1. Momentum!!!
Physics
Mr. Padilla

2. Mr. Padilla, what is momentum?
Momentum is the inertia of motion.
Newton’s 1st Law of Motion
A moving objects desire to remain moving.
Momentum is the mass of an object multiplied by its velocity.
p = mv
Units: (kg)(m/s) = (kgm)/s

3. Can we do some examples? p = mv When velocity = 0
Ted runs at a speed of 7.5m/s. If Ted has a mass of 80kg, what is his momentum?
What is the momentum of a golf ball with a mass of .01kg, rolls down a hill at a speed of 5m/s?
Can you think of a time when Ted and the golf ball will have the same momentum?
p = mv
p = (80kg)(7.5m/s)
p = 600 kgm/s
p = (.01kg)(5m/s)
p = .05 kgm/s
When velocity = 0

4. Sample Problem 6A
A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?

5. How can we change momentum?
Momentum can change by changing the objects mass or its velocity.
It is more common to change its velocity by undergoing acceleration
Since force causes acceleration, applying a force will cause a change in momentum.
How long the force is applied will determine how much the momentum is changed.

6. Impulse Impulse is force applied over time. Impulse (Δp) = Ft
Impulse is also equal to the change in momentum.
Ft = Δ(mv)
Δp = mvf - mvi
Units = Ft = Ns
Ns = kgm/s

7. Impulse –Momentum Theorem
FΔt = Δp
FΔt = mvf - mvi

8. Examples of impulse.
If you were in an out of control car, would you rather stop by hitting a brick wall or a large hay stack?
The hay stack is the smart answer.
Both will apply the same change in momentum, impulse, but the wall will do it over a shorter time, and so will exert more force. Ft is the same for both.

9. Sample Problem 6B
A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to a rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision.

10. Bouncing
Which would change velocity more, bringing a fast moving object to a stop or sending it back in the opposite direction?
For the same reasons, momentum is changed more if an object bounces.
This means there is also a greater impulse exerted.

11. Example 1
300kg of water rushes down a hill at 15m/s, it hits a dam and is brought to a stop. How much force must the dam exert to stop it in 5s?
Δ(mv) = Ft
(300kg)(15m/s) = F(5s)
4500Ns = F(5s)
900N = F

12. Example 2
If, in the last example, the water was turned and moved back the way it came at 10m/s, what would be the impulse?
What would be the force exerted?
Impulse = Δ(mv)
(300kg)(15m/s) - (300kg)(-10m/s)
4500Ns Ns
7500Ns
Impulse = Ft
7500Ns = F(5s)
1500N = F

13. Sample Problem 6C
A 2250 kg car traveling to the west slows down uniformly from 20.0 m/s to 5.00 m/s. How long does it take the car to decelerate if the force on the car is 8450 N to the east? How far does the car travel during the deceleration?

14. Conservation
If no net force or net impulse acts on a system, the momentum of that system cannot change.
When momentum, or any quantity does not change, we say it is conserved.
Law of Conservation of Momentum
In the absence of an external force, the momentum of a system remains unchanged.

15. Momentum in a System
When a cannon fires a cannon-ball, the cannon exerts a force on the cannon-ball and the cannon-ball exerts an equal and opposite force on the cannon according to Newton’s 3rd Law.
Before the firing, the system is at rest, so the momentum is zero.
After it is fired, the net momentum, or total momentum is still zero.
Look at it with calculations…

16. (m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f
Total initial momentum = total final momentum
p1i + p2i = p1f + p2f
(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f

17. When is Momentum Conserved?
If a system undergoes changes, where all forces are internal, the net momentum before and after the event are the same.
Examples:
Cars colliding
Stars exploding

18. Sample Problem 6D
A 76 kg boater, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the boater moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

19. Homework, Hooray!!!
Practice Problems
6A-D (odds only)

20. And Now…
Collisions !!!!!!!!

21. Collisions
When ever 2 objects collide and there are no external forces the net momentum of both objects before the collision equals the net momentum after the collision.

22. Types of Collisions 3 kinds
Elastic – Objects collide without being permanently deformed and with out generating heat.
Objects bounce perfectly in elastic collisions
Perfectly Inelastic – Colliding objects get tangled and coupled together.
Inelastic – objects deform during collisions, but move separately after collision

23. Perfectly Inelastic Elastic
2 objects collide and move off as one mass.
(m1v1)i + (m2v2)i = (m1+m2)vf
KE is NOT constant in inelastic collisions
Some KE is converted to sound and internal energy
Kinetic energy is conserved in perfectly elastic collisions
pi =pf and
KEi = KEf

24. Sample Problem 6E
A 1850 kg luxury sedan stopped at a traffic light is struck from the rear by a compact car with a mass of 975 kg. The two cars become entangled as a result of the collision. If the compact car was moving at a velocity of 22.0 m/s to the north before the collision, what is the velocity of the entangled mass after the collision?

25. Most collisions involve some external force.
Most external forces are negligible during a collision.
Perfectly elastic collisions are not common in the everyday world.
Heat is usually generated by any collision.
They only occur regularly on the microscopic level

26. Sample Problem 6F
Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of kg and an initial velocity of 4.00 m/s to the right. The mass of the second ball is kg, and it has an initial velocity of 3.00 m/s to the left. What is the final velocity of the composite ball of clay after the collision? What is the decrease in kinetic energy during the collision?

27. Elastic Collisions Remember: or
Kinetic energy is conserved in elastic collisions
(m1v1)i + (m2v2)i = (m1v1)f + (m2v2)f or
( ½ m1v12)i + ( ½ m2v22)i = ( ½ m1v12)f + ( ½ m2v22)f

28. Sample Problem 6G
A kg marble moving to the right at m/s makes an elastic head-on collision with a kg shooter marble moving to the left at m/s. After the collision, the smaller marble moves to the left at m/s. Assume that neither rotates before or after the collision and that both marbles are moving on a frictionless surface. What is the velocity of the kg marble after the collision?