A particle loses `25%` of its energy during collision with another identical particle at rest. the coefficient of restitution will be

Options

A. `0.25`
B. `sqrt(2)`
C. `1/sqrt(2)`
D. `0.5`

Answer

C

Explanation

By formula derived earlier,
`/_\KE=(m_(1)m_(2)v_(r)^(2))/(2(m_(1)+m_(2)))(1-e^(2))`
`(/_\KE)/(KE)=(m_(2))/(m_(1)+m_(2))(1-e^(2))` where ` KE=1/2m_(1)v^(2)`,
`m_(1)=m_(2)=m`
`implies25/100=1/4=1/2(1-e^(2))` (`:' /_\KE` of `25%` of `KE` of `m_(1)`)
`e=1/(sqrt(2))`