1. MOMENTUM! Momentum Impulse Conservation of Momentum in 1 Dimension
Conservation of Momentum in 2 Dimensions
Angular Momentum
Torque
Moment of Inertia

2. p = m v Momentum Defined p = momentum vector m = mass
v = velocity vector

3. Momentum Facts p = m v Momentum is a vector quantity!
Velocity and momentum vectors point in the same direction.
SI unit for momentum: kg · m /s (no special name).
Momentum is a conserved quantity (this will be proven later).
A net force is required to change a body’s momentum.
Momentum describes the tendency of a mass to keep going in the same direction with the same speed.
Something big and slow could have the same momentum as something small and fast.

4. Vocabulary angular momentum collision law of conservation of momentum
elastic collision
gyroscope
impulse
inelastic collision
linear momentum
momentum

5. Momentum The momentum of a ball depends on its mass and velocity.
Ball B has more momentum than ball A.

6. Momentum and Inertia
Inertia is another property of mass that resists changes in velocity; however, inertia depends only on mass.
Inertia is a scalar quantity.
Momentum is a property of moving mass that resists changes in a moving object’s velocity.
Momentum is a vector quantity.

7. Momentum Ball A is 1 kg moving 1m/sec, Ball B is 1kg at 3 m/sec.
A 1 N force is applied to deflect each ball's motion.
What happens?
Does the force deflect both balls equally?
Ball B deflects much less than ball A when the same force is applied because ball B had a greater initial momentum.

8. Calculating Momentum
The momentum of a moving object is its mass multiplied by its velocity.
That means momentum increases with both mass and velocity.
Momentum
(kg m/sec)
p = m v
Velocity (m/sec)
Mass (kg)

9. Momentum Examples
3 m /s
30 kg · m /s
10 kg
10 kg
Note: The momentum vector does not have to be drawn 10 times longer than the velocity vector, since only vectors of the same quantity can be compared in this way.
9 km /s
26º
p = 45 kg · m /s at 26º N of E
5 g

10. Comparing momentum
A car is traveling at a velocity of 13.5 m/sec (30 mph) north on a straight road. The mass of the car is 1,300 kg.
A motorcycle passes the car at a speed of 30 m/sec (67 mph). The motorcycle (with rider) has a mass of 350 kg.
Calculate and compare the momentum of the car and motorcycle.
You are asked for momentum.
You are given masses and velocities.
Use: p = m v
Solve for car: p = (1,300 kg) (13.5 m/s) = 17,550 kg m/s
Solve for cycle: p = (350 kg) (30 m/s) = 10,500 kg m/s
The car has more momentum even though it is going much slower.

11. Equivalent Momenta
Car: m = 1800 kg; v = 80 m /s p = 1.44 ·105 kg · m /s
Bus: m = 9000 kg; v = 16 m /s p = 1.44 ·105 kg · m /s
Train: m = 3.6 ·104 kg; v = 4 m /s p = 1.44 ·105 kg · m /s
continued on next slide

12. Equivalent Momenta (cont.)
The train, bus, and car all have different masses and speeds, but their momenta are the same in magnitude. The massive train has a slow speed; the low-mass car has a great speed; and the bus has moderate mass and speed. Note: We can only say that the magnitudes of their momenta are equal since they’re aren’t moving in the same direction.
The difficulty in bringing each vehicle to rest--in terms of a combination of the force and time required--would be the same, since they each have the same momentum.

13. Review Questions What is momentum equal to? A. Mass x Velocity
B. Current x Voltage
C. Force x Time
D. Frequency x Wavelength
Answer # 1: A
Answer # 2: C
Answer # 3: You can increase momentum of object by applying the greatest force possible for the longest time. To decrease the momentum of an object, decrease the mass or velocity.
Answer # 4: Yes

14. Review Questions Why is momentum a vector quantity? Mass is a vector
Velocity is a vector
Time is a vector
Answer # 1: A
Answer # 2: C
Answer # 3: You can increase momentum of object by applying the greatest force possible for the longest time. To decrease the momentum of an object, decrease the mass or velocity.
Answer # 4: Yes

15. Do Momentum Problems

16. Impulse Defined
Impulse is defined as the product force acting on an object and the time during which the force acts. The symbol for impulse is J. So, by definition:
J = F t
Example: A 50 N force is applied to a 100 kg boulder for 3 s. The impulse of this force is J = (50 N) (3 s) = 150 N · s.
Note that we didn’t need to know the mass of the object in the above example.

17. { Impulse Units proof: 1 N · s = 1 (kg · m /s2) (s) = 1 kg · m /s
J = F t shows why the SI unit for impulse is the Newton · second. There is no special name for this unit, but it is equivalent to a kg · m /s.
proof: 1 N · s = 1 (kg · m /s2) (s) = 1 kg · m /s {
Fnet = m a shows this is equivalent to a newton.
Therefore, impulse and momentum have the same units, which leads to a useful theorem.

18. Impulse - Momentum Theorem
The impulse due to all forces acting on an object (the net force) is equal to the change in momentum of the object:
Fnet t =  p
We know the units on both sides of the equation are the same (last slide), but let’s prove the theorem formally:
Fnet t = m a t = m ( v / t) t = m  v =  p

19. Stopping Time
F t = F t

20. Impulse

21. Impulse Ft = ∆mv

22. Impulse Ft = ∆mv

23. Impulse Ft = ∆mv

24. Impulse Ft = ∆mv

25. Impulse
Ft= ∆mv
F t= ∆mv

26. Impulse - Momentum Example
A 1.3 kg ball is coming straight at a 75 kg soccer player at 13 m/s who kicks it in the exact opposite direction at 22 m/s with an average force of 1200 N. How long are his foot and the ball in contact?
answer: We’ll use Fnet t =  p. Since the ball changes direction,  p = m  v = m (vf - v0) = 1.3 [22 - (-13)] = (1.3 kg) (35 m/s) = 45.5 kg · m /s. Thus, t = 45.5 / = s, which is just under 40 ms.
During this contact time the ball compresses substantially and then decompresses. This happens too quickly for us to see, though. This compression occurs in many cases, such as hitting a baseball or golf ball.

27. Fnet vs. t graph Net area =  p Fnet (N) t (s) 6
A variable strength net force acts on an object in the positive direction for 6 s, thereafter in the opposite direction. Since impulse is Fnet t, the area under the curve is equal to the impulse, which is the change in momentum. The net change in momentum is the area above the curve minus the area below the curve. This is just like a v vs. t graph, in which net displacement is given area under the curve.

28. Review Questions What is Impulse? A. Mass x Velocity
B. Current x Voltage
C. Force x Time
D. Frequency x Wavelength
Answer # 1: A
Answer # 2: C
Answer # 3: You can increase momentum of object by applying the greatest force possible for the longest time. To decrease the momentum of an object, decrease the mass or velocity.
Answer # 4: Yes

29. Review Questions What can you do to increase momentum? Increase Time
Increase Force
Increase Impulse
All of these
Answer # 1: A
Answer # 2: C
Answer # 3: You can increase momentum of object by applying the greatest force possible for the longest time. To decrease the momentum of an object, decrease the mass or velocity.
Answer # 4: Yes

30. Do Impulse Problems

31. Conservation of Momentum
The law of conservation of momentum states when a system of interacting objects is not influenced by outside forces (like friction), the total momentum of the system cannot change.
If you throw a rock forward from a skateboard, you will move backward in response.

32. Collisions in One Dimension
A collision occurs when two or more objects hit each other.
During a collision, momentum is transferred from one object to another.
Collisions can be elastic
or inelastic.

33. Collisions

34. Conservation of Momentum in 1-D
Whenever two objects collide (or when they exert forces on each other without colliding, such as gravity) momentum of the system (both objects together) is conserved. This mean the total momentum of the objects is the same before and after the collision.
(Choosing right as the + direction, m2 has - momentum.)
before: p = m1 v1 - m2 v2 v1 v2 m1 m2
m1 v1 - m2 v2 = - m1 va + m2 vb
after: p = - m1 va + m2 vb va vb m1 m2

35. Elastic collisions
Two kg billiard balls roll toward each other and collide head-on.
Initially, the 5-ball has a velocity of 0.5 m/s.
The 10-ball has an initial velocity of -0.7 m/s.
The collision is elastic and the 10-ball rebounds with a velocity of 0.4 m/s, reversing its direction.
What is the velocity of the 5-ball after the collision?

36. Elastic Collisions
You are asked for 10-ball’s velocity after collision.
You are given mass, initial velocities, 5-ball’s final velocity.
Diagram the motion, use m1v1 + m2v2 = m1v3 + m2v4
Solve for V3 :
(0.165 kg)(0.5 m/s) + (0.165 kg) (-0.7 kg)=(0.165 kg) v3 + (0.165 kg) (0.4 m/s)
V3 = -0.6 m/s

37. Directions after a collision
On the last slide the boxes were drawn going in the opposite direction after colliding. This isn’t always the case. For example, when a bat hits a ball, the ball changes direction, but the bat doesn’t. It doesn’t really matter, though, which way we draw the velocity vectors in “after” picture. If we solved the conservation of momentum equation (red box) for vb and got a negative answer, it would mean that m2 was still moving to the left after the collision. As long as we interpret our answers correctly, it matters not how the velocity vectors are drawn. v1 v2 m1 m2
m1 v1 - m2 v2 = - m1 va + m2 vb va vb m1 m2

38. Sample Problem
A crate of raspberry donut filling collides with a tub of lime Kool Aid on a frictionless surface. Which way on how fast does the Kool Aid rebound? answer: Let’s draw v to the right in the after picture.
3 (10) - 6 (15) = -3 (4.5) + 15 v v = -3.1 m/s Since v came out negative, we guessed wrong in drawing v to the right, but that’s OK as long as we interpret our answer correctly. After the collision the lime Kool Aid is moving 3.1 m/s to the left.
before
6 m/s
10 m/s
3 kg
15 kg
after
4.5 m/s v
3 kg
15 kg

39. Sample Problem 1 35 g 7 kg 700 m/s v = 0
A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passing through it, the bullet pushes the butter to the left, and the butter pushes the bullet just as hard to the right, slowing the bullet down. If the butter skids off at 4 cm/s after the bullet passes through it, what is the final speed of the bullet? (The mass of the rifle matters not.)
35 g
7 kg
v = ?
4 cm/s
continued on next slide

40. Sample Problem 1 (cont.)
Let’s choose left to be the + direction & use conservation of momentum, converting all units to meters and kilograms.
35 g
7 kg
p before = 7 (0) + (0.035) (700)
= 24.5 kg · m /s
700 m/s
v = 0
35 g
7 kg
p after = 7 (0.04) v
= v
v = ?
4 cm/s
p before = p after = v v = 692 m/s
v came out positive. This means we chose the correct direction of the bullet in the “after” picture.

41. Inelastic Collisions You are asked for the final velocity.
A train car moving to the right at 10 m/s collides with a parked train car.
They stick together and roll along the track.
If the moving car has a mass of 8,000 kg and the parked car has a mass of 2,000 kg, what is their combined velocity after the collision?
You are asked for the final velocity.
You are given masses, and initial velocity of moving train car.

42. Inelastic Collisions Diagram the problem
Use m1v1 + m2v2 = (m1v1 +m2v2) v3
Solve for v3 = m1v1 + m2v2
(m1v1 +m2v2)
v3= (8,000 kg)(10 m/s) + (2,000 kg)(0 m/s)
(8, ,000 kg)
v3= 8 m/s
The train cars moving together to right at 8 m/s.

43. Sample Problem 2 (0.035) (700) = 7.035 v v = 3.48 m/s 35 g 7 kg
Same as the last problem except this time it’s a block of wood rather than butter, and the bullet does not pass all the way through it. How fast do they move together after impact? v kg
(0.035) (700) = v v = 3.48 m/s
Note: Once again we’re assuming a frictionless surface, otherwise there would be a frictional force on the wood in addition to that of the bullet, and the “system” would have to include the table as well.

44. Bouncing
Alfred went on a date with Mabel. When Alfred dropped off Mabel after the date, he was anxious to play Angry Birds, so he forgot to kiss her on the cheek good night. She went up to her room, opened the window and threw a flower pot at Alfred. On of three things could happen.
1. The flower pot – head collision is elastic
2. The flower pot – head collision is inelastic
3. The flower pot bounces off his head
Which will hurt more?????

45. Elastic Collision
Before
After

46. Elastic Collision Alfred + Flower pot = Alfred + Flower pot
m1v1 + m2v2 = m1v3 + m2v4
100kg(0m/s) + 10kg (15 m/s) = 100kg (v3) + 10kg (0m/s)
150kgm/s = 100kg (v3)
100kg kg
1.5 m/s = v3 (elastic)

47. Inelastic Collision
Before
After

48. Inelastic Collision Alfred + Flower pot = (Alfred + Flower pot)
m1v1 + m2v2 = (m1 + m2)v3
100kg(0m/s) + 10kg(15 m/s) = (100kg + 10kg) (v3)
150kgm/s = 110kg(v3)
110kg kg
1.36 m/s = v3 (inelastic)
1.5 m/s = v3 (elastic)

49. Bouncing

50. Elastic Collision Alfred + Flower pot = Alfred + Flower pot
m1v1 + m2v2 = m1v3 + m2v4
100kg(0m/s) + 10kg(15 m/s) = 100kg(v3) + 10kg(-5m/s)
150kgm/s = 100kg(v3) – 50kgm/s
200kgm/s = 100kg(v3)
100kg kg
2.0 m/s = v3 (bouncing)
1.5 m/s = v3 (elastic)
1.36 m/s = v3 (inelastic)

51. Do Collision Problems

52. Angular Momentum
Momentum resulting from an object moving in linear motion is called linear momentum.
Momentum resulting from the rotation (or spin) of an object is called angular momentum.

53. Conservation of Angular Momentum
Angular momentum is important because it obeys a conservation law, as does linear momentum.
The total angular momentum of a closed system stays the same.

54. Calculating Angular Momentum
Angular momentum is calculated in a similar way to linear momentum, except the mass and velocity are replaced by the moment of inertia and angular velocity.
Moment of inertia
(kg m2)
Angular
momentum
(kg m/sec2)
L = I w
Angular
velocity
(rad/sec)

55. Calculating Angular Momentum
The moment of inertia of an object is the average of mass times radius squared for the whole object.
Since the radius is measured from the axis of rotation, the moment of inertia depends on the axis of rotation.

56. Gyroscopes Angular Momentum
A gyroscope is a device that contains a spinning object with a lot of angular momentum.
Gyroscopes can do amazing tricks because they conserve angular momentum.
For example, a spinning gyroscope can easily balance on a pencil point.

57. Gyroscopes
A gyroscope on the space shuttle is mounted at the center of mass, allowing a computer to measure rotation of the spacecraft in three dimensions.
An on-board computer is able to accurately measure the rotation of the shuttle and maintain its orientation in space.

58. Comparison: Linear & Angular Momentum
Linear Momentum, p
Tendency for a mass to continue moving in a straight line.
Parallel to v.
A conserved, vector quantity.
Magnitude is inertia (mass) times speed.
Net force required to change it.
The greater the mass, the greater the force needed to change momentum.
Angular Momentum, L
Tendency for a mass to continue rotating.
Perpendicular to both v and r.
A conserved, vector quantity.
Magnitude is rotational inertia times angular speed.
Net torque required to change it.
The greater the moment of inertia, the greater the torque needed to change angular momentum.