Question

A ball of mass $m$ moving with speed $u$ undergoes a head on perfect elastic collision with mass $nm$ initially at rest. The fraction of the energy transferred to the second ball by the first 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 $\frac{4n}{(1+n{)}^2}$


Using momentum conservation,
$mu+0=m{v}_1+nm{v}_2$ ... (1)
For elastic collision,
Velocity of approach = velocity of separation
$(u-0)=(v_2-v_1)$ ... (2)
From both the above equations, we get
$v_2=\frac{2u}{1+n}\Rightarrow \frac{v_2}{u}=\frac{2}{(1+n)}$
Hence, fraction of incident energy transfered.
$=\frac{\frac{1}{2}nm{v}_2^2}{\frac{1}{2}m{u}^2}=n{\left(\frac{v_2}{u}\right)}^2$
$=\frac{4n}{(1+n{)}^2}$