A ball impinges directly on another ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by the impact, the value of coefficient of restitution is

Answer

4

Explanation

Suppose `u_(1)` and `v_(1)` be the initial and final velocities of `1^(st)` 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)m v_(1)^(2)+(1)/(2)m v_(2)^(2)=(1)/(2)m v_(2)^(2)`
Loss of `KE=(1)/(2)m u_(1)^(2)-(1)/(2)m v_(2)^(2)`
According to question,
`(1)/(2)((1)/(2)m u_(1)^(2))=(1)/(2)m u_(1)^(2)-(1)/(2)mv_(2)^(2)" "` (as half of its KE is lost by impact)
`therefore u_(1)^(2)=2v_(2)^(2) or v_(2)=(u_(1))/(sqrt(2)) therefore e=(v_(2)-v_(1))/(u_(1)-u_(2))=(v_(2))/(u_(1))=(1)/(sqrt(2))`