Answer
(a) $\frac{v}{2}$
(b) II
Work Step by Step
(a) We know that the change in the potential energy of both balls is the same -- that is, $mgh=(4m)g(\frac{1}{4})h$. Similarly, the kinetic energy will also be the same but the speed of ball 2 will be different due to its greater mass. We can find the speed for ball 2 as $\frac{1}{2}mv^2=\frac{1}{2}(4m)v_2^2$
This simplifies to:
$v_2=\frac{v}{2}$
(b) We know that the statement (II) is correct -- that is, the two balls land with the same kinetic energy; therefore, the ball of mass 4m has the speed $\frac{v}{2}$.
