Momentum!!!

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. 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 = mv p = (.01kg)(5m/s) p = .05 kgm/s When velocity = 0 Can we do some examples?

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. 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 Example 1

12. 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 - -3000Ns 7500Ns Impulse = Ft 7500Ns = F(5s) 1500N = F Example 2

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. 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. 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 Perfectly Inelastic Elastic

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 0.500 kg and an initial velocity of 4.00 m/s to the right. The mass of the second ball is 0.250 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: • 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 0.015 kg marble moving to the right at 0.225 m/s makes an elastic head-on collision with a 0.030 kg shooter marble moving to the left at 0.180 m/s. After the collision, the smaller marble moves to the left at 0.315 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 0.030 kg marble after the collision?