Collisions in Two Dimensions
Collisions can take place in two dimensions. For example, soccer balls can move any which way on a soccer field, not just along a single line. Soccer balls can end up going north or south, east or west, or a combination of those. So you have to be prepared to handle collisions in two dimensions.
Sample question

In the figure, there’s been an accident at an Italian restaurant, and two meatballs are colliding. Assuming that v_{o}_{1} = 10.0 m/s, v_{o}_{2} = 5.0 m/s, v_{f}_{2} = 6.0 m/s, and the masses of the meatballs are equal, what are theta and v_{f}_{1}?
The correct answer is theta = 24 degrees and v_{f}_{1} = 8.2 m/s.

You can’t assume that these meatballs conserve kinetic energy when they collide because the meatballs probably deform from the collision. However, momentum is conserved. In fact, momentum is conserved in both the x and y directions, which means
p_{fx} = p_{ox}
and
p_{fy} = p_{oy}

Here’s what the original momentum in the x direction was:
p_{fx} = p_{ox} = m_{1}v_{o}_{1} cos 40 degrees + m_{2}v_{o}_{2}

Momentum is conserved in the x direction, so you get
p_{fx} = p_{ox} = m_{1}v_{o}_{1} cos 40 degrees + m_{2}v_{o}_{2} = m_{1}v_{f}_{1}_{x} + m_{2}v_{f}_{2} cos 30 degrees

Which means that
m_{1}v_{f}_{1}_{x} = m_{1}v_{o}_{1} cos 40 degrees + m_{2}v_{o}_{2} – m_{2}v_{f}_{2} cos 30 degrees

Divide by m_{1}:
And because m_{1} = m_{2}, this becomes
v_{f}_{1}_{x} = v_{o}_{1} cos 40 degrees + v_{o}_{2} – v_{f}_{2} cos 30 degrees

Plug in the numbers:

Now for the y direction. Here’s what the original momentum in the y direction looks like (in the downward direction):
p_{fy} = p_{oy} = m_{1}v_{o}_{1} sin 40 degrees

Set that equal to the final momentum in the y direction:

That equation turns into:
m_{1}v_{f}_{1}_{y} = m_{1}v_{o}_{1} sin 40 degrees – m_{2}v_{f}_{2} sin 30 degrees

Solve for the final velocity component of meatball 1’s y velocity:

Because the two masses are equal, this becomes
v_{f}_{1}_{y} = v_{o}_{1} sin 40 degrees – v_{f}_{2} sin 30 degrees

Plug in the numbers:

So:
v_{f}_{1}_{x} = 7.5 m/s (to the right)
v_{f}_{1}_{y} = 3.4 m/s (downward)
That means that the angle theta is
And the magnitude of v_{f}_{1} is

Practice questions

Assume that the two objects in the preceding figure are hockey pucks of equal mass. Assuming that v_{o}_{1} = 15 m/s, v_{o}_{2} = 7.0 m/s, and v_{f}_{2} = 7.0 m/s, what are theta and v_{f}_{1}, assuming that momentum is conserved but kinetic energy is not?

Assume that the two objects in the following figure are tennis balls of equal mass. Assuming that v_{o}_{1} = 12 m/s, v_{o}_{2} = 8.0 m/s, and v_{f}_{2} = 6.0 m/s, what are theta and v_{f}_{1}, assuming that momentum is conserved but kinetic energy is not?
Following are answers to the practice questions:

14 m/s, 26 degrees

Momentum is conserved in this collision. In fact, momentum is conserved in both the x and y directions, which means the following are true:
p_{fx} = p_{ox}
p_{fy} = p_{oy}

The original momentum in the x direction was
p_{fx} = p_{ox} = m_{1}v_{o}_{1} cos 40 degrees + m_{2}v_{o}_{2}

Momentum is conserved in the x direction, so
p_{fx} = p_{ox} = m_{1}v_{o}_{1} cos 40 degrees + m_{2}v_{o}_{2} = m_{1}v_{f}_{1}_{x} + m_{2}v_{f}_{2} cos 30 degrees

Solving for m_{1}v_{f}_{1}_{x} gives you:
m_{1}v_{f}_{1}_{x} = m_{1}v_{o}_{1} cos 40 degrees + m_{2}v_{o}_{2} – m_{2}v_{f}_{2} cos 30 degrees

Divide by m_{1}:
Because m_{1} = m_{2}, that equation becomes
v_{f}_{1}_{x} = v_{o}_{1} cos 40 degrees + v_{o}_{2} – v_{f}_{2} cos 30 degrees

Plug in the numbers:

Now for the y direction. The original momentum in the y direction was
p_{fy} = p_{oy} = m_{1}v_{o}_{1} sin 40 degrees

Set that equal to the final momentum in the y direction:
p_{fy} = p_{oy} = m_{1}v_{o}_{1} sin 40 degrees = m_{1}v_{f}_{1}_{y} + m_{2}v_{f}_{2} sin 30 degrees

Which turns into
m_{1}v_{f}_{1}_{y} = m_{1}v_{o}_{1} sin 40 degrees – m_{2}v_{f}_{2} sin 30 degrees

Solve for the final velocity component of puck 1’s y velocity:

Because the two masses are equal, the equation becomes
v_{f}_{1}_{y} = v_{o}_{1} sin 40 degrees – v_{f}_{2} sin 30 degrees

Plug in the numbers:

So
v_{f}_{1}_{x} = 12.4 m/s
v_{f}_{1}_{y} = 6.1 m/s
That means the angle theta is
And the magnitude of v_{f1} is


14 m/s, 12 degrees

In this situation, momentum is conserved in both the x and y directions, so the following are true:
p_{fx} = p_{ox}
p_{fy} = p_{oy}

The original momentum in the x direction was
p_{fx} = p_{ox} = m_{1}v_{o}_{1} cos 35 degrees + m_{2}v_{o}_{2}

Momentum is conserved in the x direction, so:
p_{fx} = p_{ox} = m_{1}v_{o}_{1} cos 35 degrees + m_{2}v_{o}_{2} = m_{1}v_{f}_{1}_{x} + m_{2}v_{f}_{2} cos 42 degrees

Which means:
m_{1}v_{f}_{1}_{x} = m_{1}v_{o}_{1} cos 35 degrees + m_{2}v_{o}_{2} – m_{2}v_{f}_{2} cos 42 degrees

Divide by m_{1}:
Because m_{1} = m_{2}, this becomes
v_{f}_{1}_{x} = v_{o}_{1} cos 35 degrees + v_{o}_{2} – v_{f}_{2} cos 42 degrees

Plug in the numbers:

Now for the y direction. The original momentum in the y direction was
p_{fy} = p_{oy} = m_{1}v_{o}_{1} sin 35 degrees

Set that equal to the final momentum in the y direction:
p_{fy} = p_{oy} = m_{1}v_{o}_{1} sin 35 degrees = m_{1}v_{f}_{1}_{y} + m_{2}v_{f}_{2} sin 42 degrees
Solving for m_{1}v_{f}_{1}_{y} gives you:
m_{1}v_{f}_{1}_{y} = m_{1}v_{o}_{1} sin 35 degrees – m_{2}v_{f}_{2} sin 42 degrees

Solve for the final velocity component of puck 1’s y velocity:

Because the two masses are equal, the equation becomes
v_{f}_{1}_{y} = v_{o}_{1} sin 35 degrees – v_{f}_{2} sin 42 degrees

Plug in the numbers:

So:
v_{f}_{1}_{x} = 13.4 m/s
v_{f}_{1}_{y} = 2.9 m/s
Which means the angle theta is
And the magnitude of v_{f}_{1} is
