A man of mass $m$is stationary on a stationary flat car. The car can move without friction along horizontal rails. The man starts walking with velocity $v$relative to the car. Work done by him:

If the man walks along the rails, some velocity say $V$is imparted to car also. Let $M$be the mass of car. Then from conservation of linear momentum.

 $$\begin{gathered} M.V= m\left(v-V\right) \\ \therefore V =\frac{mv}{m+M} \\ \therefore \mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{man} \\ =\frac{1}{2}\mathrm{m}{(\mathrm{v}-\mathrm{V})}^2+\frac{1}{2}{\mathrm{mV}}^2 \\ =\frac{1}{2}\left(\frac{\mathrm{mM}}{\mathrm{m}+\mathrm{M}}\right){\mathrm{v}}^2 <\frac{1}{2}{\mathrm{mv}}^2 \end{gathered}$$

Hence, option $\left(a\right)$ is correct. If the man moves normal to the rails then car will not move. Hence, work done by him in this case will be $\frac{1}{2}m{v}^2$ and option $\left(b\right)$ is also correct.