Two balls of equal mass m_1 = m_2 = 2 g move with equal…

Two balls of equal mass $m_1 = m_2 = 2\ \mathrm{g}$ move with equal speeds $v_1 = v_2 = 20\ \mathrm{cm/s}$ along mutually perpendicular paths. What is the total momentum of the two balls?

Explanation of Method

Total momentum of two balls moving at right angles.

Step 1: Each momentum

\[ p_1 = p_2 = m v = 2 \cdot 10^{-3} \cdot 0.2 = 4 \cdot 10^{-4}\ \mathrm{kg\,m/s}. \]

Step 2: Vector sum

The momenta are perpendicular, so the magnitudes combine by Pythagoras: $p = \sqrt{p_1^2 + p_2^2} = \sqrt2 \cdot 4 \cdot 10^{-4}$.

\[\begin{gathered} p \approx 5.66 \cdot 10^{-4}\ \mathrm{kg\,m/s} \\ \text{at } 45^\circ \text{ to each path} \end{gathered}\]

$p_1 = 4 \cdot 10^{-4}$ $p_2 = 4 \cdot 10^{-4}$ $p = 5.66 \cdot 10^{-4}\ \mathrm{kg\,m/s}$ $45^\circ$

Momentum is a vector, so the two perpendicular momenta add like the sides of a square: the total is the diagonal, $\sqrt2$ times one of them, pointing at 45° to each path.

Result

\[\boxed{\,\begin{gathered} p \approx 5.66 \cdot 10^{-4}\ \mathrm{kg\,m/s} \\ \text{at } 45^\circ \text{ to each path} \end{gathered}\,}\]

Remark

Check. The original result is correct. Simply adding the numbers (8 × 10⁻⁴) would be wrong, because momentum is a vector.