Body of mass M is much heavier than the other body of mass m The heavier body with speed v collides with the lighter body which was at rest initially elastically. The speed of lighter body after collision is [AIIMS 2018]

From law of conservation of momentum,

               $\mathrm{Mv} + \mathrm{m} \times 0 = {\mathrm{Mv}}_1 + {\mathrm{mv}}_2$

$\Rightarrow \mathrm{M}(\mathrm{v}-{\mathrm{v}}_1) ={\mathrm{mv}}_2$                                                     …(i)

Again, from the conservation of kinetic energy (as collision is of elastic nature),

        $\frac{1}{2}M{v}^2+\frac{1}{2}m\times 0=\frac{1}{2}M{v}_1^2+\frac{1}{2}m{v}_2^2$

$\Rightarrow M\left(v^2-v_1^2\right)=m{v}_2^2$                                                  …(ii)

On solving Eqs. (i) and (ii), we get

             $\frac{M\left(v-v_1\right)}{M\left(v+v_1\right)\left(v-v_1\right)}=\frac{m{v}_2}{m{v}_2^2}$

                                    $v_2=v+v_1$

Now, solving Eqs. (i) and (iiil, we get

                            $v_1=\frac{(M-m)v}{(M+m)}$

and                      $v_2 = \frac{2Mv}{(M + m)}$

As,                       M >>m

So,                          ${\mathrm{v}}_1=\mathrm{v} \mathrm{and} {\mathrm{v}}_2=2\mathrm{v}$

Hence, velocity of lighter body (m) is 2v.