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Chapter 7: Momentum

Chapter 7: Momentum

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Chapter 7: Momentum. Chapter 7.1: Momentum—Inertia in Motion. 22 km/h. 5225 kg. 75 km/h. 1540 kg. Momentum = mass x velocity or momentum = mv. Which has greater momentum?. Chapter 7.2: Impulse Changes Momentum. Egg drop challenge. - PowerPoint PPT Presentation

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Page 1: Chapter 7: Momentum

Chapter 7: Momentum

Page 2: Chapter 7: Momentum

Chapter 7.1: Momentum—Inertia in Motion

• Momentum = mass x velocity or momentum = mv

1540 kg

5225 kg

75 km/h

22 km/h Which has greater momentum?

Page 3: Chapter 7: Momentum

Chapter 7.2: Impulse Changes Momentum

The greater the force acting on an object, the greater its velocity and momentum.

Time is also a factor in determining momentum. How long is force applied?

Impulse = force x time interval or impulse = Ft

Impulse = change in momentum

)(mvFt

Egg drop challenge

Page 4: Chapter 7: Momentum

Chapter 7.2: Increasing Momentum

To increase momentum, force must be applied as hard as possible for as long as possible.

This is why follow through is important in a golf swing or in baseball.

Impact force hard and fast = maximum impulse

Impact measured in N. Impulse measured in N-s

Page 5: Chapter 7: Momentum

Chapter 7.2: Decreasing Momentum

To decrease the force of impact, the impact time (time during which momentum is brought to zero) must be lengthened.

Page 6: Chapter 7: Momentum

Chapter 7.2: Decreasing Momentum

To decrease the force of impact, the impact time (time during which momentum is brought to zero) must be lengthened.

Page 7: Chapter 7: Momentum

Chapter 7.2: Decreasing Momentum

Page 8: Chapter 7: Momentum

Chapter 7.2: Decreasing Momentum

Page 9: Chapter 7: Momentum

Chapter 7.2: Decreasing Momentum

Page 10: Chapter 7: Momentum

Chapter 7.2: Decreasing Momentum

Momentum (mv) is what is gained by theJumper until the chord begins to stretch.

Page 11: Chapter 7: Momentum

Chapter 7.2: Decreasing Momentum

Ft is the impulse the cord supplies to reduce the momentum to zero.

Because of the long time it takes the cord tostretch, the average force (F) on the jumperis minimal.

Momentum (mv) is what is gained by theJumper until the chord begins to stretch.

Page 12: Chapter 7: Momentum

Chapter 7.2: Decreasing Momentum

Yowza! Yowza!

Page 13: Chapter 7: Momentum

Chapter 7.3: Bouncing

Pelton Wheel

Flower pot on the head example

Page 14: Chapter 7: Momentum

Chapter 7.4: Conservation of Momentum

• Law of conservation of momentum: In the absence of an external force, the momentum of a system

remains unchanged.

The bullet gains momentum and so doesthe rifle, but the rifle-bullet system gains none.

See skateboarder example pg 94.

Page 15: Chapter 7: Momentum

Chapter 7.5: Collisions

• When objects collide in the absence of external forces, the net momentum of both objects before the collision equals the net momentum of both objects after the collision.

net momentum before collision = net momentum after collision

Elastic collision—when objects collide without being deformed permanently or generating heat.

Inelastic collision—when colliding objects become tangled or coupled together

Page 16: Chapter 7: Momentum

• The sum of the momentum vectors are the same before and after the collisions.

Chapter 7.5: Elastic Collisions

Page 17: Chapter 7: Momentum

• If mass is equal, the momentum is shared equally after the collision by each of the objects.

Chapter 7.5: Inelastic Collisions

V = 12 m/s

V = 6 m/s

V = 0

Page 18: Chapter 7: Momentum

Chapter 7.5: Inelastic Lunch

5 kg

v = 0 20 kg3 m/s

Page 19: Chapter 7: Momentum

Chapter 7.5: Inelastic Lunch

25 kg

5 kg

v = 0

2.6 m/s

(net mv) before = (net mv) after

(20 kg)(3 m/s)+(5 kg)(0 m/s) = (25 kg)(v after)

(60 kg*m/s)+(5 kg) = (25 kg)(v after)

(65 kg*m/s) = (25 kg)(v after)

2.6 m/s = (v after)

Page 20: Chapter 7: Momentum

Chapter 7.5: Inelastic Lunch

20 kg3 m/s

5 kg

2 m/s

Page 21: Chapter 7: Momentum

Chapter 7.5: Inelastic Lunch

25 kg

5 kg

v = 0

2 m/s

(net mv) before = (net mv) after

(20 kg)(3 m/s)+(5 kg)(-2 m/s) = (25 kg)(v after)

(60 kg*m/s)+(-10 kg*m/s) = (25 kg)(v after)

(50 kg*m/s) = (25 kg)(v after)

2 m/s = (v after)

Page 22: Chapter 7: Momentum

Chapter 7.5: Inelastic Lunch

20 kg3 m/s

5 kg

10 m/s

Page 23: Chapter 7: Momentum

Chapter 7.5: Inelastic Lunch

25 kg

5 kg

v = 0

1.6 m/s

(net mv) before = (net mv) after

(20 kg)(3 m/s)+(5 kg)(-20 m/s) = (25 kg)(v after)

(60 kg*m/s)+(-100 kg*m/s) = (25 kg)(v after)

(-40 kg*m/s) = (25 kg)(v after)

-1.6 m/s = (v after)

Page 24: Chapter 7: Momentum

Chapter 7.6: Momentum Vectors

side a oflength 2

Page 25: Chapter 7: Momentum

Chapter Review