2D Collision
Problem statement:
A blue ball of mass $m_1 = $ 1 kg hits another red ball, initially at rest, of mass $m_2 = $ 1.5 kg. The red ball is initially travelling along the $x$ axis with velocity $\vec v_1 = 20 \hat i$ m/s.
After the collision the blue ball is deflected an angle $\theta_1$ above the $x$ axis and has speed $V_1$. The red ball
moves with speed $V_2 = 5$ m/s at angle $\theta_2 = 30^\circ$ below the $x$ axis
Determine the speed and direction of the blue ball after the collision: $\vec V_2$.
Solution
Because there are no external forces on the system of two balls, the total momenta before and after are the same.
$${\color{blue}m_1\vec v_1} ={\color{blue}m_1 \vec V_1} + {\color{red}m_2 \vec V_2} $$
The $y$ component of $\vec v_1$ is zero, so we can write this vector equation as two equations
$$\left\{\begin{align*}
m_1 v_1 &= m_1 V_{1,x} + m_2 V_{2,x}
\\
\;
\\
0&= m_1 V_{1,y} + m_2 V_{2,y}
\end{align*}\right.$$
We can immediately solve for the $y$ components of the after-collision velocities as follows:
$$ \frac{V_{1,y}}{V_{2,y}} = \frac{m_2}{m_1}$$
or
$$V_{1,y} = \frac{m_2}{m_1}{V_{2,y} = \frac{1.5{\rm ~kg}}{1.0\rm ~kg}} \frac{5\rm ~m/s}{2} =3.75{\rm~m/s}$$
(Recall that $\cos(30^\circ) = \sqrt{3}/2$, and $\sin(30^\circ) = 1/2$.)
It is only slightly less involved to solve for $V_{1,x}$ in the first equation:
$$ V_{1,x} = \frac{m_1 v_1 - m_2 V_{2,x} }{ m_1 } = 20{\rm~m/s} - 1.5\times 5\frac{\sqrt{3}}{2}{\rm~m/s} = 13.5{\rm~m/s} $$
The total speed of mass 1 after the collision is 14 m/s at an angle of $\arctan(3.75/13.5) = 15.5^\circ$.
The kinetic energy before the collision was 200 J, after 100 J + 18.75 J = 118.75 J
N. Alberding, 2013