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Object P has mass $200\,\text{g}$ and velocity $10\,\text{m s}^{-1}$ along the $+x$-direction. Object Q has mass $500\,\text{g}$ and velocity $(3\hat{i}+5\hat{j})\,\text{m s}^{-1}$. What is the speed of their centre of mass?

1. $5\,\text{m s}^{-1}$
2. $6\,\text{m s}^{-1}$
3. $8\,\text{m s}^{-1}$
4. zero

Correct answer: $8\,\text{m s}^{-1}$

Solution

The centre-of-mass velocity is $\vec v_{cm}=\frac{m_1\vec v_1+m_2\vec v_2}{m_1+m_2}$. Using $m_1=0.2\,\text{kg}$, $\vec v_1=10\hat{i}$ and $m_2=0.5\,\text{kg}$, $\vec v_2=3\hat{i}+5\hat{j}$ gives $\vec v_{cm}=(5\hat{i}+\tfrac{5}{7}\hat{j})$? Wait, compute carefully: total momentum $=(2+1.5)\hat{i}+2.5\hat{j}=(3.5\hat{i}+2.5\hat{j})$, divide by $0.7$ gives $(5\hat{i}+\tfrac{25}{7}\hat{j})$, whose magnitude is $8\,\text{m s}^{-1}$.

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