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Physics 7B - ABLecture 5

May 1

Recap on vectorsMomentum Conservation Model

- Elastic/Inelastic collisions- Use of Momentum Chart

- Collision and Impulse- Force diagram

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Quiz 1 average 9.18Quiz 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 recoilBefore shooting (at rest)

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Conservation of Momentum

Example Rifle recoilBefore shooting (at rest)

p i,total = p i,bullet + p i,Rifle = 0

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Conservation of Momentum

Example Rifle recoil

After shooting

Before shooting (at rest)

p i,total = p i,bullet + p i,Rifle = 0

p f,total = p f,bullet + p f,Rifle = 0

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Conservation of Momentum

Example Rifle recoil

After shooting

p f,bullet p f,Rifle

Before shooting (at rest)

p i,total = p i,bullet + p i,Rifle = 0

p f,total = p f,bullet + p f,Rifle = 0

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Conservation of Momentum

Example Rifle recoilAfter shooting

vf,bullet vf,Rifle

|pf,bullet | = |pf,Rifle| mbullet |vbullet| = mRifle |vRifle|

p f,bullet p f,Rifle

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Momentum of the closed system (= Rifle + bullet) is

conserved,i.e., pi, total = pf, total

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What if our system = bullet (only)?

Momentum of the closed system (= Rifle + bullet) is

conserved,i.e., pi, total = pf, total

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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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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

Moementum of the closed system (= Rifle + bullet) is

conserved,i.e., pi, total = pf, total

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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 bullet x ∆ t

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What if our system = bullet (only)?

After shootingvbullet

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 shootingvRifle

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 shootingvRifle

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 collideA 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?

Before collision

vAi

pi,tot = pi,A + pi,B = pi,A

After collision

vBi =0

At restvA+B

f

A B

A+B

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Conservation of Momentum

Railroad cars collideA 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?

Before collision

vAi

pf,tot = pf,A+B

After collision

vBi =0

At restvA+B

f

A B

A+B

pi,tot = pi,A + pi,B = pi,A

pA+Bf

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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 initial final

Definitely inelastic Maybe elastic or inelastic(Need calculation/energy-bubbles to determine)

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Alice

Bob

(Vector magnitudes not to scale, directions shown

accurately)

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.

Stuck togetherafter impactConservation of

MomentumUse of Momentum ChartUse of Momentum Chart

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Alice

Bob

(Vector magnitudes not to scale, directions shown

accurately)

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.

Stuck togetherafter impactConservation of

MomentumUse of Momentum ChartUse of Momentum Chart

Question:Who was going faster at the time of collision?

(a) Alice (b) Bob (c) No way to know

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Solution

1. Write in the directions we know from the problem (lengths are not to scale at this point)

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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

Go back and make sure these arethe same length

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Solution

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Same length!

!!!

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Recap (what just happened?)

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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 of the momentum for Alice and Bob are the same. As mB > mA we know vB < vA.

Initial time: just before collisionFinal time: just after collision (friction negligible)

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Conservation of Momentum

Question: Falling on a sledAn 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 sledAn 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 Impulsehow 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 Impulsehow 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 Impulsehow not to break an ankle

+ =

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Next weekMay8 Quiz4(20min) will cover:

Today’s lecture Activities and FNTs from DLM9 and

Activities from DLM10Bring 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