Transcript Document

Physics 7B - AB
Lecture 5
May 1
Recap on vectors
Momentum Conservation Model
- Elastic/Inelastic collisions
- Use of Momentum Chart
- Collision and Impulse
- Force diagram
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Quiz 1 average 9.18
Quiz 1 Re-evaluation Request Due
May 8 (next Thursday)
Quiz 2 graded and being returned
this week, Solution+Rubrics on the
web site
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To describe the motion of objects, we use
several vector quantities such as…
• Position vector R
e.g. Rinitial, Rfinal
• Displacement vector ∆R = Rfinal – Rinitial
• Velocity vector v = dr/dt
• Acceleration vector a = dv/dt
• Force vector F
How are these vectors related to
each other??
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To describe the motion of objects, we use
several vector quantities such as…
• Position vector R vs Displacement vector ∆R
∆R = Rfinal – Rinitial
• Displacement vector ∆R vs Velocity vector v = dr/dt
• Velocity vector v = dr/dt vs Acceleration vector a = dv/dt
• Acceleration vector a = dv/dt vs Force vector F
How are these vectors related to
each other??
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Conservation of Momentum
Example Rifle recoil
Before shooting (at rest)
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Conservation of Momentum
Example Rifle recoil
Before shooting (at rest)
p i,total = p i,bullet + p i,Rifle = 0
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Conservation of Momentum
Example Rifle recoil
Before shooting (at rest)
p i,total = p i,bullet + p i,Rifle = 0
After shooting
p f,total = p f,bullet + p f,Rifle = 0
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Conservation of Momentum
Example Rifle recoil
Before shooting (at rest)
p i,total = p i,bullet + p i,Rifle = 0
After shooting
p f,bullet
p f,Rifle
p f,total = p f,bullet + p f,Rifle = 0
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Conservation of Momentum
Example Rifle recoil
vf,bullet
After shooting
p f,bullet
vf,Rifle
p f,Rifle
|pf,bullet | = |pf,Rifle|
mbullet |vbullet| = mRifle |vRifle|
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Momentum of the closed system (= Rifle + bullet)
is conserved,i.e., pi, total =
pf, total
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Momentum of the closed system (= Rifle + bullet)
is conserved,i.e., pi, total =
pf, total
What if our system = bullet (only)?
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•
•
When does momentum of something
change??
… when a force F acts on the something
during a time interval
e.g. A bat hits a baseball
change in momentum is called: Impulse
Impulse Is related to the net external force in
the following way:
Net Impulseext = ∆ p = ∫ ∑ Fext(t)dt
Approximate a varying force as an average force acting
during a time interval ∆t
Net Impulseext = ∆ p = ∑ Fave.ext x ∆ t
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Moementum of the closed system (= Rifle +
bullet) is conserved,i.e., pi, total =
pf, total
What if our system = bullet (only)?
During the gun powder explosion
Exploding gun powder/Rifle system exerts force on the bullet.
The bullet exerts force on the Exploding gun powder /Rifle
system. (Newton’s 3rd law = Every action has an equal and
opposite reaction)
FExploding gun powder/Rifle on the bullet = – Fbullet on Exploding gun powder/Rifle
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What if
our system = bullet (only)?
During the gun powder explosion
Net Impulseext on the bullet = ∆ pbullet = ∫ ∑ Fext(t)dt
Approximate a varying force as an average force acting during a
time interval ∆t
Net Impulseext on the bullet = ∆ pBullet = ∑ Fave.ext x ∆ t =
= Fave.Exploding gun powder/Rifle on the bullet14 x ∆ t
What if
our system = bullet (only)?
vbullet
After shooting
pbullet
Pf, bullet = pi, bullet + ∆pbullet
∆ pbullet = ∑ Fave.ext x ∆ t = Fave.Exploding gun powder/Rifle on the bullet x ∆ t
∆pbullet is non zero because the bullet system interacted with the
(external) gun powder/Rifle system
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What if
our system = Rifle (only)?
After shooting
vRifle
pRifle
Net Impulseext on the Rifle = ∆ pRifle = ∫ ∑ Fext(t)dt
Approximate a varying force as an average force acting during
a time interval ∆t
Net Impulseext on the bullet =∆ pRifle = ∑ Fave.ext x ∆ t =
= Fave.bullet on the Exploding gun powder/Rifle x ∆ t
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What if
our system = Rifle (only)?
After shooting
vRifle
pRifle
pf, Rifle = pi, Rifle + ∆pRifle
∆ pRifle = ∑ Fave.ext x ∆ t = Fave.bullet on Exploding gun powder/Rifle x ∆ t
∆pRifle is non zero because the Rifle/Exploding gun powder
system interacted with the (external) bullet system
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Conservation of Momentum
Railroad cars collide
A 10,000kg railroad car A, traveling at a speed of
24m/s strikes an identical car B, at rest. If the car
lock together as a result of the collision, what is
their common speed afterward?
vAi
i =0
v
B
Before
A
B
collision
pi,tot = pi,A + pi,B = pi,A
After
collision
A+B
At rest
vA+Bf
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Conservation of Momentum
Railroad cars collide
A 10,000kg railroad car A, traveling at a speed of
24m/s strikes an identical car B, at rest. If the car
lock together as a result of the collision, what is
their common speed afterward?
vAi
i =0
v
B
Before
A
B
At rest
collision
pi,tot = pi,A + pi,B = pi,A
vA+Bf
After
A+B
collision
f
p
pf,tot = pf,A+B
A+B
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Collisions
Inelastic collisions:
Elastic collisions:
• Momentum conserved regardless
• Total energy conserved regardless
• In inelastic collisions, KE is transferred to other types of
energy
initial
final
Definitely inelastic
initial
final
Maybe elastic or inelastic
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(Need calculation/energy-bubbles to determine)
Conservation of
Momentum
Stuck together
after impact
Alice
Use of Momentum Chart
You are a CSI at the scene of a car
crash. The drivers, Alice and Bob, are
unharmed but each claims the other was
speeding.
Bob
(Vector magnitudes not to scale,
directions shown accurately)
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Stuck together
after impact
Conservation of
Momentum
Alice
Use of Momentum Chart
Looking at the car make, you discover
that Bob’s car has twice the mass of
Alice’s car. As shown, the cars travelled
at roughly a 45 degree angle from the
point of impact. The ground is flat.
Question:
Who was going faster
at the time of collision?
(a) Alice
(b) Bob
Bob
(Vector magnitudes not to scale,
directions shown accurately)
(c) No way to know
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Solution
Go back and make sure these are
the same length
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1. Write in the directions we know from the problem
(lengths are not to scale at this point)
2. We know this is a closed momentum system.
3. Use last row to find direction of initial momentum.
4. This means that |pi, Alice| = |pi, Bob| so that the system
initial momentum is at 45o
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Solution
Same length!
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!!!
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Recap (what just happened?)
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Initial time: just before collision
Final time: just after collision (friction negligible)
The final momentum of the system is pointing
, therefore the initial momentum of
the system must point
. We only get the initial momentum right if the magnitude o
the momentum for Alice and Bob are the same. As mB > mA we know vB < v25A.
Conservation of Momentum
Question: Falling on a sled
An empty sled is sliding on a frictionless
ice when Dan drops vertically from a tree
above onto the sled. When he lands, the
sled will;
(a) Speed up
(b) Slow down
(c) Keep the same speed
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Conservation of Momentum
Question: Falling on a sled
An empty sled is sliding on a frictionless
ice when Dan drops vertically from a tree
above onto the sled. When he lands, the
sled will;
(a) Speed up
(b) Slow down
(c) Keep the same speed
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Collisions and Impulse
how not to break a leg
Question: Why is it a good idea to
bend your knee when landing after
jumping from some height?
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Collisions and Impulse
how not to break a leg
Question: Why is it a good idea to
bend your knee when landing after
jumping from some height?
Hint:
Net Impulseext = ∆ p
= ∑ Fave.ext x ∆ t
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Collisions and Impulse
how not to break an ankle
+
=
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Next week
May8 Quiz4(20min) will cover:
Today’s lecture
Activities and FNTs from DLM9 and Activities
from DLM10
Bring Calculator!
Closed-book, formulas will be provided.
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Be sure to write your name, ID number & DL section!!!!!
1
MR 10:30-12:50
Dan Phillips
2
TR 2:10-4:30
Abby Shockley
3
TR 4:40-7:00
John Mahoney
4
TR 7:10-9:30
Ryan James
5
TF 8:00-10:20
Ryan James
6
TF 10:30-12:50
John Mahoney
7
W 10:30-12:50
Brandon Bozek
7
F 2:10-4:30
Brandon Bozek
8
MW 8:00-10:20
Brandon Bozek
9
MW 2:10-4:30
Chris Miller
10 MW 4:40-7:00
Marshall Van Zijll
11 MW 7:10-9:30
Marshall Van Zijll
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