Answer
The total momentum of the system has a magnitude of $199kgm/s$ and is directed $46.8^o$ north of due east.
Work Step by Step
We take due east to be the $+x$ direction and due north to be the $+y$ direction. We examine the collision in x and y components separately.
a) x Component:
Object A, whose mass is $m=17kg$ and who has initial horizontal velocity $\vec{v}_{0x}=+8m/s$, collides with object B, whose mass is $M=29kg$ and has initial horizontal velocity $\vec{V}_{0x}=0$ (because it was moving due north).
After the collision, 2 objects stick together and have mass $M+m=46kg$ and horizontal velocity $\vec{V_{fx}}$
The total linear momentum is conserved. Therefore, $$M\vec{V}_{0x}+m\vec{v}_{0x}=(M+m)\vec{V_{fx}}$$ $$\vec{V_{fx}}=\frac{0+m\vec{v}_{0x}}{M+m}=+2.96m/s$$
The horizontal total momentum of the system is $\vec{p_x}=(M+m)\vec{V_{fx}}=+136.16kg.m/s$
b) y Component:
Object A has initial vertical velocity $\vec{v}_{0y}=0$ while object B has initial vertical velocity $\vec{V}_{0y}=+5m/s$.
After the collision, 2 objects stick together and have mass $M+m=46kg$ and vertical velocity $\vec{V_{fy}}$
The total linear momentum is conserved. Therefore, $$M\vec{V}_{0y}+m\vec{v}_{0y}=(M+m)\vec{V_{fy}}$$ $$\vec{V_{fy}}=\frac{0+M\vec{V}_{0y}}{M+m}=+3.15m/s$$
The vertical total momentum of the system is $\vec{p_y}=(M+m)\vec{V_{fy}}=+144.9kg.m/s$
The total momentum of the system is $\vec{p}=136.16i+144.9j$
- Magnitude: $p=\sqrt{136.16^2+144.9^2}=199kg.m/s$
- Direction: take $\theta$ to be the angle $\vec{p}$ makes with $+x$ axis $$\tan\theta=\frac{144.9}{136.16}=1.064$$ $$\theta=46.8^o$$
So the total momentum of the system is directed $46.8^o$ north of due east.
