Collisions in Two Dimensions

Astrophysics For Dummies

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_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_f₁_x + m₂v_f₂ cos 30 degrees
4. Which means that
  
  m₁v_f₁_x = m₁v_o₁ cos 40 degrees + m₂v_o₂ – m₂v_f₂ cos 30 degrees
5. Divide by m₁:
  
  And because m₁ = m₂, this becomes
  
  v_f₁_x = 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_f₁_y = 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_f₁_y = v_o₁ sin 40 degrees – v_f₂ sin 30 degrees
12. Plug in the numbers:
13. So:
  
  v_f₁_x = 7.5 m/s (to the right)
  
  v_f₁_y = 3.4 m/s (downward)
  
  That means that the angle theta is
  
  And the magnitude of v_f₁ is

Practice questions

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?
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:

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_f₁_x + m₂v_f₂ cos 30 degrees
4. Solving for m₁v_f₁_x gives you:
  
  m₁v_f₁_x = m₁v_o₁ cos 40 degrees + m₂v_o₂ – m₂v_f₂ cos 30 degrees
5. Divide by m₁:
  
  Because m₁ = m₂, that equation becomes
  
  v_f₁_x = 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_f₁_y + m₂v_f₂ sin 30 degrees
9. Which turns into
  
  m₁v_f₁_y = 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_f₁_y = v_o₁ sin 40 degrees – v_f₂ sin 30 degrees
12. Plug in the numbers:
13. So
  
  v_f₁_x = 12.4 m/s
  
  v_f₁_y = 6.1 m/s
  
  That means the angle theta is
  
  And the magnitude of v_f1 is
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_f₁_x + m₂v_f₂ cos 42 degrees
4. Which means:
  
  m₁v_f₁_x = m₁v_o₁ cos 35 degrees + m₂v_o₂ – m₂v_f₂ cos 42 degrees
5. Divide by m₁:
  
  Because m₁ = m₂, this becomes
  
  v_f₁_x = 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_f₁_y + m₂v_f₂ sin 42 degrees
  
  Solving for m₁v_f₁_y gives you:
  
  m₁v_f₁_y = 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_f₁_y = v_o₁ sin 35 degrees – v_f₂ sin 42 degrees
11. Plug in the numbers:
12. So:
  
  v_f₁_x = 13.4 m/s
  
  v_f₁_y = 2.9 m/s
  
  Which means the angle theta is
  
  And the magnitude of v_f₁ is

About This Article

About the book author:

Dr. Steven Holzner has written more than 40 books about physics and programming. He was a contributing editor at PC Magazine and was on the faculty at both MIT and Cornell. He has authored Dummies titles including Physics For Dummies and Physics Essentials For Dummies. Dr. Holzner received his PhD at Cornell.