1. Momentum and Impulse

2. Section 9.1 Objectives Define the momentum of an object.
Determine the impulse given to an object.
Calculate the momentum and impulse of an object.

3. Let’s start with everyday language
What do you say when a sports team is on a roll?
They may not have the lead but they may have ___________
MOMENTUM
A team that has momentum is hard to stop.

4. What is Momentum?
An object with a lot of momentum is also hard to stop
Momentum = p = mv
Units: kg∙m/s
m=mass
v=velocity
Momentum is also a vector (it has direction)

5. Let’s practice
A 1200 kg car drives west at 25 m/s for 3 hours. What is the car’s momentum?
Identify the variables:
1200 kg = mass
25m/s, west = velocity
3 hours = time
P = mv = 1200 x 25 = kg m/s, west

6. 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

7. How hard is it to stop a moving object?
To stop an object, we have to apply a force over a period of time.
This is called Impulse
Impulse = FΔt Units: N∙s
F = force (N)
Δt = time elapsed (s)

8. How hard is it to stop a moving object?
Using Newton’s 2nd Law we get
FΔt= mΔv
Which means
Impulse = change in momentum

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. Why does an egg break or not break?
An egg dropped on a tile floor breaks, but an egg dropped on a pillow does not. Why?
FΔt= mΔv
In both cases, m and Δv are the same.
If Δt goes up, what happens to F, the force?
Right! Force goes down. When dropped on a pillow, the egg starts to slow down as soon as it touches it. A pillow increases the time the egg takes to stops.

11. Practice Problem
A 57 gram tennis ball falls on a tile floor. The ball changes velocity from -1.2 m/s to +1.2 m/s in 0.02 s. What is the average force on the ball?
Identify the variables:
Mass = 57 g = kg
Δvelocity = +1.2 – (-1.2) = 2.4 m/s
Time = 0.02 s
using FΔt= mΔv
F x (0.02 s) = (0.057 kg)(2.4 m/s)
F= 6.8 N

12. Car Crash
Would you rather be in a head on collision with an identical car, traveling at the same speed as you, or a brick wall?
Assume in both situations you come to a complete stop in the same amount of time.
Take a guess

13. Car Crash (cont.) Everyone should vote now
Raise one finger if you think it is better to hit another car, two if it’s better to hit a wall and three if it doesn’t matter.
And the answer is…..

14. Car Crash (cont.) The answer is… It Does Not Matter! Look at FΔt= mΔv
In both situations, Δt, m, and Δv are the same! The time it takes you to stop depends on your car, m is the mass of your car, and Δv depends on how fast you were initially traveling.

15. Egg Drop connection FΔt= mΔv
How are you going to use this in your egg drop?
Which of these variables can you control?
FΔt= mΔv
Which of them do you want to maximize, which do you want to minimize
(note: we are looking at the force on the egg. Therefore, m represents the egg mass, not the entire mass of the project)

16. Section 9.2 Objectives
Relate Newton’s third law to conservation of momentum in collisions. 
Identify conditions that momentum is conserved. 
Solve conservation of momentum problems.

17. Law of Conservation of Momentum
pi = pf
The initial momentum = the final momentum
pi,A + pi,B = pf,A + pf.B
The sum of initial momentums = the sum of final momentums 17

18. Conditions for Conservation of Momentum
Closed system
Objects don’t enter or leave
Isolated system
No external forces act on the system 18

19. Conservation of Momentum19

21. 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

22. 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

23. 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 doesn’t matter)
35 g
7 kg
v = ?
4 cm/s
continued on next slide

24. 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.

25. Sample Problem 2 35 g p before = 7 (0) + (0.035) (700)
= 24.5 kg · m /s
7 kg
700 m/s
v = 0
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?
(0.035kg)+(7kg) = 7.035kg v kg
p before = p after kg.m/s = (7.035kg) 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.

26. Sample Problem 3
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

27. Section 9.2 Objectives
Explain how the law of conservation of momentum can be applied to collisions in two dimensions
Solve two dimensional collision problems using vector addition

28. 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

29. 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.

30. 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

31. 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.

32. 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

33. From a State Standards Test
Copyright © 2004 California Department of Education.

34. From a State Standards Test
Copyright © 2004 California Department of Education.

35. From the State Standards Test
Copyright © 2004 California Department of Education.