Lecture series for Conceptual Physics 8 th Ed.. Momentum p 82 momentum =mass x velocity p =m x v...
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Transcript of Lecture series for Conceptual Physics 8 th Ed.. Momentum p 82 momentum =mass x velocity p =m x v...
![Page 1: Lecture series for Conceptual Physics 8 th Ed.. Momentum p 82 momentum =mass x velocity p =m x v Impulse p83 impulse =force x time I =F x t Together:](https://reader035.fdocuments.us/reader035/viewer/2022062311/5a4d1b257f8b9ab059996dd6/html5/thumbnails/1.jpg)
Lecture series for Conceptual Physics 8th Ed.
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Momentum p 82
momentum = mass x velocity
p = m x v
Impulse p83
impulse = force x timeI = F x t
Together:
F x t = m x vImpulse = change in momentum
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Case 1 : Increasing Momentum p85
Increasing the force…hitting it harder.
and / orIncreasing the time of contact.
The force on the golf ball builds up to a huge amount then decreases.
We’ll use the average force.
F t
F t
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Case 2: Decreasing Momentum Over a Long Time p85
Long means gentle, gradual, softly.
A large momentum can be reduced by a small force
in a long time.
A large momentum can by reduced by a large force in a short time.
Not so gently.
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m v = F t
Case 3: Decreasing Momentum Over a Short Time
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A flower pot falling on your head is bad enough. But,if the pot bounces, it’ll really hurt!
First, there’s the impulse required to stop the pot.
Then, there’s the impulse in the other direction to throw it back.
When something bounces, it hits twice as hard.
Figure 5-8 the Pelton water wheel.
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Conservation of Momentum p 88
The momentum before a collision
equals the momentum after a collision.
BEFORE:p = m v = m x 0 = 0
AFTER:m (-v) + m v = 0
Momentum is conserved…it is the same before and after.
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Collisions p 90 pbefore = pafter
These are elastic collisions
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More on Collisions
Notice that these box cars stick together.
These are inelastic collisions.
pbefore = pafter
(m x 10 m/s)before = (2m x v)after
vafter = 5 m/s
Before
During
After
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More inelastic collisions p91
Before: m v + m (-v) = 0
After: m 0 + m 0 = 0
Before: m vA + m vB = k
After: v (m + m) = k
Fig 5-12 p91
m v + m (-v) = m 0 + m 0 m vA + m vB = v (m + m)
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1 m/s
m = 5kg
4 m/s
m = 1 kg
Mv + m(-v) = (M+m)v
(5kg)(1m/s) + (1kg)(-4m/s) = (5kg + 1kg)v
v = 1/6 = 0.17 m/s
What’s the velocity of the fish after lunch?
We’ll do the more complicated collisions on the board.
The End