1. Two Dimensional Collisions
Unit A Momentum

2. Collision Simulator

3. 2-D Collisions and Conservation of Momentum25 65 B1
vA = 4.20 m/s
vA’=3.80 m/s
vB’=1.80 m/s A1

4. Vector Resolution
- Resolution breaks a vector into two perpendicular components so that vectors may be added together.
1) m/s, 157° +y –y –x
+x 0
29.6 m/s, 157
opposite
adjacent  *

5. Adding Perpendicular Vectors
1) Add: 4.71 m/s S to 3.62 m/s W
Always draw a diagram, x-component first!
adjacent
3.62 m/s W ? 0 
opposite
4.71 m/s S
add 180 only when and
are both negative *

6. 2) Add: 47.2 m/s E to 19.6 m/s N
3) Add:  159 m/s, 270 to 298 m/s, 0 *

7. 4) Add:  m/s W to  m/s N *

8. Danger: Momentum is always conserved (in isolated system).
Velocity is NOT conserved.
Ek may or may not be conserved.
Always find answer based on initial or final momentum, not velocity or Ek.

9. Two-Dimensional Collisions
Example 1: Compute the southern and western momenta of a 74.6 kg cyclist travelling at 6.35 m/s S 39.0 W.
39 N E W S N E W S *

10. E/W momentum is conserved so:
Example 2: a) Calculate the recoil velocity of a 1975 kg wheel mounted Howitzer if it lobs a 35.4 kg shell at a muzzle velocity of 394 m/s E 27.5 up.
E/W momentum is conserved so: E UP
27.5 E UP *

11. b)  What happens to the up/down momentum of the shell?
- Since the shell gains upward momentum, the Howitzer/Earth gains an equal amount of downward momentum! *

12. Example 3: A 1. 06 t Mazda going at 46. 0 km/h E collides with a 1
Example 3: A 1.06 t Mazda going at 46.0 km/h E collides with a 1.85 t Chrysler going 29.0 km/h N at an icy intersection. Right after the collision, what is the velocity of the entangled wreckage? E N
Chrysler
Mazda E N *

13. *

14. Example:

15. Initial p

16. Initial p

17. Final p

18. Final p

19. Final p

20. Red Velocity

21. Conservation of Momentum Example 2: Intersection Collision
A 40.0 kg object is travelling south at a velocity of 2.8 m/s when it collides with a 6.0 kg object travelling east at 3.0 m/s. If these two objects stick together upon collision, calculate v.
4 kg
6 kg  *

22. Example 3: Glancing Collision
A 4.0 kg object (A) is moving east at an unknown velocity when it collides with a 6.1 kg (B) stationary obkect. After the collision, the 4.0 kg object has a v = 2.8 m/s E 32 N and the 6.1 kg object has a v = 1.5 m/s E 41 S. Calculate
4 kg
6.1 kg
32 
41 
2.8 m/s
1.5 m/s
v = ? *

23. Example 4: Glancing Collision
The diagram shows two identical billiard balls before and after a glancing collision. Calculate A
10 m/s B
5 m/s
60 *

24. *