Answer
The lighter fragment gains 250 J of kinetic energy.
Work Step by Step
We can find the speed of the heavier fragment.
$K = \frac{1}{2}m_B~v_B^2$
$v_B = \sqrt{\frac{(2)(100~J)}{5.0~kg}}$
$v_B = 6.32~m/s$
We can use conservation of momentum to find the speed of the lighter fragment. Since the initial momentum was zero, both fragments will have the same magnitude of momentum.
$m_A~v_A = m_B~v_B$
$v_A = \frac{m_B~v_B}{m_A}$
$v_A = \frac{(5.0~kg)(6.32~m/s)}{2.0~kg}$
$v_A = 15.8~m/s$
We can find the kinetic energy of the lighter fragment.
$K = \frac{1}{2}m_A~v_A^2$
$K = \frac{1}{2}(2.0~kg)(15.8~m/s)^2$
$K = 250~J$
The lighter fragment gains 250 J of kinetic energy.
