Answer
- Magnitude: $p=4.24kg.m/s$
- Direction: $\vec{p}$ is directed $45^o$ south of due east.
Work Step by Step
We take due east to be $+x$ direction and due south to be $+y$ direction.
The magnitude of the momentum of each arrow is $p_x=p_y=mv=30\times0.1=3kg.m/s$
The total momentum of this two-arrow system is $\vec{p}=(3kg.m/s)i+(3kg.m/s)j$
- Magnitude: $p=\sqrt{3^2+3^2}=4.24kg.m/s$
- Direction: take $\theta$ to be the angle $\vec{p}$ makes with due east; we have $$\tan\theta=\frac{3}{3}=1$$ $$\theta=45^o$$
So $\vec{p}$ is directed $45^o$ south of due east.
