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Do Now:What makes the shuttle go UP?

Objectives:Utilize IMPULSE to calculate:Force – time – change in velocity

Home work:Page 233:#’s 1 – 5 allPage 250-52:#’s 47, 51, 55, 60, 64

Momentum

LESSON ONE

Momentumρ = mv

You must know how to do these actions:1.Calculate time needed to slow or accelerate an object to a certain velocity

using applied Force with known masses. IMPULSE (F)(Δt) = Δρ

2.Find velocities of objects that collide in inelastic collisions

1. (stick together, car crash, clay and balls etc.)

3.Find velocities of objects that collide in elastic collisions (

1. do NOT stick together, billiard balls)

4.Using the conservation of momentum, find velocities of objects using just masses, velocities.

5.Find velocities and directions of objects that collide in two dimensions.

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1. IMPULSE (F)(Δt) = Δρ = mΔv Δm Δv

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Before the collision, the baseball moves toward the bat. During the collision, the baseball is squashed against the bat.

After the collision, however, the baseball moves at a higher velocity away from the bat, and the bat continues in its path, but at a slower velocity.

How are the velocities of the ball, before and after the collision, related to the force acting on it?

Newton’s second law describes how the velocity of an object is changed by a net force acting on it.

The change in velocity of the ball must have been caused by the force exerted by the bat on the ball.

The force changes over time.

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Newton’s second law of motion, F = ma, can be rewritten by using the definition of acceleration as the change in velocity divided by the time needed to make that change.

It can be represented by the following equation:

F = ma = m

vt

Multiplying both sides of the equation by the time interval, t, results in the following equation:

Ft = mv

Impulse, or Ft, is the product of the average force on an object and the time interval over which it acts. Impulse is measured in newton-seconds.

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What does an air bag do from an “IMPULSE” point of view?

IMPULSELESSON ONE SUMMARY

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Using the Impulse-Momentum Theorem to Save Lives

A large change in momentum occurs only when there is a large impulse.

A large impulse can result either from a large force acting over a short period of time or from a smaller force acting over a long period of time.

What happens to the driver when a crash suddenly stops a car?

An impulse is needed to bring the driver’s momentum to zero.

An air bag reduces the force by increasing the time interval during which it acts.

It also exerts the force over a larger area of the person’s body, thereby reducing the likelihood of injuries.

Momentum

LESSON TWOJ

Elastic events

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Do Now:What is the impulse needed to accelerate a 5.00 Kg mass from rest to 10.0 m/s?

Objectives:Utilize ELASTIC to calculate:Force – time – change in velocity

Home work: Due Tuesday Page 233: #’s 1 – 25 allPage 250-52:#’s 47, 51, 55, 60, 6450, 70, 72, 73, 74

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Two-Particle Collisions

Although it would be simple to consider the bat as a single object, the ball, the hand of the player, and the ground on which the player is standing are all objects that interact when the baseball player hits the ball.

A much easier system is the collision of two balls of unequal mass and velocity.

During the collision of the two balls, each briefly exerts a force on the other. Despite the differences in sizes and velocities of the balls, the forces they exert on each other are equal and opposite, according to Newton’s third law of motion.

These forces are represented by the following equation.

FB on A = -FA on B

Momentum

LESSON TWO

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Elastic Collisionstwo masses stay separateKinetic Energy is conserved

MomentumLESSON TWO

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Use the result of Newton’s third law of motion –FA on B = FB on A.

PA2 = -FA on B Δt + PA1

Add the momenta of the two balls.

THIS IS IT!!!!

m1v1 + m2v2 = m1v'1 + m2v

'2

This shows that the sum of the momenta of the balls is the same before and after the collision.

That is, the momentum gained by ball 2 is equal to the momentum lost by ball 1.

If the system is defined as the two balls, the momentum of the system is constant.

MomentumLESSON TWO

What is happening??

Newton’s Cradle: Dominique Toussaint on wikipedia

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1.

2.

3.

4.

5.

6.

7.

List ELASTIC “Events”

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F Δt = Δm Δv

F = Δm Δv / Δt = 3.95 (7.91) / 1.00 = 31.2445 N

Wt = mg = 3.95 * 9.8 = 38.71 N …..

It won’t get off the ground!

LESSON THREEDO NOW: A 4.00 Kg rocket launches from rest. 0.05 Kg of fuel is exhausted @ – 625 m/s for 1.00 sec.

How far up does the rocket travel?

Home work: Due MONDAY Jan 14Page 233: #’s 1 – 5 all WX:3.3 SM:1.0 RX: 0.0 ZH:0.0 RC: 0.0

Page 240: #’s 19 – 21 all WX:3.6 SM:1.0 RX: 0.0 ZH:0.0 RC: 0.0

Page 250-52:#’s 47, 51, 55, 60, 64 WX:3.3 SM:1.0 RX: 0.0 ZH:0.0 RC: 0.0

50, 70, 72, 73, 74 WX:3.3 SM:1.0 RX: 0.0 ZH:0.0 RC: 0.0

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IN-Elastic EventMomentum

LESSON THREEJan 16

Do Now:Two balls collide. The 1st ball has a mass of 3.00 Kg and an initial velocity of 2.00 m/s @ 0.00º The 2nd ball has a mass of 9.00 Kg and an initial velocity of 0.00 m/sAfter the collision:The 1st ball has a velocity of 0.5 m/s @ 0.00º What is the velocity of the second ball if it too moves off @ 0.00º ?

Objectives:Utilize IN-ELASTIC to calculate:Force – time – change in velocityHome work:Page 238: #’s 13 – 18 allPage 243 #22Page 252-53: #’s: 76, 77 (add part “b” the bullet imbeds itself in the block)

Page 252-53: #’s: 86, 90 Page 255: # 2- 5, 7

m1v1 + m2v2 = (m1+ m2)v‘3

In - Elastic

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m1v1 + m2v2 = (m1+ m2)v‘3

#1 Two cars collide…and they stick together. The 1st car has a mass of 1875 Kg and an initial velocity of 23.00 m/s @ 0.00º The 2nd car has a mass of 1025 Kg and an initial velocity of 17.00 m/s @ 0.00º After the collision: What is the velocity of the two cars if they both move off @ 0.00º ?

#2 Two cars collide…and they stick together. The 1st car has a mass of 1875 Kg and an initial velocity of 23.00 m/s @ 0.00º The 2nd car has a mass of 1025 Kg and an initial velocity of 17.00 m/s @ 180.0º After the collision: What is the velocity (speed and direction) of the two cars?

#3 If the cars in #2 took 1.00 seconds to collide and stick together, What was the force on the SECOND Car? What was the acceleration felt by a 60.00 Kg passenger in car number 2?

Momentum

LESSON THREE

QUIZ

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1.

2.

3.

4.

5.

6.

7.

List IN - ELASTIC “Events”

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Momentum

LESSON Four

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Two-Two-Dimensional Dimensional CollisionsCollisions

Up until now, you have looked at momentum in one dimension only. The law of conservation of momentum holds for all closed systems with no external forces. It is valid regardless of the directions of the particles before or after they interact.

Now you will look at momentum in two dimensions.

For example, billiard ball A strikes stationary ball B. Consider the two balls to be the system.

The original momentum of the moving ball is pA1; the momentum of the stationary ball is zero. Therefore, the momentum of the system before the collision is equal to pA1.

Momentum

LESSON Four

Home work Page 242:Example Problem # 4

Page 243:#’s 22– 25 all

Do NowDo Now::What is the initial momentum of ball What is the initial momentum of ball #1 if the mass is 10.4Kg and the #1 if the mass is 10.4Kg and the velocity is + 12.45 m/s?velocity is + 12.45 m/s?

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After Collision

m1

m2

v1'

v2'

After the collision, both balls are moving and have momenta. According to the law of conservation of momentum, the initial momentum equals the vector sum of the final momenta, so pA1 = pA2 + pB2.

The equality of the momentum before and after the collision also means that the sum of the components of the vectors before and after the collision must be equal.

Therefore, the sum of the final

y-components must be zero.

pA2y + pB2y = 0

They are equal in magnitude but have opposite signs. The sum of the horizontal components is also equal.

pA1 = pA2x + pB2x

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DO NOW In Class:

A 2.00-kg ball, A, is moving at a speed of 5.00 m/s. It collides with a stationary ball, B, of the same mass. After the collision, A moves off in a direction 30.0º to the left of its original direction. Ball B moves off in a direction 90.0º to the right of ball A's final direction. How fast are they moving after the collision?

PHET Momentum Billiard BallsSolution: 1. Sketch the before and after states

Before

A BvA1

vB1=0

A

B

After

vA2

vB2

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2. Draw a momentum vector diagram.

Note that pA2 and pB2 form a 90º angle.

Note that ALL the InitialInitial momentum is in the X DIRECTION

90ºpA2

30º

pB2

p2

3. Perform calculations:

Determine InitialInitial momenta: FinalFinal momenta

pIAx + pIAy + pIBx + pIBy = pFAx + pFAy + pFBx + pFBy

pIAx = mavIAX = (2.00 kg)(5.00m/s) = 10.0 kg•m pFAx = mavFAx Cos(30)

pIAy = 0 pFBx = mavFBx Cos(-60)

pIBx = 0 pFAy = mbvFAy SIN (30)

pIBy = 0 pFBy = mbvFBy SIN (-60)

•pA1 pB1

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2. Two Equations Two Variables

3. Solve for vFA in terms of vFB

pIx = 10 = mavFAx Cos(30) + mavFBx Cos(-60)

pIy = 0 = mbvFAy SIN (30) + mbvFBy SIN (-60)

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HOMEWORK TEST Due NOW In Class:

A 5.00-kg ball, A, is moving at a speed of 10.00 m/s. It collides with a stationary ball, B, of the same mass. After the collision, A moves off in a direction 60.0º to the left of its original direction. Ball B moves off in a direction 30.0º to the right of ball A's final direction. How fast are they moving after the collision?

PHET Momentum Billiard BallsSolution: 1. Sketch the before and after states

Before

A BvA1

vB1=0

A

B

After

vA2

vB2

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2. Draw a momentum vector diagram.

Note that pA2 and pB2 form a 90º angle.

3. Perform calculations:

Determine initial momenta:

pA1 = mava1 = (5.00 kg)(10.00m/s) = 50.0 kg•m/s {X-Dir}

pB1 = 0

Find p2:

pA ‘ + pB‘ = 50.0 kg•m/s X direction + 0 Y direction

•pA1 pB1

60 deg

30 deg

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Two Dimensional IN CLASS

“Initial”ρx = M1 (V1)(Cos{Θ }) + M2 (V2)(Cos{Θ })

“Initial”ρy = M1 (V1)(Sin{Θ }) + M2 (V2)(Sin{Θ })

Two Equations two Unknowns

Page 253 # 79

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