Question

A body of mass $2m$ moving with velocity $v$ makes a head on elastic collision with another body of mass $m$ which is initially at rest. Loss of kinetic energy of the colliding body ) (mass $2m$) is

• A. 1/9 of its initial kinetic energy
• B. 1/6 of its initial kinetic energy
• C. 1/4 of its initial kinetic energy
• D. 1/2 of its initial kinetic energy
• E. 8/9 of its initial kinetic energy

Answer 1

Answer: (E)

Initial $K.E$ of ball of mass $2m=K_1$
$=\frac{1}{2}\times 2m\times V^2$
$=m{V}^2$
Collision is elastic so both $K.E$ and momentum are conserved. Let velocities of balls are $V_1$ and $V_2$ after collision.

So, $KE$ is conserved
$\frac{1}{2}(2m)V^2=\frac{1}{2}(2m)V_1^2+\frac{1}{2}m{V}_2^2$
$\Rightarrow V^2=V_1^2+\frac{1}{2}{V}_2^2...(i)$
And, momentum is conserved
$(2m)V+m(0)=2m\left(V_1\right)+m{V}_2$
$\Rightarrow 2V=2V_1+V_2...(ii)$
Now,
$V_2=2\left(V-V_1\right)$
Put this value in Eq. (i), we get
$V^2=V_1^2+\frac{1}{2}\times 4{\left(V-V_1\right)}^2$
$\Rightarrow 3V_1^2-4V{V}_1+V^2=0$
$\Rightarrow 3{\left(\frac{V_1}{V}\right)}^2-4\left(\frac{V_1}{V}\right)+1=0$
or $\frac{v_1}{v}=-\frac{-(-4)\pm \sqrt{16-12}}{2\times 3}$
$\Rightarrow \frac{v_1}{v}=\frac{4\pm 2}{2\times 3}$
$\Rightarrow V_1=V$ (Not possible)
or $V_1=\frac{1}{3}V$
So, final $K.E$ of ball of mass $2m$,
$k_2=\frac{1}{2}(2m)\left(V_1^2\right)=\frac{1}{2}\times 2m\times \frac{V^2}{9}=\frac{1}{9}\left(k_1\right)$
Hence, loss of $K.E.$ of Ist ball
$=K_1-\frac{1}{g}{K}_1=\frac{8}{9}{K}_1$