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

2. Chapter 9: Linear Momentum and Collisions
The linear momentum of a particle of mass m and velocity v is defined as
The linear momentum is a vector quantity.
It’s direction is along v.
The components of the momentum of a particle:

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 is directly proportional to both mass and speed.
Something big and slow could have the same momentum as something small and fast.

4. Momentum Examples p = 45 kg · m /s at 26º N of E 3 m /s 30 kg · m /s
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

5. 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

6. Equivalent Momenta (cont.)
The train, bus, and car all have different masses and speeds, but their momenta are the same in magnitude.
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.

7. 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.

8. { 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.
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.

9. 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

10. Stopping Time
F t = F t
Imagine a car hitting a wall and coming to rest. The force on the car due to the wall is large (big F ), but that force only acts for a small amount of time (little t ). Now imagine the same car moving at the same speed but this time hitting a giant haystack and coming to rest. The force on the car is much smaller now (little F ), but it acts for a much longer time (big t ). In each case the impulse involved is the same since the change in momentum of the car is the same. Any net force, no matter how small, can bring an object to rest if it has enough time. A pole vaulter can fall from a great height without getting hurt because the mat applies a smaller force over a longer period of time than the ground alone would.

11. 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.

12. 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.

13. As long as there are no external forces acting on a system of particles, collisions between the particles will exhibit conservation of linear momentum.
This means that the vector sum of the momenta before collision is equal to the vector sum of the momenta of the particles afterwards.

14. Conservation of linear momentum

15. 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

16. 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

17. Simple Examples of Head-On Collisions
(Elastic)(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One at rest initially has twice the mass.
Collision between two objects. One not at rest initially has twice the mass.

18. Simple Examples of Head-On Collisions
(Totally Inelastic Collision, only Momentum Conserved)
Collision between two objects of the same mass. One mass is at rest.
Collision between two objects. One at rest initially has twice the mass.
Collision between two objects. One not at rest initially has twice the mass.

19. Elastic and inelastic collisions in one dimension
Momentum is conserved in any collision, elastic and inelastic.
Mechanical Energy is only conserved in elastic collisions.
Perfectly inelastic collision: After colliding, particles stick together. There is a loss of energy (deformation).
Elastic collision: Particles bounce off each other without loss of energy.
Inelastic collision: Particles collide with some loss of energy, but don’t stick together.

20. Perfectly inelastic collision of two particles
(Particles stick together)
Notice that p and v are vectors and, thus have a direction (+/-)
There is a loss in energy Eloss

21. For elastic collisions in one dimension:
Suppose we know the initial masses and velocities.
Then:
(9.20)
(9.21)
Note, that these are pretty specialized equations, (elastic collision in one dimension, known initial velocities, and masses)

22. Black board example 9.2
Two carts collide elastically on a frictionless track. The first cart (m1 = 1kg) has a velocity in the positive x-direction of 2 m/s; the other cart (m = 0.5 kg) has velocity in the negative x-direction of 5 m/s.
Find the speed of both carts after the collision.
What is the speed if the collision is perfectly inelastic?
How much energy is lost in the inelastic collision?

23. Black board example 9.3 and demo Determining the speed of a bullet
A bullet (m = 0.01kg) is fired into a block (0.1 kg) sitting at the edge of a table. The block (with the embedded bullet) flies off the table (h = 1.2 m) and lands on the floor 2 m away from the edge of the table.
a.) What was the speed of the bullet?
b.) What was the energy loss in the bullet-block collision? (skip)
vb = ?
h = 1.2 m
x = 2 m

24. Motion of a System of Particles.
Newton’s second law for a System of Particles
The center of mass of a system of particles (combined mass M) moves like one equivalent particle of mass M would move under the influence of an external force.

25. Center of mass Center of mass for many particles:
Black board example 9.6
Where is the center of mass of this arrangement of particles.
(m3 = 2 kg; m1 = m2 = 1 kg)?
Velocity of the center of mass:
Acceleration of the center of mass:

26. A rocket is shot up in the air and explodes.
Describe the motion of the center of mass before and after the explosion.

27. A method for finding the center of mass of any object.
Hang object from two or more points.
Draw extension of suspension line.
Center of mass is at intercept of these lines.

28. Impulse (change in momentum)
A change in momentum is called “impulse”:
During a collision, a force F acts on an object, thus causing a change in momentum of the object:
For a constant (average) force:
Think of hitting a soccer ball: A force F acting over a time Dt causes a change Dp in the momentum (velocity) of the ball.

29. Black board example 9.6
A soccer player hits a ball (mass m = 440 g) coming at him with a velocity of 20 m/s. After it was hit, the ball travels in the opposite direction with a velocity of 30 m/s.
What impulse acts on the ball while it is in contact with the foot?
The impact time is 0.1s. What average force is the acting on the ball?
How much work was done by the foot? (Assume and elastic collision.) (skip)

30. 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

31. 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.

32. 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.

33. Proof of Conservation of Momentum
The proof is based on Newton’s 3rd Law. Whenever two objects collide (or exert forces on each other from a distance), the forces involved are an action-reaction pair, equal in strength, opposite in direction. This means the net force on the system (the two objects together) is zero, since these forces cancel out. F M F m
force on M due to m
force on m due to M
For each object, F = (mass) (a) = (mass) (v / t ) = (mass v) / t = p / t. Since the force applied and the contact time is the same for each mass, they each undergo the same change in momentum, but in opposite directions. The result is that even though the momenta of the individual objects changes, p for the system is zero. The momentum that one mass gains, the other loses. Hence, the momentum of the system before equals the momentum of the system after.

34. Conservation of Momentum applies only in the absence of external forces!
In the first two sample problems, we dealt with a frictionless surface. We couldn’t simply conserve momentum if friction had been present because, as the proof on the last slide shows, there would be another force (friction) in addition to the contact forces. Friction wouldn’t cancel out, and it would be a net force on the system.
The only way to conserve momentum with an external force like friction is to make it internal by including the tabletop, floor, or the entire Earth as part of the system. For example, if a rubber ball hits a brick wall, p for the ball is not conserved, neither is p for the ball-wall system, since the wall is connected to the ground and subject to force by it. However, p for the ball-Earth system is conserved!

35. Black board example 9.1
(similar to blocks and spring HW problem)
You (100kg) and your skinny friend (50.0 kg) stand face-to-face on a frictionless, frozen pond. You push off each other. You move backwards with a speed of 5.00 m/s.
What is the total momentum of the you-and-your-friend system?
What is your momentum after you pushed off?
What is your friends speed after you pushed off?
How much energy (work) did you and your friend expend?(skip)

36. Sample Problem 3
An apple is originally at rest and then dropped. After falling a short time, it’s moving pretty fast, say at a speed V. Obviously, momentum is not conserved for the apple, since it didn’t have any at first. How can this be?
answer: Gravity is an external force on the apple, so momentum for it alone is not conserved. To make gravity “internal,” we must define a system that includes the other object responsible for the gravitational force--Earth. The net force on the apple-Earth system is zero, and momentum is conserved for it. During the fall the Earth attains a very small speed v. So, by conservation of momentum:
apple m V F v
Earth M F
m V = M v

37. Sample Problem 4
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

38. Conservation of Momentum in 2-D
To handle a collision in 2-D, we conserve momentum in each dimension separately Choosing down & right as positive:
before: px = m1 v1 cos1 - m2 v2 cos2
py = m1 v1 sin1 + m2 v2 sin2 m2 m1 2 1 v2 v1
after: px = -m1 va cosa + m2 vb cos b
py = m1 va sina + m2 vb sin b m1 m2
 b a vb va
Conservation of momentum equations:
m1 v1 cos1 - m2 v2 cos2 = -m1 va cosa + m2 vb cos b
m1 v1 sin1 + m2 v2 sin 2 = m1 va sina + m2 vb sin b

39. Conserving Momentum w/ Vectors
BEFORE p1 m2 m1 2 1
p 2
p before p1
p 2
p a m1 m2
AFTER
p after
 b a
p a
p b
p b
This diagram shows momentum vectors, which are parallel to their respective velocity vectors. Note p1 + p 2 = p a + p b and p before = p after as conservation of momentum demands.

40. Exploding Bomb A e c m A c m e after before
A bomb, which was originally at rest, explodes and shrapnel flies every which way, each piece with a different mass and speed. The momentum vectors are shown in the after picture.
continued on next slide

41. Exploding Bomb (cont.)
Since the momentum of the bomb was zero before the explosion, it must be zero after it as well. Each piece does have momentum, but the total momentum of the exploded bomb must be zero afterwards. This means that it must be possible to place the momentum vectors tip to tail and form a closed polygon, which means the vector sum is zero.
If the original momentum of the bomb were not zero, these vectors would add up to the original momentum vector.

42. Two-dimensional collisions (Two particles)
Conservation of momentum:
Split into components:
If the collision is elastic, we can also use conservation of energy.

43. Velocity Components in Projectile Motion
(In the absence of air resistance.)
Note that the horizontal component
of the velocity remains the same if air
resistance can be ignored.

44. Example of Non-Head-On Collisions
(Energy and Momentum are Both Conserved)
Collision between two objects of the same mass. One mass is at rest.
If you vector add the total momentum after collision,
you get the total momentum before collision.

45. 2-D Sample Problem
152 g
A mean, old dart strikes an innocent mango that was just passing by minding its own business. Which way and how fast do they move off together?
before
40
34 m/s
0.3 kg
5 m/s
Working in grams and taking left & down as + :
152 (34) sin 40 = 452 v sin
152 (34) cos 40 (5) = 452 v cos
after
Dividing equations :
= tan
452 g
 =  
Substituting into either of the first two equations : v
v = 9.14 m/s

46. Alternate Solution
40
Shown are momentum vectors (in g m/s). The black vector is the total momentum before the collision. Because of conservation of momentum, it is also the total momentum after the collisions. We can use trig to find its magnitude and direction. 
5168 p
40
1500
Law of Cosines : p2 =  5168  1500 cos 40
p = g m/s
Dividing by total mass : v = ( g m/s) / (452 g) = 9.14 m/s
sin 
sin 40
Law of Sines : =
 = 
1500
Angle w/ resp. to horiz. = 40   53.49

47. Comments on Alternate Method
Note that the alternate method gave us the exact same solution.
This method can only be used when two objects collide and stick, or when one object breaks into two. Otherwise, we’d be dealing with a polygon with more sides than a triangle.
In using the Law of Sines (last step), the angle involved (ß) is the angle inside the triangle. A little geometry gives us the angle with respect to the horizontal.

48. Black board example 9.5
Accident investigation. Two automobiles of equal mass approach an intersection. One vehicle is traveling towards the east with 29 mi/h (13.0 m/s) and the other is traveling north with unknown speed. The vehicles collide in the intersection and stick together, leaving skid marks at an angle of 55º north of east. The second driver claims he was driving below the speed limit of 35 mi/h (15.6 m/s).
13.0 m/s
??? m/s
Is he telling the truth?
What is the speed of the “combined vehicles” right after the collision?
How long are the skid marks (mk = 0.5)?

49. ROTATIONAL INERTIA & ANGULAR MOMENTUM

50. For every type of linear quantity we have a rotational quantity that does much the same thing
Linear Quantities
Speed
Force
Mass
Momentum
Distance
Rotational Quantities
Rotational (Angular) Speed
Torque
Rotational Inertia
Angular Momentum
Angle

51. Rotational Inertia(I)
AKA (not really but could be) Rotational Mass
Resistance to change in rotational motion
Objects that are rotating about an axis tend to stay rotating, objects not rotating tend to remain at rest, unless an outside torque is applied
A torque is required to change the status of an object’s rotation
It’s the rotational equivalent to mass,
Harder to give an ang. acc. to an object w/ a larger I

52. Moment of Inertia I = m r 2 I =  mi ri 2 = m1 r12 + m2 r22
Any moving body has inertia.
The more inertia or rotational inertia a body has, the harder it is to change its linear/rotational motion.
Single point-like mass m
System of masses r m2 Q m1 r1
I = m r 2 r2 Q
I =  mi ri 2
= m1 r12 + m2 r22

53. Moment of Inertia Example
Two merry-go-rounds have the same mass and are spinning with the same angular velocity. One is solid wood (a disc), and the other is a metal ring. Which has a bigger moment of inertia relative to its center of mass? r r   m m
answer: I is independent of the angular speed. Since their masses and radii are the same, the ring has a greater moment of inertia. This is because more of its mass is farther from the axis of rotation. Since I is bigger for the ring, it would more difficult to increase or decrease its angular speed.

54. The big idea Rotational Inertia depends on mass and radius
If either one of these is large, then rotational inertia is large, and object will be harder to rotate
Different types of objects have different equations for rotational inertia
But all equations have m and r2 in them.

57. Rotational Inertia (cont.)
Some objects have more rotational inertia than others
Objects with mass closer to axis of rotation are easier to rotate, b/c it they have less rotational inertia
If the mass is farther away from the axis, then object will have more rotational inertia, and will therefore be harder to rotate

58. Angular Momentum, L
Depends on linear momentum and the distance from a particular point.
If r and v are  then the magnitude of angular momentum w/ resp. to point Q is given by L = r p = m v r.
In this case L points out of the page. If the mass were moving in the opposite direction, L would point into the page.
Unit: kg  m2 / s
A torque is needed to change L, just a force is needed to change p.
Anything spinning has angular has angular momentum. The more it has, the harder it is to stop it from spinning. v r m Q

59. Angular Momentum: General Definition
If r and v are not  then the angle between these two vectors must be taken into account. The general definition of angular momentum is given by a vector cross product:
L = r  p
This formula works regardless of the angle. As you know from our study of cross products, the magnitude of the angular momentum of m relative to point Q is: L = r p sin = m v r sin. In this case, by the right-hand rule, L points out of the page. If the mass were moving in the opposite direction, L would point into the page. v  r m Q

60. 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.

61. Angular Acceleration  =   / t
Angular acceleration occurs when a spinning object spins faster or slower.
 =   / t
Note how this is very similar to a =  v / t for linear acceleration.
Ex: If a wind turbine spinning at 21 rpm speeds up to 30 rpm over 10 s due to a gust of wind, its average angular acceleration is 9 rpm / 10 s. This means every second it’s spinning 9 revolutions per minute faster than the second before. Let’s convert the units:
9 rpm
9 rev / min
9 rev
9  (2  rad) = = =
= rad / s2
10 s
10 s
min  10 s
(60 s)  10 s

62. Torque & Angular Acceleration
Newton’s 2nd Law, as you know, is Fnet = m a
The 2nd Law has a rotational analog:  net = I 
A force is required for a body to undergo acceleration. A “turning force” (a torque) is required for a body to undergo angular acceleration.
The bigger a body’s mass, the more force is required to accelerate it. Similarly, the bigger a body’s rotational inertia, the more torque is required to accelerate it angularly.
Both m and I are measures of a body’s inertia (resistance to change in motion).

63. Linear Momentum & Angular Momentum
Recall, angular momentum’s magnitude is given by
L = m v r r must be perpendicular v m
So, if a net torque is applied, angular velocity must change, which changes angular momentum. r
proof:  net = r sinθFnet = rperpm a
= rperp m  v / t =  L / t
So net torque is the rate of change of angular momentum, just as net force is the rate of change of linear momentum.
From the formula v = r , we get
L = m v rperp= m r (r ) = m rperp 2  = I 

64. Why does a tightrope walker carry a long pole?
The pole is usually fairly heavy and by carrying it, he creates a lot of mass far away from the axis of rotation
This increases his rotational inertia
And therefore makes it harder for him to rotate/tip over

65. Sports Connection Running Gymnastics/Diving
When you run you bend your legs to reduce your rotational inertia
Gymnastics/Diving
Pull body into tight ball to achieve fast rotation

66. Angular Momentum “inertia of rotation”
Ang. Momentum= Rotational Inertia X Rotational Speed
L=Iω

67. Conservation of Angular Momentum
If no outside torque is being applied, then total angular momentum in a system must stay the same
This means, if you decrease radius, you increase rotational speed
Increase radius, then rotational speed decreases
I – represents rotational inertia
ω -represents angular speed

68. Sports Connection… Ice skating
Skater starts out in slow spin with arms and legs out
Skater pulls arms and legs in tight to body
Skater is then spinning much fast (higher rotational speed)
Gymnastics (pummel horse or floor routine)
Small radius to achieve fast rotational speed during moves, increase radius when low rotational speed is desired (during landing)

69. Do cats violate physical law?
Video
No rotate their tail one way, so that their body rotates the other so that their feet are facing the ground and they land on their feet.
This combined with their flexibility all them to almost always land on their feet 69

70. Helicopter tail rotor failure

71. Universe Connection
Rotating star shrinks radius…. What happens to rotational speed??
Goes way up….. Spins very fast
Rotating star explodes outward…. What happens to rotational speed??
Goes way down … spins much slower