chapter 7: momentum
DESCRIPTION
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 PresentationTRANSCRIPT
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?
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
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
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.
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.
Chapter 7.2: Decreasing Momentum
Chapter 7.2: Decreasing Momentum
Chapter 7.2: Decreasing Momentum
Chapter 7.2: Decreasing Momentum
Momentum (mv) is what is gained by theJumper until the chord begins to stretch.
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.
Chapter 7.2: Decreasing Momentum
Yowza! Yowza!
Chapter 7.3: Bouncing
Pelton Wheel
Flower pot on the head example
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.
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
• The sum of the momentum vectors are the same before and after the collisions.
Chapter 7.5: Elastic Collisions
• 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
Chapter 7.5: Inelastic Lunch
5 kg
v = 0 20 kg3 m/s
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)
Chapter 7.5: Inelastic Lunch
20 kg3 m/s
5 kg
2 m/s
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)
Chapter 7.5: Inelastic Lunch
20 kg3 m/s
5 kg
10 m/s
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)
Chapter 7.6: Momentum Vectors
side a oflength 2
Chapter Review