Answer
a) $\theta=73^o$
b) $v_f=4.27m/s$
Work Step by Step
- The 50-kg skater's momentum before the collision is $p_1=mv=50\times3=150kg.m/s$
- The 70-kg skater's momentum before the collision is $p_2=mv=70\times7=490kg.m/s$
We take due east to be the $+x$ direction and due south to be the $+y$ direction. So, in unit-vector notation, we have $\vec{p_1}=150i$ and $\vec{p_2}=490j$
According to the principle of conservation of linear momentum, if we call the momentum of the system after the collision $\vec{p_f}$, we have $$\vec{p_f}=\vec{p_1}+\vec{p_2}=150i+490j (kg.m/s)$$
After the collision, the system of 2 skaters has mass $M=50+70=120kg$, so the velocity of the system is $$\vec{v_f}=\frac{\vec{p_f}}{M}=1.25i+4.08j (m/s)$$
a) We have $$\tan\theta=\frac{4.08}{1.25}=3.264$$ $$\theta=73^o$$
b) The speed is $v_f=\sqrt{1.25^2+4.08^2}=4.27m/s$
