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

Given : $m_1=m$ $m_2=nm$

Let the velocities of $A$ and $B$ after the collision be $v_1$ and $v_2$ respectively.

initial velocity of A before collision is $v$ and that of B is zero.

Initial kinetic energy of the system $E=\frac{1}{2}m{v}^2$

Using $v_2=\frac{(m_2-e{m}_1)u_2+(1+e)m_1{u}_1}{m_1+m_2}$

$v_2=\frac{(nm-1\times m)\left(0\right)+(1+1)m\left(v\right)}{m+nm}$ $⟹v_2=\frac{2v}{(1+n)}$

Thus Kinetic energy of B after the collision $E_B^{\prime }=\frac{1}{2}\left(nm\right)v_2^2$

$E_B^{\prime }=\frac{1}{2}\left(nm\right)\frac{4v^2}{(1+n{)}^2}$

Fraction of total kinetic energy retained by B $\frac{E_B^{\prime }}{E}=\frac{4n}{(n+1{)}^2}$