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Q.
A body of mass $2\,m$ 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 $2\,m$) is
Initial $K.E$ of ball of mass $2 \,m=K_{1}$
$=\frac{1}{2} \times 2 \,m \times V^{2}$
$=m V^{2}$
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
$\frac{1}{2}(2 m) V^{2}=\frac{1}{2}(2 m) V_{1}^{2}+\frac{1}{2} m V_{2}^{2}$
$\Rightarrow V^{2}=V_{1}^{2}+\frac{1}{2} V_{2}^{2}\,\,\,...(i)$
And, momentum is conserved
$(2 m) V +m(0)=2 m\left(V_{1}\right)+m V_{2}$
$\Rightarrow 2 V=2 V_{1}+V_{2}\,\,\,...(ii)$
Now,
$V_{2}=2\left(V-V_{1}\right)$
Put this value in Eq. (i), we get
$V^{2}=V_{1}^{2}+\frac{1}{2} \times 4\left(V-V_{1}\right)^{2}$
$\Rightarrow 3 V_{1}^{2}-4 V V_{1}+V^{2}=0$
$\Rightarrow 3\left(\frac{V_{1}}{V}\right)^{2}-4\left(\frac{V_{1}}{V}\right)+1=0$
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 $2\, m$,
$k_{2}=\frac{1}{2}(2 m)\left(V_{1}^{2}\right)=\frac{1}{2} \times 2 m \times \frac{V^{2}}{9}=\frac{1}{9}\left(k_{1}\right)$
Hence, loss of $K.E.$ of Ist ball
$=K_{1}-\frac{1}{g} K_{1}=\frac{8}{9} K_{1}$
