Question

A ball of mass 'm' moving with a speed 'u' under goes a head-on elastic collision with a ball of mass (nm) initially at rest. The fraction of the incident energy transferred to the heavier ball is

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

Answer 1

Answer: (D)

$\frac{4n}{(1+n{)}^2}$

We know that
$v_{2f}=\left(\frac{m_2-m_1}{m_1+m_2}\right)v_{2i}+\frac{2m_1}{(m_1+m_2)}{v}_{1i}$
$m_i=m,m_2=nm,v_{1i}=u,v_{2i}=0$
$\Rightarrow v_{2f}=\frac{2}{(1+n)}u$
K.E of the sationary particle after collision
$=\frac{1}{2}nm{\left[\frac{2}{(n+1)}u\right]}^2$
Kinetic energy of particle of mass m before collision
$=\frac{1}{2}m{u}^2$
So fraction of K.E. transferred
$=\frac{n}{u^2}\left[\frac{4u^2}{(n+1{)}^2}\right]=\frac{4n}{(1+n{)}^2}$