastic)

Options

A. `(e-1)/(2)`
B. `(e)/(2)`
C. `(e+1)/(2)`
D. `e`

Answer

C

Explanation

Given, mass of sphere = m
According to the question, we can draw the following diagram

Applying the conservation law of momentum, we get
`mv+m(0)=mv_(1)+mv_(2)`
`mv=mv_(1)+mv_(2)`
`mv=m(v_(1)+v_(2))`
`v=v_(1)+v_(2)`
`v=v_(1)+v_(2)`
`rArr" "v_(1)=v-v_(2)` . . .(i)
We know that, coefficient of restitution
`=("Relative velocity of separation")/("Relative velocity of approach")`
or `e=(v_(2)-v_(1))/(v-0)" or"ev=v_(2)-v_(1)`
From Eq. (i), we get
`rArr" "ev=v_(2)-(v-v_(2))`
`rArr" "ev=v_(2)-v+v_(2)`
`rArr" "ev=2v_(2)-vrArrev+v=2v_(2)`
`rArr" "v(1+e)=2v_(2)`
`rArr" "(v_(2))/(v)=(e+1)/(2)`