Question

A ball of mass $m$ moving with speed $u$ undergoes a head-on elastic collision with a ball of mass $n m$ 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. $\frac{2 n}{(1+n)^{2}}$
• D. $\frac{4 n}{(1+n)^{2}}$

Answer 1

Answer: (D)

As the collision is elastic, we can find
$v_{1}=\frac{1-n}{1+n} u, v_{2}=\frac{2 u}{1+n}$

Hence, the required fraction is
$\frac{\frac{1}{2} n m v_{2}^{2}}{\frac{1}{2} m u^{2}}=n\left(\frac{v_{2}}{u}\right)^{2}$
$=\frac{4 n}{(1+n)^{2}}$