Question

A 9.00 kg object, travelling South with a speed of 8.50 m/s, is moving towards a 12.0 kg

object that is moving East with a speed of 6.00 m/s. Determine the total momentum

Answer 1

The total momentum is the vector sum of all momenta. Assume that y-axis points northward, y-axis looks eastward:

$$\vec{p}_x=\sum^n_{i=1} m_i\vec{v}_{xi}=12\cdot6=72\text{ kg}\cdot\text{m/s},\\ \space\\ \vec{p}_y=\sum^n_{i=1} m_i\vec{v}_{yi}=9\cdot(-8.5)=-76.5\text{ kg}\cdot\text{m/s}.$$

We can find the magnitude of the total momentum:

$$p=\sqrt{p_x^2+p_y^2}=\sqrt{72^2+(-76.5)^2}=105\text{ kg}\cdot\text{m/s}.$$

The angle above x-axis (north of east) is

$$\theta=\text{atan}\frac{\vec{p_y}}{\vec{p_x}}=-46.7^\circ,$$

or 46.7° below x-axis (below east axis), or 46.7° south of east.