A ball of mass 'm' moving with speed 'u' undergoes a head-on elastic collision with a ball of mass 'nm' initially at rest. Find the fraction of the incident energy transferred to the second ball.

Options

A. `n/(1+n)`
B. `n/((1+n)^(2))`
C. `(2n)/((1+n)^(2))`
D. `(4n)/((1+n)^(2))`

Answer

D

Explanation

As the collision in elastic we can find
`v_(1)=(1-n)/(1+n)u, v_(2)=(2u)/(1+n)`

Hence the required fraction i
`(1/2nmv_(2)^(2))/(1/2"mu"^(2))=n(v_(2)/u)^(2)=(4n)/((1+n)^(2))`