Angel Tree Project

• We will adopt needy children for Christmas. All I am asking from you is that you RAISE a minimum of $20. All money needs to be collected by Wednesday December 4.

• If you make me happy, I will return the favor.

• Class that raises the most per student will earn an extra bonus!

Momentum

The secret of collisions & explosions

Who Pushes Who

• Rin Tin Tin and the Refrigerator meet at the 50 yard line

Mass 20 Kg

Speed 17 m/s

Mass 160 Kg

Speed 2 m/s

Who pushes who over the 50 yard line?

Momentum = Mass x Velocity

Whoever has the most momentum = mv wins

Mass 20 Kg

Speed 17 m/s

Mass 160 Kg

Speed 2 m/s

Momentum = 340 Kg m/s Momentum = 320 Kg m/s

Momentum = mv

• Mass times velocity!

• Velocity is a vector but here we can usually think of it as speed

• Example: A speeding car has a mass of 1000 Kg and a speed of 20 m/s. What is its momentum?

• mv = 1000kg x 20 m/s = 20,000 kg m/s

Unit of Momentum

Kg m/s

Jump to conservation

Momentum is a Vector

• p = mv• Force is required to change the momentum of

an object. Newton stated his 2nd law:

F = p/t The rate of change of momentum of a body

equals the net force exerted on it Equivalent to F = ma

Proof of equivalence of two forms of Newton’s 2nd Law

F = p/t = (mv –mv0)/t =

m(v - v0)/ t = mv/t = maQ: Which form of the law is more general? (which includes the possibility that the mass could change?)

Example

• Water leaves a hose at a rate of 1.5 kg/s with speed 20 m/s and hits a car without splashing back. What force is exerted by the water on the car?

F = p/t = (p final - pinitial )/ t = (0 – 30kg m/s)/1sec = -30N

Momentum is Conserved

• The total momentum of an isolated system of bodies remains the same

• “isolated” means net external force is zero• Momentum before = momentum after• m1v1 + m2v2 = m1v1’ + m2v2’• Applies to all interactions, especially

collisions, explosion-like events, and “dumpings”

• Closely related to Newton’s 3rd law

Law of Conservation of Momentum

• Momentum before = momentum after

• m1v1 + m2v2 = m1v1’ + m2v2’

Apostrophe thingee is pronounced “prime”

Impulse

• Impulse = F t = p

• Impulse is product of force and time during which force acts

• Impulse equals change of momentum

• F is usually non uniform and time interval is usually short

Momentum Lab• Set target car at the 50 cm mark, with photogate A at the 40 cm mark and

photogate B at the 60 cm mark.• Place moving car with rubber band onto track with launcher. Pull back as

far as you can and release moving car, record time A and time B on chart.• Calculate the speed of moving car before collision and speed of target car

after collision.• Redo trials with 6 different marble combinations. Fill in chart.• Calulate ∆p to show how how momentum was conserved in each of the

collisions.

Trial Time A Time B Sp A Sp B Mass A Mass B Pa Pb ∆ p

1

2

3

4

5

6

7

Two Kinds of Collision

• Inelastic - sticking Examples: glue balls fly into each other,air track

gliders with clay

• Elastic – bouncing• Examples: hard balls or protons collide

Courtesy St. Mary College Physics Department

Courtesy Deer Vally HS Espace Academy

Dumping Example

• A 10 kg cart rolls on a frictionless track at speed v. Suddenly 10 kg of rocks are dropped straight down into the cart. What happens to its speed? How come?

Answer v/2 ; momentum mv is conserved, m is doubled so v must be halved.

Courtesy Easyhaul Cart Inc.

Explain This

• A rock falls to earth. Is momentum conserved? Include earth in your explanation

• Does the earth really come up to meet the rock?

Railroad Cars CollideInelastically(stick)

• A 10,000 kg railcar moving 2.4 m/s hits and sticks with an identical car at rest. What is the final speed of the two cars?

m1v1 + m2v2 =

(10,000 kg) (2.4 m/s) + (10,000 kg) (0m/s) =

24,000 kg m/s + 0 kg m/s = (m1 + m2)v’

so v’ = 24,000 kg m/s / 10,000 kg + 10,000 kg

v’ = 24,000 kg m/s /20,000 kg = 1.2 m/s

Elastic Example• A 3 kg glider moves to the right at 4.0 m/s and

collides elastically with a 1 kg glider moving to the left at 1.0 m/s. If the first glider moves back toward the left at 1.0 m/s, how fast does the second glider move to the right?

Recoil of a pistol (an explosion-like event)• What is the recoil velocity of a 1 kg pistol that

shoots a .02 kg bullet at 400 m/s?

Initial momentum = 0

& mBvB + mpvp =

(0.02kg x 400 m/s) + 1kg x vp =

vp = - 8 m/s Q: Does the shooter recoil too?

Skip Think and Solve # 63 & # 65

New example• A 3 kg glider moves to the right at 4.0 m/s and

collides inelastically with a 1 kg glider at rest. What is the final speed of the two joined together gliders?

m1v1 + m2v2 = (m1 + m2) vf

12 kg m/s + 0 kg m/s = (4kg) (vf)

vf = 3 m/s

Next example• A 3 kg glider moves to the right at 4.0 m/s and

collides inelastically with a 1 kg glider moving to the right at 1.0 m/s. What is the final speed of the two joined together gliders?

m1v1 + m2v2 = (m1 + m2) vf

12 kg m/s + 1kg m/s = 13 kg m/s = (4.0 kg) (vf)

vf = 3.25 m/s

Next example• A 3 kg glider moves to the right at 4.0 m/s and

collides inelastically with a 1 kg glider moving to the right at 2.0 m/s. What is the final speed of the two joined together gliders?

m1v1 + m2v2 = (m1 + m2) vf

12 kg m/s + 1 kg x 2 m/s = 14 kg m/s = 4.0 kg vf

vf = 3.50 m/s

Next example

• A 3 kg glider moves to the right at 4.0 m/s and collides inelastically with a 1 kg glider moving to the left at 5.0 m/s. What is the final speed of the two joined together gliders?

m1v1 + m2v2 = (m1 + m2) vf

12 kg m/s - 1kg x 5 m/s = 7 kg m/s = 4.0 kg vf

vf = 1.75 m/s

Next example

• A 3 kg glider moves to the right at 4.0 m/s and collides inelastically with a 1 kg glider moving to the left at 20.0 m/s. What is the final speed of the two joined together gliders?

m1v1 + m2v2 = (m1 + m2) vf

12 kg m/s - 1kg x 20 m/s = -8 kg m/s = 4.0 kg vf

vf = -2.00 m/s

Sled Collision

• Kids on a sled, total mass 100kg move to the right at 4.0 m/s. They collide inelastically with other kids, mass 150 kg, moving to the left at 2.5 m/s. Find the final velocity of the two sleds.

solution

m1v1 + m2v2 = (m1+

m2)v’

100Kg x 4m/s –150Kg x 2.5m/s =

400 Kg m/s – 375 Kg m/s =

25 Kg m/s = 250 Kg x Vf

Vf = 0.1 m/s

Sled Collision in reverse• Kids on a sled, total mass 100kg move

to the right at unknown speed v. They collide inelastically with other kids, mass 150 kg, moving to the left at 2.5 m/s. The final velocity of the two sleds is 0.2 m/s. Find v

Explosion like Event

• If a stationary student on a skateboard throws a rock with momentum 10 kg m/s, what momentum will the student get?

Answer; -10 kg m/s (Newton’s 3rd Law)

Think and Solve63) m1v1 + m2v2 = m1v1’ + m2v2’

Let mass of flat car = m Then mass of diesel engine = 4m

v1 = 5 km/h v2 = 0

4 m X 5 km/h + 0 = 5 m X Vf

20 m = 5m X Vf

Vf = 4 km/h

65) A) 5 kg X 1 m/s + 0 = 6 kg X Vf

Vf = 5/6 m/s

B) 5 kg X 1 m/s – 1 kg X 4 m/s = 6 kg X Vf

5 kg m/s – 4 kg m/s = 6 kg X Vf

1 kg m/s = 6 kg Vf

Vf = 1/6 m/s

Inelastic Collision Lab• To solve for momentum of the system: m1v1 + m2v2= (m1+m2)v

• Car 2 is at rest so to solve for v = m1v1/ (m1+m2+mputty)

• Use the rubber band launcher and release the car down the ramp so it will go at least 60 cm. Place a photogate there, calculate the velocity of the car at that spot, then using the mass of .060 kg for the car, calculate its momentum.

• Then, calculate (theoretically) the velocity an inelastic collision between the original car and another car at the 50 cm mark will create. Use the following equation to solve for v. v= .06(v1)/.13

• Do the lab as instructed above and calculate the velocity of the inelastic collision by placing putty on the second car and placing it on the 50 cm mark, do all the necessary calculations.

• Compare your actual v with your predicted v.

Trial Mass Time A Speed A Momentum

One car, no collision .060 kg

Two cars + putty, inelastic collision (.06+.06+ .01)

.130 kg

“Impulse” Why should you

• Bend your knees when you land?

• Pull back when the baseball enters

your mitt?

• Follow through when you swing?

• Not walk into a punch?

(like Mike Tyson did)

You Predict

• Two gliders of equal mass collide elastically. The first is moving with speed v. The second is at rest. What happens?

First one stops, second moves off at speed v

But why? Find the answer yourself and get extra credit

Energy Conservation in Elastic and Inelastic Collisions• Elastic – kinetic energy is conserved as

well as momentum and total energy• Inelastic – kinetic energy is not

conserved – some energy turns into heat

• Elastic – bounce• Completely inelastic - stick

Ballistic Pendulum

• A bullet of mass m is fired into a block of wood of mass M suspended from a string. The bullet remains in the block which rises a height h. What was the speed of the bullet? Show that

• v = (2gh)1/2(m + M)/m ghor 2/

h

Collisions in Two Dimensions

• Remember momentum p is a vector

• x and y components are conserved separately

• What is the total vertical momentum?

1

2