1/3/11 Welcome Back!!!! Today: Review concept of work and power through lab activity.

1/3/11 Welcome Back!!!!

Today: Review concept of work and power through lab activity.

1/4/11

Today: Review concept of work, energy, and momentum through lab activity.

If you did not turn in power lab, place it in blue sorter now!

Those of you who missed the test on momentum can take it in class today or tomorrow.

1/4/11

Today: Finish review of concept of work, energy, and momentum through lab activity.

When completed, place it in blue sorter.

Those of you who missed the test on momentum can take it in class today or tomorrow.

Momentum and Collisions

http://www.physicsclassroom.com/Class/momentum/momtoc.html

Think about this:

Which does more damage in striking a tree, an F-150 or a Mini-Cooper?

Is this always true?What other information do you need

to determine your response?What term do you think describes

this?

Define Inertia

The property of any body to resist changes in its state of motion.

The measure of Inertia is:MassWhich of Newton’s Laws is this

associated with?First Law

Momentum is inertia in motion

The linear momentum of an object of mass m moving with velocity v is defined as the product of the mass and the velocity. Momentum is represented by the symbol p.

Momentum p = mv

Where p is momentum in kgm/s m is mass in kg v is velocity in m/s

Is momentum scalar or vector? Vector (direction matching that of the

velocity) SI units are kilogram x meters per second

(kgm/s)

p = mv

Determine the momentum of a 60-kg halfback moving eastward at 9 m/s.

540 kg m/s

Example 1

Determine the momentum of a F 150 truck moving northward at 45 mph. Assume a weight 5000 pounds.

Remember the SI unit for momentum in kg * m/s

1 mile = 1.6 km1 kg = 2.2 pounds

45400 kg * m/s

Example 2

How fast (in mph) would a Mini Cooper (2500 pounds) need to be traveling to have the same momentum as the truck in example 1? 1 mile = 1.6 km

1 kg = 2.2 pounds90 mph you could do this in your head, just look at respective weights!

Example 3

Calculate the momentum of the Titanic, of mass 4.2 x 107 kg, moving at 14 knots (1 knot = 1.852 km/h).

3.02 x 108 kg * m/s

12/13 Almost done…..

Goal: Discuss Impulse Momentum Theory and Conservation of Momentum

Tomorrow you will be working on problem solving

Wednesday we will consider special situations

Thursday we will have a test.

How can you change momentum of an object? Change the velocity. What term describes a change in velocity? Acceleration How do you change the velocity, ie cause

acceleration? Apply a Net force.

As force increases, what happens to momentum?

It increases. Will the change be instantaneous? No. It takes time. As time increases, what happens to

momentum? It increases.

The quantity of force applied during a time interval is called Impulse

• Impulse = F∆t• As impulse increases what happens to the

change in momentum?• It increases• What happens to change in momentum if

the impulse decreases?• It decreases

Impulse-Momentum Theorem

The expression is called the impulse-momentum theorem.

Impulse = Change in Momentum F∆t = m∆v = ∆p = mvf – mvi What part of the equation describes the impulse? FΔt Is impulse a scalar or vector quantity? Vector quantity in the direction of the force. What units is Impulse is measured in? Ns

ptF Δ=Δ

In other words

A net force, F, applied to an object for a certain time interval, t, will cause a change in the object’s momentum equal to the product of the force and time interval.

In simple terms, a small force acting for a long time can produce the same change in momentum as a large force acting for a short time.

Increasing Momentum

Teeing Off a golf ball or Swinging at a baseball

How would you increase the momentum?

Apply the greatest force for as long as possible

In other words: FOLLOW THROUGH!!!

Impulse = F∆t

Calculate the impulse when an average force of 10N is exerted upon a cart for 2.5 seconds.

25N*s

Answer example 4 and 5

Example 4A hockey puck has a

mass of 0.115 kg and is at rest. A hockey player makes a shot, exerting a constant force of 30.0 N on the puck for 0.16 s. With what speed does it head toward the goal?

Example 4

F∆t = m∆v Solve for ∆v (vf – vi) ∆v = F∆t/m ∆v = (30 N)(0.16sec)/0.0115kg ∆v = 41.7 m/s How would the velocity of the puck

change if the hockey player didn’t follow through with the shot?

Velocity would be less giving goalie more time to react and block the shot .

Example 5

If the momentum of the NASA space shuttle as it leaves the atmosphere is 3.75 x 108 kg m/s, and its mass is 75,000 kg, what is its speed?

(3.75 x 108 kg m/s) / 75,000 kg5000 m/s

Decreasing “Momentum”

You lose control of your car. You can hit a wall or a haystack. Which do you choose? Why?

Your momentum will be decreased by same impulse with either choice since momentum = impulse.

So what is different? Remember impulse is Force x time Hitting the haystack increases the time,

thus decreasing the Force

Effect of Collision Time Upon the Force….or why a boxer “rides” the punch

Example 6

A 2200-kg sport utility vehicle traveling at 26 m/s can be stopped in 21 s by gently applying the brakes, in 5.5 s in a panic stop, or in 0.22 s if it hits a concrete wall. What is the momentum in all of these situations? What average force is exerted on the SUV in each of these stops?

Example 6 2200 kg at 26m/s F∆t = m∆v

Determine momentum mv = (2200kg)(0m/s - 26m/s) = -57200 kg*m/s Why is momentum negative? We are stopping. Set momentum equal to impulse and solve for

Force At 21sec: F(21s) = -57200 kg*m/s F = (-57200 kg*m/s)/21s = -2720 N

Example 6

• At 5.5sec: F(5.5s) = -57200 kg*m/s• F = (-57200 kg*m/s)/5.5s = -10400 N

• At .22sec: F(.22) = -57200 kg*m/• F = (-57200 kg*m/s)/.22s = -260000 N

Think…..

When a dish falls, will the impulse be less if it lands on a carpet than if it lands on a hard floor?

No. The impulse will be the same for either surface because the same momentum change occurs for each. Force is less on carpet because of greater time for momentum change.

Momentum and Multiple ObjectsMomentum and Multiple Objects

In the last section, we looked at the momentum of objects one at a time. This section we will observe more than one object, and how they interact.

Take for example a game of billiards…

What happens…

…when one billiard ball is knocked into another?

Several Outcomes

…but all have something in common.

Either the cue ball stops completely or slows down, and,

The ball it hits goes from being still to moving at some speed.

If…

If the two balls were BOTH moving, there would be even more possible outcomes, such as…

The cue ball would bounce back while the other ball went forward,

The cue ball knocks the other forward, and it keeps going forwards, too,

The cue ball and the other ball roll backwards.

AND, what if…

If you imagine playing billiards with different sized balls, you get even more possible outcomes!

What if the ball you were hitting with the cue ball was a bowling ball? What would happen?

Conservation of Momentum

In all of those examples, momentum is conserved.

Conservation of momentum means that in any collision, the total amount of momentum remains constant.

In other words, when two objects collide, the momentum of object A plus the momentum of object B initially is equal to the momentum of object A plus the momentum of object B at the end.

Most Generally:

The total momentum of all objects interacting with one another remains constant regardless of the nature of forces between the objects.

How to calculate?

Compare the total momentum of two objects before and after they interact.

The momentum of each object changes before and after an interaction, but the total momentum of the two objects together remains constant.

Law of Conservation of Momentum

The law of conservation of momentum states: The momentum of any closed, isolated system does not change.

During any event, such as an explosion or a collision, individual parts of the system of objects involved may experience changes in momentum.

However, the total momentum of the system before the event must equal the total momentum of the system after the event.

To solve conservation of momentum problems, use the formula:

The sum of the momenta before the collision equals the sum of the momenta after the collision.

afterbefore pp

p1 + p2 = p’1 + p’2

'22

'112211 vmvmvmvm

p = momentum before collision (kg m/s)

p’ = momentum after collision (kg m/s)

m1 = mass of object 1 (kg)

v1 = velocity of object 1 before the collision (m/s)

m2 = mass of object 2 (kg)

v2 = velocity of object 2 before the collision (m/s)

v1’ = velocity of object 1 after the collision (m/s)

v2’ = velocity of object 2 after the collision (m/s)

p1 + p2 = p’1 + p’2

Can be further extended to :

A Two ice skaters—one guy and one girl—are standing, facing each other on a frozen pond. They push off of each other, and the guy, who has a mass of 64.2kg, travels backwards at a velocity of 5.6m/s. If the girl has a mass of 49.3 kg, what will be her velocity in the opposite direction?

Insert one of these to solve for mass

Example 7

A 5.0 kg bowling ball with a velocity of 6.0 m/s strikes a 1.5 kg standing pin squarely. If the ball continues on at a velocity of 3.0 m/s what will be the velocity of the pin after the collision?

Example 8 REWRITE ANSWER IN long equation set up

Before the collision: (5kg)(6m/s) = 30 kg*m/s (1.5kg)(0m/s) = 0 kg*m/s Total p = 30 kg*m/s + 0 kg*m/s = 30 kg*m/s

After the collision: Total p = 30 kg*m/s (From before collision) (5kg)(3m/s) + (1.5kg)(? m/s) = 30 kg*m/s (? m/s) = 30 kg*m/s – [(5kg)(3m/s)]/ (1.5kg) 10 m/s

Example 9

A 5 kg bowling ball is rolling in the gutter towards the pins at 2.4 m/s. A second bowling ball with a mass of 6 kg is thrown in the gutter and rolls at 4.6 m/s. It eventually hits the smaller ball and the 6 kg ball slows to 4.1 m/s. What is the resulting velocity of the 5 kg ball?

Example 9

Before the collision: (5kg)(2.4m/s) = 12 kg*m/s (6kg)(4.6m/s) = 27.6 kg*m/s Total p = 12 kg*m/s + 27.6 kg*m/s = 39.6 kg*m/s

After the collision: Total p = 39.6 kg*m/s (From before collision) (6kg)(4.1m/s) + (5kg)(? m/s) = 39.6 kg*m/s (? m/s) = 39.6 kg*m/s – [(6kg)(4.1m/s)]/(5kg) 3m/s

Example 10

Two people are practicing curling. The red stone is sliding on the ice towards the west at 5.0 m/s and has a mass of 17.0 kg. The blue stone has a mass of 20.0 kg and is stationary. After the collision, the red stone moves east at 1.25 m/s. Calculate the momentum and velocity of the blue stone after the collision.

Example 10

Before the collision: (17kg)(-5m/s) = -85.0 kg*m/s (20kg)(0m/s) = 0 kg*m/s Total p = kg*m/s + 0 kg*m/s = -85.0 kg*m/s

After the collision: Total p = - 85.0 kg*m/s (From before collision) (17kg)(1.25m/s) + (20kg)(? m/s) = - 85kg*m/s (? m/s) = -85.0 kg*m/s – [(17kg)(1.25m/s)]/ (20kg) -5.31m/s blue stone p = -106 kg*m/s

Sample Problem 9

4 m/s

The Astronaut’s Game

Momentum Conserved…

Momentum is conserved in collisions, like the billiard balls we talked about.

Momentum is also conserved in two objects pushing away from each other.

What happens if two ice skaters facing each other push?

The Ice Skaters

They begin at rest (total momentum of zero), and they end going in opposite directions (the total momentum still adds to zero).

12/15

Please pick up your work from yesterday

Today’s Goal: Discuss different types of conservation of momentum problems.

A billiard ball approaches a cushioned edge of a billiard table with momentum, p. After the collision with the cushion, it bounces straight back with the same amount of momentum in the opposite direction. What is the impulse on the ball?

= -p –p = -2p

Sample Problem A

A 76kg man is standing at rest in a 45kg boat. When he gets out of the boat, he steps out with a velocity of 2.5m/s to the right (onto the dock). What is the velocity of the boat?

m1v1 + m2v2 = m1v1’ + m2v2’

[m1v1 + m2v2 - m1v1’]/m2 = v2’

0 + 0 – (76 kg x 2.5 m/s)/45 kg = -4.22 m/s

Sample Problem B

A 63.0kg astronaut is on a spacewalk when his tether to the shuttle breaks. He is able to throw a 10.0kg oxygen tank away from the shuttle with a velocity of 12.0m/s. Assuming he started from rest, what is his velocity?

m1v1 + m2v2 = m1v1’ + m2v2’

[m1v1 + m2v2 – m2v2’]/m1 = v1’

0 + 0 – (10 kg x 12 m/s)/63 kg = -1.90 m/s

Sample Problem C

A 6.50kg bowling ball is moving at a velocity of 3.5m/s when it hits a 0.30kg billiard ball going 1.3m/s in the opposite direction. If the bowling ball stops moving after the collision, what is the velocity of the billiard ball?

m1v1 + m2v2 = m1v1’ + m2v2’

[m1v1 + m2v2 - m1v1’]/m2 = v2’[(6.5 kg x 3.5 m/s) + (.3kg x -1.3m/s) -0]/.3 kg = 74.5 m/s

Sample Problem D

A 65-kg ice skater traveling at 6.0 m/s runs head on into an 85-kg skater traveling straight forward at 4.5 m/s. At what speed and in what direction are the ice skaters traveling if they move joined together after the collision?

m1v1 + m2v2 = (m1 + m2)v3

(m1v1 + m2v2 )/(m1 + m2) = v3

Sample Problem E

A 4.0-kg object traveling westward at 25 m/s hits a 15-kg object at rest. The 4.0-kg object bounces eastward at 8.0 m/s. What is the speed and direction of the 15-kg object?

Does this seem familiar?

The idea of momentum being equal—especially when two objects are pushing away from each other—may seem familiar.

The conservation of momentum equation is derived from Newton’s 3rd Law.

Newton’s 3rd law says…

Newton and Momentum

From last section:Ft = p, or (as we’ll use it) Ft = mvf - mvi

If we consider two forces, we’ll call them F1 and F2, we have that:

F1t = m1v1,f – m1v1,i …and…

F2t = m2v2,f – m2v2,i

Since these are the only two forces at work in the collision, Newton’s 3rd Law says they must be equal and opposite, so:

F1 = -F2

Newton and Momentum

Since they are in the same collision, the time is exactly the same, so:

F1t = -F2t

Which means (substituting from momentum):m1v1,f – m1v1,i = -(m2v2,f – m2v2,i)

Rearrange the equation, and voila!

m1v1,i + m2v2,i = m1v1,f + m2v2,f

In real collisions…

In this chapter (and this book altogether) we treat force as constant in collisions.

The reality is that force is not constant, but varies throughout a collision.

Think of two rubber balls hitting each other…at first, the balls apply force to each other, but the more contact, the more they are compressed and the more force they push with, until their velocities are reversed—then the force decreases until they lose contact and the force exerted is zero.

Average Force

Because of that, when we calculate impulse, we generally use average force over the collision.

Collisions

One type of collision we have not considered is one in which two objects stick together and move with a common velocity after colliding.

This is called a perfectly inelastic collision.

In a perfectly inelastic collision,

m1v1,i + m2v2,i = (m1+ m2)vf

(essentially this is the same equation; there is only one final velocity because the two objects have to move at the same speed)

Elastic or Inelastic?!?

What does the word elastic mean in physics?

When we think of something as elastic it tends to keep its shape—when bent or stretched it returns to normal.

In a collision, elastic or inelastic refers to whether or not an object is deformed during the collision.

Inelastic Collisions

In an inelastic collision, kinetic energy is not constant.

Some energy is converted to sound, some is converted to internal energy as the objects deform.

Sample Problem

Two clay balls collide head on in a perfectly inelastic collision. The first ball has a mass of 0.500kg and an initial velocity of 4.00m/s to the right. The second ball is 0.250kg, and has an initial velocity of 3.00m/s to the left. What is the final velocity of the two balls after they collide and stick together, and how much kinetic energy is lost in the collision?

Practice Problem #3

A 0.25kg arrow with a velocity of 12m/s west strikes a 6.8kg target (standing still). What is the final velocity of the target and arrow? What is the decrease in kinetic energy?

Elastic Collisions

In elastic collisions, the objects collide and return back to their original shape.

The overall kinetic energy in an elastic collision is conserved.

This means that:

KE1,i + KE2,i = KE1,f + KE2,f

Practice Problem #4

A 0.015kg marble moving to the right at 0.225m/s makes an elastic head on collision with a 0.030 marble moving left at 0.180m/s. After the collision, the smaller marble moves to the left at a velocity of 0.315m/s.

A) Find the final velocity of the bigger marble.

B) Verify by comparing kinetic energy before and after collision.

In reality…

In real life, there are very, very few collisions that are perfectly inelastic or elastic.

Most collisions lose energy in the form of sound, heat, etc.

The third category of collisions is called inelastic collisions; collisions in which objects bounce off of each other, but kinetic energy is not conserved.

Inelastic Collisions

For the problems we will solve, we will consider all collisions either perfectly inelastic or elastic.