Explanation

(d) Work done by retarding force = change in kinetic energy
$- fs = 0 - \frac{1}{2}mv^{2}$
$\therefore fs = \frac{1}{2}mv^{2}$
Since, retarding force $f$ and kinetic energy of both bodies are same so, stopping distance will be same. Hence, (a) is correct.
$a = \frac{f}{m}$
$\therefore\ v^{2} = u^{2} - 2as$
Or $0 = u^{2} - 2as$
$\therefore u^{2} = 2as = \frac{2sf}{m}$
$\therefore u \propto \frac{1}{\sqrt{3}}$
It means more is the mass, less is the initial speed
$\therefore\ v = u - at$
Or $0 = u - at$
$\therefore\ \ t = \frac{u}{a} = \frac{u}{\frac{f}{m}} = \frac{um}{f}$
$\therefore\ \ t \propto m$
Hence, more is the mass, less is the time of stopping. Hence, (b) is correct
$\because\ KE = K = \frac{p^{2}}{2m}$
$\therefore p^{2} \propto m$
It means more is the mass, more is the momentum
Hence, (c) is correct.