Answer
The amount of kinetic energy that is added to the system during the explosion is $~~\frac{1}{6}~m~v^2$
Work Step by Step
We can use conservation of momentum to find the velocity of the more massive piece after the explosion:
$p_f = p_i$
$(\frac{3}{4}~m)(v_f)+(\frac{1}{4}~m)(0) = m~v$
$(\frac{3}{4}~m)(v_f) = m~v$
$(\frac{3}{4})(v_f) = v$
$v_f = \frac{4}{3}~v$
We can find the initial kinetic energy:
$K_i = \frac{1}{2}mv^2$
We can find the kinetic energy after the explosion:
$K_f = \frac{1}{2}(\frac{3}{4}~m)(\frac{4}{3}v)^2+ \frac{1}{2}(\frac{1}{4}~m)(0)^2$
$K_f = \frac{1}{2}(\frac{4}{3}~m~v^2)+0$
$K_f = \frac{2}{3}~m~v^2$
We can find the increase in kinetic energy:
$\Delta K = K_f-K_i$
$\Delta K = \frac{2}{3}~m~v^2-\frac{1}{2}~m~v^2$
$\Delta K = \frac{1}{6}~m~v^2$
The amount of kinetic energy that is added to the system during the explosion is $~~\frac{1}{6}~m~v^2$.
