1. Aim: How do we solve collisions in two dimensions?

2. Two Dimensional Inelastic Collision
In a two dimensional collision, the momentum must be conserved in each dimension.
In the x-direction, momentum is conserved.
In the y-direction, momentum is conserved.

3. Conservation of Momentum in Two Dimensions
Perfectly Inelastic Collisions m1v1ix + m2v2ix = (m1 + m2)vfx m1v1iy + m2v2iy = (m1 + m2)vfy

4. Write Up
Using conservation of momentum, sketch the path taken by the mass m1 after the collision. Explain in a sentence how you determined this.
M1 will move in a northeast direction since momentum must be conserved in both the vertical and horizontal directions

5. Two Dimensional Collision Thought Question
Two objects that are moving along an xy plane on a frictionless floor collide. Assume that they form a closed, isolated system. The following table gives some of the momentum components (in kilogram-meters per second) before and after the collision. What are the missing values?
Situation 1: for B After px = -2, for B after py =4
Situation 2: for D Before px =11, for C After py =1
Situation 3: for E Before py=-2, for E After px =4

6. Sample Problems
1. A 1325 kg car, C, moving north at 27.0 m/s, collides with a 2165 kg car, D moving east at 11.0 m/s. The two cars are stuck together. In what direction and with what speed do they move after the collision?

8. Sample Problems
2. A 1345 kg car moving east at 15.7 m/s is struck by a 1923 kg car moving north. They are stuck together and move with an initial velocity of 14.5 m/s at Θ=63.5◦. Was the north-moving car exceeding the 20.1 m/s speed limit?

10. Conservation of Momentum in Two Dimensions
Elastic and Inelastic Collisions m1v1ix + m2v2ix = m1v1fx + m2v2fx m1v1iy + m2v2iy = m1v1fy +m2v2fy

11. Elastic Collisions Remember elastic collisions conserve kinetic energy, but kinetic energy is a scalar quantity so we just write m1(v1i)2 + m2(v2i)2 =m1(v1f)2 +m2(v2f)2

12. 3. A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.5 x 105 m/s and makes a collision with the second proton. After the collision, one proton moves off at an angle of 37 degrees to the original direction of motion and the second deflects at an angle of ф to the same axis. Find the final speeds of the two protons and the angle ф.
2.1 x 105 m/s, 53 degrees

15. Sample Problem
4. A billiard ball of mass = kg moving with a horizontal velocity of 2 m/s strikes a second ball, initially at rest, of the same mass. As a result of the collision, the first ball is deflected off at an angle of 30◦ with a speed of 1.20 m/s.
Write down the equations for the components of momentum in the x and y-direction separately.
Solve these equations for speed of the second ball and angle Θ. Assume collision is elastic.

16. Thought Question 2 a) Path 2 b) Path 1 c) Path 3
A hockey puck A has an initial momentum of 5 kg m/s before colliding with stationary puck B. Which path will puck A take after the collision if the x component of puck B has an x component of
5 kg m/s
More than 5 kg m/s
Less than 5 kg m/s
a) Path 2 b) Path 1 c) Path 3

17. 5. Two masses, m=1 kg and M=5 kg collide as shown below.
Determine the speed of both masses after the collision.
Is the collision an elastic or inelastic collision?

18. Solution to Number 5
a) Conservation of Momentum in the x-direction 1(20)=5v2f cos37 v2f = 5 m/s Conservation of Momentum in the y-direction 0=-1v1f +5v2f sin37 0=-1v1f +5(5)sin37 v1f =15 m/s b) We must see if kinetic energy is conserved The kinetic energy before the collision can be found by using KE=1/2 (1)(20)2=200 J The kinetic energy after the collision can be found by using KE=1/2 (1)(15)2 + ½ (5)(5)2=175 J Kinetic Energy is not conserved so the collision must be inelastic