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

1. In the figure, there’s been an accident at an Italian restaurant, and two meatballs are colliding. Assuming that v_o1 = 10.0 m/s, v_o2 = 5.0 m/s, v_f2 = 6.0 m/s, and the masses of the meatballs are equal, what are theta and v_f1?

   

   The correct answer is theta = 24 degrees and v_f1 = 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

      p_fx = p_ox

      and

      p_fy = p_oy

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

      p_fx = p_ox = m₁v_o1 cos 40 degrees + m₂v_o2

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

      p_fx = p_ox = m₁v_o1 cos 40 degrees + m₂v_o2 = m₁v_f1x + m₂v_f2 cos 30 degrees

   4. Which means that

      m₁v_f1x = m₁v_o1 cos 40 degrees + m₂v_o2 – m₂v_f2 cos 30 degrees

   5. Divide by m₁:

      

      And because m₁ = m₂, this becomes

      v_f1x = v_o1 cos 40 degrees + v_o2 – v_f2 cos 30 degrees

   6. Plug in the numbers:

      

   7. 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 = m₁v_o1 sin 40 degrees

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

      

   9. That equation turns into:

      m₁v_f1y = m₁v_o1 sin 40 degrees – m₂v_f2 sin 30 degrees

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

       

   11. Because the two masses are equal, this becomes

       v_f1y = v_o1 sin 40 degrees – v_f2 sin 30 degrees

   12. Plug in the numbers:

       

   13. So:

       v_f1x = 7.5 m/s (to the right)

       v_f1y = 3.4 m/s (downward)

       That means that the angle theta is

       

       And the magnitude of v_f1 is

       

Practice questions

1. Assume that the two objects in the preceding figure are hockey pucks of equal mass. Assuming that v_o1 = 15 m/s, v_o2 = 7.0 m/s, and v_f2 = 7.0 m/s, what are theta and v_f1, 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 v_o1 = 12 m/s, v_o2 = 8.0 m/s, and v_f2 = 6.0 m/s, what are theta and v_f1, assuming that momentum is conserved but kinetic energy is not?

   

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:

      p_fx = p_ox

      p_fy = p_oy

   2. The original momentum in the x direction was

      p_fx = p_ox = m₁v_o1 cos 40 degrees + m₂v_o2

   3. Momentum is conserved in the x direction, so

      p_fx = p_ox = m₁v_o1 cos 40 degrees + m₂v_o2 = m₁v_f1x + m₂v_f2 cos 30 degrees

   4. Solving for m₁v_f1x gives you:

      m₁v_f1x = m₁v_o1 cos 40 degrees + m₂v_o2 – m₂v_f2 cos 30 degrees

   5. Divide by m₁:

      

      Because m₁ = m₂, that equation becomes

      v_f1x = v_o1 cos 40 degrees + v_o2 – v_f2 cos 30 degrees

   6. Plug in the numbers:

      

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

      p_fy = p_oy = m₁v_o1 sin 40 degrees

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

      p_fy = p_oy = m₁v_o1 sin 40 degrees = m₁v_f1y + m₂v_f2 sin 30 degrees

   9. Which turns into

      m₁v_f1y = m₁v_o1 sin 40 degrees – m₂v_f2 sin 30 degrees

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

       

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

       v_f1y = v_o1 sin 40 degrees – v_f2 sin 30 degrees

   12. Plug in the numbers:

       

   13. So

       v_f1x = 12.4 m/s

       v_f1y = 6.1 m/s

       That means the angle theta is

       

       And the magnitude of v_f1 is

       

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:

      p_fx = p_ox

      p_fy = p_oy

   2. The original momentum in the x direction was

      p_fx = p_ox = m₁v_o1 cos 35 degrees + m₂v_o2

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

      p_fx = p_ox = m₁v_o1 cos 35 degrees + m₂v_o2 = m₁v_f1x + m₂v_f2 cos 42 degrees

   4. Which means:

      m₁v_f1x = m₁v_o1 cos 35 degrees + m₂v_o2 – m₂v_f2 cos 42 degrees

   5. Divide by m₁:

      

      Because m₁ = m₂, this becomes

      v_f1x = v_o1 cos 35 degrees + v_o2 – v_f2 cos 42 degrees

   6. Plug in the numbers:

      

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

      p_fy = p_oy = m₁v_o1 sin 35 degrees

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

      p_fy = p_oy = m₁v_o1 sin 35 degrees = m₁v_f1y + m₂v_f2 sin 42 degrees

      Solving for m₁v_f1y gives you:

      m₁v_f1y = m₁v_o1 sin 35 degrees – m₂v_f2 sin 42 degrees

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

      

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

       v_f1y = v_o1 sin 35 degrees – v_f2 sin 42 degrees

   11. Plug in the numbers:

       

   12. So:

       v_f1x = 13.4 m/s

       v_f1y = 2.9 m/s

       Which means the angle theta is

       

       And the magnitude of v_f1 is