Physics - Center of Mass Question with Solution | TestHub

The velocity of the center of mass of a system remains constant if no external forces act on the system.

Therefore, the velocity of the center of mass after the collision is the same as its velocity before the collision. Let's take the direction of $v_1$ as positive. Then $\overrightarrow{v_1}=5\hat{i}\text{ m/s}$ and $\overrightarrow{v_2}=-2\hat{i}\text{ m/s}$. The velocity of the center of mass before collision is:$V_{CM}=\frac{m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}}{m_1+m_2}=\frac{(2\text{ kg})(5\text{ m/s})+(3\text{ kg})(-2\text{ m/s})}{2\text{ kg}+3\text{ kg}}$ $V_{CM}=\frac{10-6}{5}\text{ m/s}=\frac{4}{5}\text{ m/s}=0.8\text{ m/s}$. Since no external forces act, the velocity of the center of mass remains $0.8\text{ m/s}$ after the collision.