Question

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

• A. $\frac{n}{1+n}$
• B. $(\frac{n}{1+n}{)}^2$
• C. $\frac{2n}{1+n^2}$
• D. $\frac{4n}{(1+n{)}^2}$

Answer 1

The correct option is D $\frac{4n}{(1+n{)}^2}$

By law of conservation of momentum

$mu=m{u}_1+nm{u}_2\Rightarrow u=u_1+n{u}_1+n{u}_2\to \left(1\right)$

By definition of co-efficient of restitition,

$u-0=u_2-u_1\to \left(2\right)$

From $\left(1\right)\&\left(2\right)$

$u-2\frac{2u}{n+1}\to \left(3\right)$

Fraction of energy transfered $E=\frac{\frac{1}{2}{m}_2{u}_2^2}{\frac{1}{2}{m}_1{u}_1^2}\therefore E=\frac{\frac{1}{2}nm(\frac{2u}{n+1}{)}^2}{\frac{1}{2}m{u}^2}=\frac{4n}{(n+1{)}^2}$