Physics - Center of Mass Question with Solution | TestHub

Immediately after the 1st collision the right body moves back with speed $v_0$ while the left body still moves forward with $v_0$. Since the situation is symmetric, both bodies will come to rest simultaneously at maximum compression. If $x_0$ is the maximum compression,$2\times \frac{1}{2}m{v}_0^2=\frac{1}{2}k{x}_0^2\therefore x_0=v_0\sqrt{\frac{2m}{k}}=0.36\sqrt{\frac{2\times 0.5}{16}}=0.09\text{m}=9\text{cm}$

so (A) is correct. After this the spring re-expands; the right body returns to the wall with speed $v_0$ exactly when the spring is again unstretched, so it hits the wall a second time when the spring is unstretched, and (B) is correct. Just before the 2nd collision the left body is moving away from the wall with speed $v_0$; the elastic bounce simply reverses the right body's velocity, so finally both bodies move away from the wall with the same velocity $v_0$, and (D) is correct. The wall therefore reverses the total momentum:$\therefore \Delta \overrightarrow{P}=2m{v}_0-\left(-2m{v}_0\right)=4m{v}_0=4\times 0.5\times 0.36=0.72\text{kg m/s (away from the wall)}$

so (C) is also correct.