Velocity from total momentum

Problem

A system of two masses: $m_1 = 2\;\text{kg}$, $m_2 = 5\;\text{kg}$. The total momentum is $\vv{p} = (14;\, 0)\;\text{kg\,m/s}$. Knowing that $\vv{v}_2 = (2;\, 0)\;\text{m/s}$, find $\vv{v}_1$ and comment.

Solution

Total momentum is the vector sum of the contributions from the two masses: $\vv{p} = m_1\,\vv{v}_1 + m_2\,\vv{v}_2$ Isolating the unknown $\vv{v}_1$: $m_1\,\vv{v}_1 = \vv{p} - m_2\,\vv{v}_2$ Substituting the values (work by components; here everything is along $x$): $2\,\vv{v}_1 = (14;\, 0) - 5\cdot(2;\, 0) = (14;\, 0) - (10;\, 0) = (4;\, 0)$ $\vv{v}_1 = \frac{(4;\, 0)}{2} = (2;\, 0)\ \mathrm{m/s}$ Comment: it turns out $\vv{v}_1 = \vv{v}_2 = (2;\, 0)\;\text{m/s}$. Both masses move with the same velocity, which is that of the centre of mass: $\vv{v}_\text{cm} = \frac{\vv{p}}{m_1 + m_2} = \frac{(14;\, 0)}{7} = (2;\, 0)\ \mathrm{m/s}$ The system translates rigidly: there is no relative motion between the two masses. $\ev{\vv{v}_1 = (2;\, 0)\ \mathrm{m/s}\ =\ \vv{v}_\text{cm}}$

Links

Topics: Momentum and collisions Concepts: Momentum · Centre of mass Skills: Symbolic setup