Source: High school physics (Chinese)
Problem Sets:
Problem
Two blocks are in contact with a light, unattached spring between them. One block has a mass 3 times the other. The spring is compressed, storing 100 J of elastic potential energy. The blocks are released simultaneously from rest.
Let the masses be $m$ and $M=3m$. Let their final velocities be $v$ and $V$. The system is isolated, so momentum and energy are conserved. Conservation of momentum (initial momentum is zero):
$$mv + MV = 0 \implies mv = -3mV \implies v = -3V$$The kinetic energies are $K_m = \frac{1}{2}mv^2$ and $K_M = \frac{1}{2}MV^2 = \frac{1}{2}(3m)V^2$. The ratio of kinetic energies is:
$$\frac{K_m}{K_M} = \frac{\frac{1}{2}m(-3V)^2}{\frac{1}{2}(3m)V^2} = \frac{9mV^2}{3mV^2} = 3 \implies K_m = 3K_M$$Conservation of energy (potential energy is converted to kinetic energy):
$$E_p = K_m + K_M = 100 \text{ J}$$Substituting $K_m = 3K_M$:
$$3K_M + K_M = 100 \text{ J} \implies 4K_M = 100 \text{ J} \implies K_M = 25 \text{ J}$$ $$K_m = 3K_M = 3(25 \text{ J}) = 75 \text{ J}$$