Transcript Document

Lecture series for Conceptual Physics 8th Ed.
Momentum
p 82
momentum = mass x velocity
p= mxv
Impulse
p83
Together:
impulse = force x time
I= Fxt
Impulse = change in momentum
F x t = m x Dv
Case 1 : Increasing Momentum
p85
Increasing the force…hitting it harder.
and / or
Increasing the time of contact.
F
Ft
t
The force on the golf ball builds up to a huge amount then decreases.
We’ll use the average force.
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.
m Dv = F
t
Case 3:
Decreasing
Momentum Over a
Short Time
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.
Conservation of Momentum
The momentum
before a collision
BEFORE:
p=mv=mx0=0
equals
p 88
the momentum
after a collision.
AFTER:
m (-v) + m v = 0
Momentum is conserved…it is the same before and after.
Collisions
p 90
pbefore = pafter
These are elastic collisions
More on Collisions
Before
Notice that these
box cars stick
together.
During
After
These are inelastic collisions.
pbefore = pafter
(m x 10 m/s)before = (2m x v)after
vafter = 5 m/s
More inelastic collisions p91
Fig 5-12 p91
Before: m v + m (-v) = 0
Before:
After: m 0 + m 0 = 0
After:
m v + m (-v) = m 0 + m 0
m vA + m vB = k
v (m + m) = k
m vA + m vB = v (m + m)
What’s the velocity of the fish after lunch?
m = 1 kg
m = 5kg
1 m/s
4 m/s
Mv + m(-v) = (M+m)v
(5kg)(1m/s) + (1kg)(-4m/s) = (5kg + 1kg)v
v = 1/6 = 0.17 m/s
We’ll do the more complicated collisions on the board.
The End