A ball collides impinges directly on a similar ball at rest. The first ball is brought to rest after the impact. If half of the kinetic energy is lost by impact, the value of coefficient of restitution `( e)` is

Answer

4

Explanation

Suppose `u_(1)` and `v_(1)` be the initial and final velocities of 1st ball and `u_(2)` and `v_(2)` be the similar quantities for ball 2. Here, `u_(2)=0` and `v_(1)=0`
`E_(i)=(1)/(2) m u_(1)^(2)+(1)/(2)m u_(2)^(2)=(1)/(2) m u_(1)^(2)," "E_(f)=(1)/(2),v_(1)^(2)+(1)/(2)mv_(2)^(2)=(1)/(2)mv_(2)^(2)`
Loss of KE `=(1)/(2)m u_(1)^(2)-(1)/(2)mv_(2)^(2)`
According to eqation,
`(1)/(2)((1)/(2)m u_(1)^(2))=(1)/(2) m u_(1)^(2)-(1)/(2) m v_(2)^(2)" (as half of its KE is lost by impact)"`
`therefore" "u_(1)^(2)=2v_(2)^(2)" or "v_(2)=(u_(1))/(sqrt2)" "therefore" "e=(v_(2)-v_(1))/(u_(1)-u_(2))=(v_(2))/(u_(1))=(1)/(sqrt2)`