Answer
- Magnitude: $p=322kg.m/s$
- Direction: $\vec{p}$ is directed $15.7^o$ north of east.
Work Step by Step
Take due east to be $+x$ direction and due north to be $+y$ direction.
- The momentum of an 85-kg jogger heading due east is $$\vec{p_1}=(mv)i=(85\times2)i=(170kg.m/s)i$$
- The momentum of a 55-kg jogger heading $32^o$ north of east is $$\vec{p_2}=(mv\cos32)i+(mv\sin32)j$$ $$\vec{p_2}=(55\times3\cos32)i+(55\times3\sin32)j$$ $$\vec{p_2}=(140kg.m/s)i+(87.4kg.m/s)j$$
The sum of the 2 momenta is $$\vec{p}=\vec{p_1}+\vec{p_2}=(310kg.m/s)i+(87.4kg.m/s)j$$
- Magnitude: $p=\sqrt{310^2+87.4^2}=322kg.m/s$
- Direction: take $\theta$ to be the angle $\vec{p}$ makes with $+x$ axis $$\tan\theta=\frac{87.4}{310}=0.282$$ $$\theta=15.7^o$$
$\vec{p}$ is directed $15.7^o$ north of east.
