By Steven Holzner

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

  1. 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.

    1. 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

    2. Here’s what the original momentum in the x direction was:

      pfx = pox = m1vo1 cos 40 degrees + m2vo2

    3. Momentum is conserved in the x direction, so you get

      pfx = pox = m1vo1 cos 40 degrees + m2vo2 = m1vf1x + m2vf2 cos 30 degrees

    4. Which means that

      m1vf1x = m1vo1 cos 40 degrees + m2vo2 m2vf2 cos 30 degrees

    5. Divide by m1:

      image1.jpg

      And because m1 = m2, this becomes

      vf1x = vo1 cos 40 degrees + vo2vf2 cos 30 degrees

    6. Plug in the numbers:

      image2.jpg

    7. 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

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

      image3.jpg

    9. That equation turns into:

      m1vf1y = m1vo1 sin 40 degrees – m2vf2 sin 30 degrees

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

      image4.jpg

    11. Because the two masses are equal, this becomes

      vf1y = vo1 sin 40 degrees – vf2 sin 30 degrees

    12. Plug in the numbers:

      image5.jpg

    13. 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

Practice questions

  1. Assume that the two objects in the preceding figure are hockey pucks of equal mass. Assuming that vo1 = 15 m/s, vo2 = 7.0 m/s, and vf2 = 7.0 m/s, what are theta and vf1, assuming that momentum is conserved but kinetic energy is not?

  2. Assume that the two objects in the following figure are tennis balls of equal mass. Assuming that vo1 = 12 m/s, vo2 = 8.0 m/s, and vf2 = 6.0 m/s, what are theta and vf1, assuming that momentum is conserved but kinetic energy is not?

    image8.jpg

Following are answers to the practice questions:

  1. 14 m/s, 26 degrees

    1. Momentum is conserved in this collision. In fact, momentum is conserved in both the x and y directions, which means the following are true:

      pfx = pox

      pfy = poy

    2. The original momentum in the x direction was

      pfx = pox = m1vo1 cos 40 degrees + m2vo2

    3. Momentum is conserved in the x direction, so

      pfx = pox = m1vo1 cos 40 degrees + m2vo2 = m1vf1x + m2vf2 cos 30 degrees

    4. Solving for m1vf1x gives you:

      m1vf1x = m1vo1 cos 40 degrees + m2vo2m2vf2 cos 30 degrees

    5. Divide by m1:

      image9.jpg

      Because m1 = m2, that equation becomes

      vf1x = vo1 cos 40 degrees + vo2vf2 cos 30 degrees

    6. Plug in the numbers:

      image10.jpg

    7. Now for the y direction. The original momentum in the y direction was

      pfy = poy = m1vo1 sin 40 degrees

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

      pfy = poy = m1vo1 sin 40 degrees = m1vf1y + m2vf2 sin 30 degrees

    9. Which turns into

      m1vf1y = m1vo1 sin 40 degrees – m2vf2 sin 30 degrees

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

      image11.jpg

    11. Because the two masses are equal, the equation becomes

      vf1y = vo1 sin 40 degrees – vf2 sin 30 degrees

    12. Plug in the numbers:

      image12.jpg

    13. So

      vf1x = 12.4 m/s

      vf1y = 6.1 m/s

      That means the angle theta is

      image13.jpg

      And the magnitude of vf1 is

      image14.jpg

  2. 14 m/s, 12 degrees

    1. In this situation, momentum is conserved in both the x and y directions, so the following are true:

      pfx = pox

      pfy = poy

    2. The original momentum in the x direction was

      pfx = pox = m1vo1 cos 35 degrees + m2vo2

    3. Momentum is conserved in the x direction, so:

      pfx = pox = m1vo1 cos 35 degrees + m2vo2 = m1vf1x + m2vf2 cos 42 degrees

    4. Which means:

      m1vf1x = m1vo1 cos 35 degrees + m2vo2m2vf2 cos 42 degrees

    5. Divide by m1:

      image15.jpg

      Because m1 = m2, this becomes

      vf1x = vo1 cos 35 degrees + vo2vf2 cos 42 degrees

    6. Plug in the numbers:

      image16.jpg

    7. Now for the y direction. The original momentum in the y direction was

      pfy = poy = m1vo1 sin 35 degrees

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

      pfy = poy = m1vo1 sin 35 degrees = m1vf1y + m2vf2 sin 42 degrees

      Solving for m1vf1y gives you:

      m1vf1y = m1vo1 sin 35 degrees – m2vf2 sin 42 degrees

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

      image17.jpg

    10. Because the two masses are equal, the equation becomes

      vf1y = vo1 sin 35 degrees – vf2 sin 42 degrees

    11. Plug in the numbers:

      image18.jpg

    12. So:

      vf1x = 13.4 m/s

      vf1y = 2.9 m/s

      Which means the angle theta is

      image19.jpg

      And the magnitude of vf1 is

      image20.jpg