A particle of mass `m` moving with some velocity on a smooth surface strikes another particle of mass `eta m` at rest , elasticity head - on. What fraction of incident `K.E.` is transferred to the particle of mass `eta m`.

Explanation

`m u + eta m xx 0 = mv_(1) + eta m v_(2)` (i)
`v_(1) + u = v_(2) + 0` (ii)
Solving `v_(2) = (2 u)/(1 + eta)`
Now `K.E.` of particle of mass `eta m = (1)/(2) eta mv_(2)^(2)`
Incident `K.E. = (1)/(2) mu^(2)`
Fraction of incident `K.E.` transferred to `eta m` is
`(1//2 eta mv_(2)^(2))/(1//2 m u^(2)) = eta ((v_(2))/(u))^(2) = ( 4 eta)/(( 1 + eta)^(2))`