Collisions in Two Dimensions

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 vo1 = 10.0 m/s, vo2 = 5.0 m/s, vf2 = 6.0 m/s, and the masses of the meatballs are equal, what are theta and vf1?
image0.jpg

The correct answer is theta = 24 degrees and vf1 = 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
pfx = pox
and
pfy = poy
Here’s what the original momentum in the x direction was:
pfx = pox = m1vo1 cos 40 degrees + m2vo2
Momentum is conserved in the x direction, so you get
pfx = pox = m1vo1 cos 40 degrees + m2vo2 = m1vf1x + m2vf2 cos 30 degrees
Which means that
m1vf1x = m1vo1 cos 40 degrees + m2vo2 – m2vf2 cos 30 degrees
Divide by m1:
image1.jpg
And because m1 = m2, this becomes
vf1x = vo1 cos 40 degrees + vo2 – vf2 cos 30 degrees
Plug in the numbers:
image2.jpg
Now for the y direction. Here’s what the original momentum in the y direction looks like (in the downward direction):
pfy = poy = m1vo1 sin 40 degrees
Set that equal to the final momentum in the y direction:
image3.jpg

That equation turns into:
m1vf1y = m1vo1 sin 40 degrees – m2vf2 sin 30 degrees
Solve for the final velocity component of meatball 1’s y velocity:
image4.jpg
Because the two masses are equal, this becomes
vf1y = vo1 sin 40 degrees – vf2 sin 30 degrees
Plug in the numbers:
image5.jpg
So:
vf1x = 7.5 m/s (to the right)
vf1y = 3.4 m/s (downward)
That means that the angle theta is
image6.jpg

And the magnitude of vf1 is
image7.jpg