1. Conservation Laws Conservation of Momentum II

2. Two Dimensional Explosions
Momentum is conserved overall and it is also conserved in both the x and y directions.
Use the same equations and techniques that we used in the previous lesson. Only this time solve everything twice.
First solve the explosion in the x-direction using only x velocities.
Then solve the explosion in the y-direction using only y velocities.

3. Example 1
A 5 kg mass explodes into three pieces. A 1 kg fragment moves in the -y direction at 6 m/s. A 2 kg fragment moves in the -x direction at 4 m/s. Determine the speed and direction of the third fragment.
This splits into three fragments. We need the modify the equations.
5 kg
Initially the problem begins with a 5 kg mass.
It never tells you what the mass is originally doing.
Assume the simplest case.
It must be at rest , v0 = 0 . This means v0x = 0 and v0y = 0
Now take a look at what happens as a result of the explosion.

4. Example 1
A 5 kg mass explodes into three pieces. A 1 kg fragment moves in the -y direction at 6 m/s. A 2 kg fragment moves in the -x direction at 4 m/s. Determine the speed and direction of the third fragment.
The second and third sentences tell us part of what happens during the explosion. They are not specific about what happens to the third fragment.
2 kg
-4 m/s
2 kg
1 kg
-6 m/s
Before exploding the total mass was 5 kg.
The mass of the third fragment must be 2 kg .

5. Example 1 Solve the x-direction Solve the y-direction 1 kg -6 m/s 2 kg
A 5 kg mass explodes into three pieces. A 1 kg fragment moves in the -y direction at 6 m/s. A 2 kg fragment moves in the -x direction at 4 m/s. Determine the speed and direction of the third fragment.
1 kg
-6 m/s
2 kg
-4 m/s
Solve the x-direction
+3 m/s
+4 m/s
Solve the y-direction

6. Example 1
A 5 kg mass explodes into three pieces. A 1 kg fragment moves in the -y direction at 6 m/s. A 2 kg fragment moves in the -x direction at 4 m/s. Determine the speed and direction of the third fragment.
Now Solve for the velocity and direction of the third fragment.
2 kg
+4 m/s
+3 m/s
5 m/s
37o
Or, you could recognize this one as a triangle and save some work.
The calculations to the left would be necessary if the right triangle wasn’t easy.

7. Two Dimensional Collisions
Momentum is conserved overall and it is also conserved in both the x and y directions.
Use the same equations and techniques that we used in the previous lesson. Only this time solve everything twice.
First solve the collision in the x-direction using only x velocities.
Then solve the collision in the y-direction using only y velocities.

8. Example 2 First let’s imaging what the motion looks like 5 m/s 2 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
First let’s imaging what the motion looks like
1 kg
2 kg
5 m/s

9. Example 2 First let’s imaging what the motion looks like 5 m/s 2 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
First let’s imaging what the motion looks like
2 kg
5 m/s
1 kg

10. Example 2 First let’s imaging what the motion looks like 5 m/s 2 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
First let’s imaging what the motion looks like
2 kg
5 m/s
1 kg

11. Example 2 First let’s imaging what the motion looks like 5 m/s 2 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
First let’s imaging what the motion looks like
2 kg
5 m/s
1 kg

12. Example 2 First let’s imaging what the motion looks like ? m/s 2 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
First let’s imaging what the motion looks like
2 kg
? m/s
1 kg
5 m/s

13. Example 2 First let’s imaging what the motion looks like ? m/s 2 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
First let’s imaging what the motion looks like
2 kg
? m/s
1 kg
5 m/s
© RJansen

14. Example 2 First let’s imaging what the motion looks like ? m/s 2 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
First let’s imaging what the motion looks like
2 kg
? m/s
1 kg
5 m/s

15. Example 2 ? m/s First let’s imaging what the motion looks like 2 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
2 kg
? m/s
First let’s imaging what the motion looks like
We also only need to look at just the instant before the collision and the instant right after the collision.
Why? It was constant velocity before and constant velocity after.
1 kg
5 m/s

16. Example 2 Let’s move this over to get some room for the equations.
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
Let’s move this over to get some room for the equations.
2 kg
? m/s
1 kg
5 m/s

17. Components Example 2 ? m/s I see vectors at angles. v1y 
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
2 kg
? m/s
1 kg
5 m/s
I see vectors at angles.
v1x
v1y 
What is the secret first step to solve this problem?
Components
37o
4 m/s
3 m/s
We have a lot of information about the 1 kg mass (mass 2). Solving for the velocity components does not take much effort in this case.
The 2 kg mass (mass 1) has unknown components. If we solve for these then we can find the overall velocity and angle of this mass.

18. Example 2 Solve in the x-direction ? m/s v1y  2 kg v1x 4 m/s 1 kg 37o
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
Solve in the x-direction
2 kg
? m/s
1 kg
5 m/s
37o
4 m/s
3 m/s
v1x
v1y 

19. Example 2 Solve in the x-direction 2 kg v1x 3 m/s 5 m/s 4 m/s 1 kg
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
Solve in the x-direction
2 kg
1 kg
5 m/s
4 m/s
v1x
3 m/s

20. Example 2 Solve in the x-direction Now Solve in the y-direction ? m/s
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
Solve in the x-direction
2 kg
? m/s
1 kg
5 m/s
37o
4 m/s
3 m/s
v1y 
Now Solve in the y-direction

21. Example 2 Solve in the x-direction Now Solve in the y-direction
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
Solve in the x-direction
2 kg
1 kg
3 m/s
v1y
1.5 m/s
Now Solve in the y-direction
Remember, down is negative

22. Example 2 Solve in the x-direction Now Solve in the y-direction ? m/s
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
Solve in the x-direction
1.5 m/s
2 kg
? m/s
1 kg
5 m/s
37o
4 m/s
3 m/s 
Now Solve in the y-direction

23. Example 2 Solve in the x-direction Now Solve for v1 and 
A 2 kg mass moving at 5 m/s in the +x-direction strikes a 1 kg ball at rest. The collision is slightly off center. After the collision the 1 kg ball moves with a speed of 5 m/s at an angle of 37o below the x-axis. What is the speed and direction of the 2 kg ball?
Solve in the x-direction
2 kg
? m/s
1.5 m/s 
3 m/s
Now Solve for v1 and 
Now Solve in the y-direction

24. Ballistic Pendulum L vb
A moving bullet strikes a stationary block, that is held by a string.
The collision is perfectly inelastic.
The combined bullet and block become the mass of a pendulum.
The pendulum converts kinetic energy into potential energy.

25. Ballistic Pendulum L vb

26. Ballistic Pendulum L vb

27. Ballistic Pendulum L vb

28. Ballistic Pendulum L vb

29. Ballistic Pendulum L vb

30. Ballistic Pendulum L
vbB

31. Ballistic Pendulum L
vbB h

32. Ballistic Pendulum L
vbB h

33. Ballistic Pendulum L
vbB h

34. Ballistic Pendulum L
vbB h

35. Ballistic Pendulum L
vbB h

36. Ballistic Pendulum L
vbB h

37. Ballistic Pendulum L h

38. Ballistic Pendulum L vb This involves two phases
It starts with the velocity of a bullet

39. Ballistic Pendulum L vb This involves two phases
It starts with the velocity of a bullet vb

40. Ballistic Pendulum L vb This involves two phases
It starts with the velocity of a bullet vb

41. Ballistic Pendulum L vb This involves two phases
It starts with the velocity of a bullet vb

42. Ballistic Pendulum L vb This involves two phases
It starts with the velocity of a bullet vb

43. Ballistic Pendulum L vb This involves two phases
It starts with the velocity of a bullet vb

44. Ballistic Pendulum L vbB This involves two phases
It starts with the velocity of a bullet
It ends with the bullet and block moving together
Phase I: Perfectly inelastic collision takes place
The block is at rest initially

45. Ballistic Pendulum L vbB This involves two phases
The second phase begins with the bullet and block having speed.
The ending values of phase 1 become the beginning values of phase 2

46. Ballistic Pendulum L This involves two phases
vbB
This involves two phases
The second phase begins with the bullet and block having speed.
© RHJasen

47. Ballistic Pendulum L This involves two phases
vbB
This involves two phases
The second phase begins with the bullet and block having speed.

48. Ballistic Pendulum L This involves two phases
vbB
This involves two phases
The second phase begins with the bullet and block having speed.

49. Ballistic Pendulum L This involves two phases
vbB
This involves two phases
The second phase begins with the bullet and block having speed.

50. Ballistic Pendulum L This involves two phases
vbB
This involves two phases
The second phase begins with the bullet and block having speed.

51. Ballistic Pendulum L This involves two phases
vbB
This involves two phases
The second phase begins with the bullet and block having speed.

52. Ballistic Pendulum L h vbB v bB= 0 θ This involves two phases
The second phase begins with the bullet and block having speed. θ h
It ends with the bullet and block having height

53. Ballistic Pendulum L h vbB v bB= 0 However, the height is not given θ
The length and angle are known
Use the right triangle

54. Ballistic Pendulum However, we solve the problem backwards
The problem starts with a bullet having momentum mvb
We know the angle θ and we want the speed of the bullet vb
It collides perfectly inelastically with a block at rest
The kinetic energy from the collision turns into potential energy
The height results in the string having an angle θ