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7.1 Momentum

- Momentum- inertia in motion
- Inertia- reluctance to change motion (velocity)
- =mass (kg) x velocity (m/s)

- p=mv (kg m/s)
- Vector quantity (direction matters!)

- Rev Q. 1-2

7.2 Impulse Changes Momentum

- The impulse on an object is the average net force exerted on the object multiplied by the time interval over which the force acts.
- Impulse = FDt

- The impulse on an object is also equal to the change in momentum of the object.
- FDt = pf -pi
- F = (pf -pi)/Dt

Changing p

- Case 1: Increasing Momentum
- For the greatest velocity increase, apply a large force for a long time
- Ex: Slam your car’s accelerator for as long as possible
- BIG F x BIG Dt = BIG impulse

- Case 2: Decreasing Momentum
- For the greatest (safest) velocity decrease, apply a negative force for a long time
- Ex: Gently press your car’s break for as long as possible
- negative F x BIG Dt = BIG impulse

Example Problem: Change in Mass

- Can a sports car ever have the same momentum as a vehicle three times its size?
- YES, if the velocity of the larger vehicle is 1/3 of the smaller car.
- m1v1 = (3m1)(1/3v1)

Example Problem: Average Force

- A 2200kg vehicle traveling at 26m/s can be stopped in 21s by gently applying the brakes. It can be stopped in 3.8s if the driver slams on the brakes. It can be stopped in 0.22s if it hits a brick wall. What is the average force over these three time intervals?

Average Force over 22s

- For 21s, m=2200kg, Dt=21s, Dv=26m/s
- Plugging this all into our new force equation
- F = (mvf -mvi)/Dt

- F = ((2200kg)(0m/s) –(2200kg)(26m/s))/21s
- F = -2.7x103N

Average Force over 3.8s

- For 3.8s, replace only the change in time!
- F = ((2200kg)(0m/s) –(2200kg)(26m/s))/3.8s
- F = -1.5x104N

Average Force over 0.22s

- For 0.22s, replace only the change in time!
- F = ((2200kg)(0m/s) –(2200kg)(26m/s))/0.22s
- F = -2.6x105N
- Rev Q. 3-10
- Con Dev 7-1 1-5

7.3 Bouncing

- Bouncing involves a reversal of direction, which increases momentum change
- Because impulse is a change in momentum and momentum involves v, the reversal in direction plays a big part in increasing momentum!
- Physics Physlets P8.2
- Rev Q. 11-13

7.4 Conservation of Momentum

- According to Newton’s Third Law of motion and the law of conservation of momentum, the forces exerted by colliding objects on each other are equal in magnitude and opposite in direction.
- In terms of momentum, Newton’s Second Law can be written as ΣF = Dp/Dt
- That it, the rate of change of momentum equals the net applied force.
- Momentum is conserved in a closed, isolated system.
- Spbefore = Spafter

Conservation of p

- The law of conservation of momentum can be used to explain the propulsion of rockets.
- Vector analysis is used to solve momentum-conservation problems in two dimensions.
- Physics PhysletsI8.4
- Rev Q. 14-15
- Con Dev 7-1 q.6

7.5 Collisions (Elastic)

- Objects collide without deforming (absorbing force) or generating heat
- Two poolballs bouncing off of each other
- p1i + p2i = p1f + p2f
- m1v1i + m2v2i = m1v1f+ m2v2f
- Physics Phylets P8.6-7

Collisions (Inelastic)

- Objects stick together
- Two train cars coupling together
- pi1 + pi2 = pf
- m1v1i + m2v2i = (m1+m2)vf
- Physics Physlets E8.3-4 P8.5, P8.10
- Rev Q. 16-17

7.6 Momentum Vectors

- Using the Tail-to-Tip method, we can add momentum vectors just as we did velocity, acceleration, and forces
- Momentum is still conserved
- Firecracker falling, then exploding
- Two cars colliding in an intersection (one traveling N, the other traveling E)
- New path is NE

- Rev Q. 18

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