Question

A ball of mass m makes perfectly elastic head-on collision with a ball of mass nm which is initially at rest. Show that the fractional transfer of energy by the first ball is 4n/(1 + n)${}^2$. Deduce the value of n for which the transfer is maximum.

Answer 1

For perfectly elastic head on collision $e=1$,

$mv=m{v}_1+nm{v}_2,\left(Conservationofmeomentum\right)v-0=v_2-v_{1}i.e.,v_1=v_2-v(e=1)mv=-mv+m{v}_2(1+n)v_2=\frac{2v}{1+n}$
$K.E.\propto m{v}^2$
The fraction of the
incident energy, transferred to the heavier ball is$=\frac{{nm\left(\frac{2v}{1+n}\right)}^2}{m{v}^2}=\frac{4n}{{(1+n)}^2}$
differentiat w.r.t. n. to find maxima.
$\int \frac{4n}{{(1+n)}^2}=02n(1+n)={(1+n)}^{2}2n=1+nn=1$