Work and Energy

An Introduction

Monday,October 27, 2008

Work Energy Theorem

Kinetic Energy Definition

Announcements

� Test repair begins during lab sections

on Tuesday. Attendance is

MANDATORY unless you have a

perfect score!

Work

� Work tells us how much a force or combination of forces changes the energy of a system.

� Work is the bridge between force (a vector) and energy (a scalar).

� W = F ∆r cos θ

� F: force (N)

� ∆r : displacement (m)

� θ: angle between force and displacement

Units of Work

� SI System: Joule (N m)

� 1 Joule of work is done when 1 N acts on a

body moving it a distance of 1 meter

� British System: foot-pound

� (not used in Physics B)

� cgs System: erg (dyne-cm)

� (not used in Physics B)

� Atomic Level: electron-Volt (eV)

Force and direction of motion

both matter in defining work!

� There is no work done by a force if it causes no displacement.

� Forces can do positive, negative, or zerowork. When an box is pushed on a flat floor, for example…

� The normal force and gravity do no work, since they are perpendicular to the direction of motion.

� The person pushing the box does positive work, since she is pushing in the direction of motion.

� Friction does negative work, since it points opposite the direction of motion.

Conceptual Checkpoint� Question: If a man holds a 50 kg box at arms

length for 2 hours as he stands still, how much work does he do on the box?

Conceptual Checkpoint� Question: If a man holds a 50 kg box at arms

length for 2 hours as he walks 1 km forward, how much work does he do on the box?

Conceptual Checkpoint� Question: If a man lifts a 50 kg box 2.0 meters,

how much work does he do on the box?

Work and Energy

Work changes mechanical energy!

� If an applied force does positive workon a system, it tries to increasemechanical energy.

� If an applied force does negative work, it tries to decrease mechanical energy.

The two forms of mechanical energy are called potential and kinetic energy.

Sample problem

Jane uses a vine wrapped around a pulley to lift a 70-kg Tarzan to a tree house 9.0 meters above the ground.

a)How much work does Jane do when she lifts Tarzan?

b)How much work does gravity do when Jane lifts Tarzan?

Sample problemJoe pushes a 10-kg box and slides it across the floor at constant velocity of

3.0 m/s. The coefficient of kinetic friction between the box and floor is 0.50.

a) How much work does Joe do if he pushes the box for 15 meters?

b) How much work does friction do as Joe pushes the box?

Sample problemA father pulls his child in a little red wagon with constant speed. If the father pulls with a force of 16 N for 10.0 m, and the handle of

the wagon is inclined at an angle of 60o above the horizontal, how

much work does the father do on the wagon?

Kinetic Energy

� Energy due to motion

� K = ½ m v2

� K: Kinetic Energy

� m: mass in kg

� v: speed in m/s

� Unit: Joules

Sample problemA 10.0 g bullet has a speed of 1.2 km/s.

a) What is the kinetic energy of the bullet?

b) What is the bullet’s kinetic energy if the speed is halved?

c) What is the bullet’s kinetic energy if the speed is doubled?

The Work-Energy Theorem

� The net work due to all forces equals the change in the kinetic energy of a

system.

� Wnet = ∆K

�Wnet: work due to all forces acting on an object

�∆K: change in kinetic energy (Kf – Ki)

Sample problemA 15-g acorn falls from a tree and lands on the ground 10.0 m below with a speed of 11.0 m/s.

a) What would the speed of the acorn have been if there had been no air resistance?

b) Did air resistance do positive, negative or zero work on the acorn? Why?

Sample problemA 15-g acorn falls from a tree and lands on the ground 10.0 m below

with a speed of 11.0 m/s.

c) How much work was done by air resistance?

d) What was the average force of air resistance?

Tuesday,

October 29, 2008

Work done by variable forces

Announcements

� Test Repair begins TODAY!!!

� 4th period in Mr. Perkins’ room

� 5th period in Dr. Bertrand’s room

Constant force and work

� The force shown is a constant force.

� W = F∆r can be used to calculate the work done by this force when it moves an object from xa to xb.

� The area under the curve from xa to xb can also be used to calculate the work done by the force when it moves an object from xa to xb

F(x)

xxa xb

Variable force and work

� The force shown is a variable force.

� W = F∆r CANNOT be used to calculate the work done by this force!

� The area under the curve from xa to xb can STILL be used to calculate the work done by the force when it moves an object from xa to xb

F(x)

xxa xb

Springs

� When a spring is stretched or compressed from its equilibrium

position, it does negative work, since the spring pulls opposite the

direction of motion.

� Ws = - ½ k x2

� Ws: work done by spring (J)

� k: force constant of spring (N/m)

� x: displacement from equilibrium (m)

� The force doing the stretching does positive work equal to the

magnitude of the work done by the spring.

� Wapp = - Ws = ½ k x2

Springs: stretching

m

m

x

0

Fs = -kx (Hooke’s Law)

100

0

-100

-200

200F(N)

0 1 2 3 4 5x (m)

Ws = negative area= - ½ kx2

Fs

Fs

Sample problem

It takes 180 J of work to compress a certain spring 0.10 m.

a) What is the force constant of the spring?

b) To compress the spring an additional 0.10 m, does it take 180 J,more than 180 J, or less than 180 J? Verify your answer with a calculation.

Sample problemA vertical spring (ignore its mass) whose spring constant is

900 N/m, is attached to a table and is compressed 0.150 m.

a) What speed can it give to a 0.300 kg ball when released?

b) How high above its original position (spring compressed) will the ball fly?

Wednesday,

October 29, 2008

More work by variable forces

Announcements

� Exam repair continues…

� HW due today, pass to back of room

Sample Problem

� How much work is done by the force shown when it acts on an object and pushes it from x = 0.25 m to x = 0.75 m?

Sample Problem

� How much work is done by the force shown

when it acts on an object and pushes it from

x = 2.0 m to x = 4.0 m?

Power

� Power is the rate of which work is done.

� P = W/∆t

� W: work in Joules

� ∆t: elapsed time in seconds

� When we run upstairs, t is small so P is big.

� When we walk upstairs, t is large so P is small.

Unit of Power

� SI unit for Power is the Watt.

� 1 Watt = 1 Joule/s

� Named after the Scottish engineer

James Watt (1776-1819) who

perfected the steam engine.

� British system

� horsepower

� 1 hp = 746 W

How We Buy Energy…

� The kilowatt-hour is a commonly used unit by the electrical power company.

� Power companies charge you by the kilowatt-hour (kWh), but this not power, it is really energy consumed.

� 1 kW = 1000 W

� 1 h = 3600 s

� 1 kWh = 1000J/s • 3600s = 3.6 x 106J

Sample problemA record was set for stair climbing when a man ran up the 1600 steps of the Empire State Building in 10 minutes and 59 seconds. If the height gain of each step was 0.20 m, and the man’s mass was 70.0 kg, what was his average power output during the climb? Give your answer in both watts and horsepower.

Sample problemCalculate the power output of a 1.0 g fly as it walks straight up a window pane at 2.5 cm/s.

Thursday,

October 30, 2008

Force Types

Announcements

Force types

� Forces acting on a system can be divided into two types according to how they affect potential energy.

� Conservative forces can be related to potential energy changes.

� Non-conservative forces cannot be related to potential energy changes.

� So, how exactly do we distinguish between these two types of forces?

Conservative forces� Work is path independent.

� Work can be calculated from the starting and ending points only.

� The actual path is ignored in calculations.

� Work along a closed path is zero.� If the starting and ending points are the same, no work is

done by the force.

� Work changes potential energy.

� Examples:� Gravity

� Spring force

� Conservation of mechanical energy holds!

Non-conservative forces

� Work is path dependent.

� Knowing the starting and ending points is not sufficient to calculate the work.

� Work along a closed path is NOT zero.

� Work changes mechanical energy.

� Examples:

� Friction

� Drag (air resistance)

� Conservation of mechanical energy does not hold!

Potential energy

� Energy of position or configuration

� “Stored” energy

� For gravity: Ug = mgh

� m: mass

� g: acceleration due to gravity

� h: height above the “zero” point

� For springs: Us = ½ k x2

� k: spring force constant

� x: displacement from equilibrium position

Conservative forces and

Potential energy

� Wc = -∆U

� If a conservative force does positive work on a

system, potential energy is lost.

� If a conservative force does negative work, potential energy is gained.

� For gravity

� Wg = -∆Ug = -(mghf – mghi)

� For springs

� Ws = -∆Us = -(½ k xf2 – ½ k xi

2)

More on paths and

conservative forces.

Q: Assume a conservative force moves an object along the various paths. Which two works are equal?

A: W2 = W3

(path independence)

Q: Which two works, when added together, give a sum of zero?

A: W1 + W2 = 0

or

W1 + W3 = 0

(work along a closed path is zero)

Friday,

October 31, 2008

Conservative Forces and Potential Energy

Announcements

Sample problem

A B

D C

A box is moved in the closed path shown.

a) How much work is done by gravitywhen the box is moved along the path A->B->C?

b) How much work is done by gravity when the box is moved along the path A->B->C->D->A?

mg

∆r

Sample problem

A B

D C

A box is moved in the

closed path shown.

b) How much work is

done by gravity

when the box is

moved along the

path A->B->C->D-

>A?

mg

∆r

Sample problem A box is moved in the closed path shown.

a) How much work is done by gravity when the box is moved along the path A->B->C?

b) How much work is done by gravity when the box is moved along the path A->B->C->D->A?

Solutiona) WG = 0 + F∆r

WG = 0 –mgh = -mgh

b) WG = 0 -mgh + 0 + mgh

= 0

The work in b) is zero because

work along a closed path is

zero for any conservative

force.

Sample problemA box is moved in the closed

path shown.

a) How much work would

be done by friction if the

box were moved along

the path A->B->C?

b) How much work is done

by friction when the box

is moved along the path

A->B->C->D->A?

Solutiona) Wf = -µkmgd - µkmgd

Wf = -2µkmgd

b) Wf = -µkmgd - µkmgd -

µkmgd - µkmgd

= -4 µkmgd

Because friction is a

nonconservative force,

work along the closed

path in b) is not zero.

Sample problem (#8.6)

As an Acapulco cliff diver drops to the water from a height of 40.0 m, his gravitational potential energy decreases by

25,000 J. How much does the diver weigh?

Friday, October 31, 2008

Conservation of Mechanical Energy

+ Pendulums

Announcements

• Turn in HW: Ch 7 (26, 27, 29)

• Lab next week: Pendulums and Conservation of Energy

Sample problem

If 60.0 J of work are required to stretch a spring from a 2.00 cm elongation to a 5.00 cm elongation, how much

additional work is needed to stretch it from a 5.00 cm

elongation to a 8.00 cm elongation?

Law of Conservation of

Energy

� In any isolated system, the total energy remains constant.

� Energy can neither be created nor

destroyed, but can only be

transformed from one type of energy

to another.

Law of Conservation of

Mechanical Energy

� E = K + U = Constant

� K: Kinetic Energy (1/2 mv2)

� U: Potential Energy (gravity or spring)

� ∆E = ∆U + ∆K = 0

� ∆K: Change in kinetic energy

� ∆U: Change in gravitational or spring potential energy

Conservation of Energy

Simulation

http://phet.colorado.edu/simulations/sims.php?sim=Energy_Skate_Park

Energy Skate Park

Sample problem (#8.15)A 0.21 kg apple falls from a tree to the ground, 4.0 m below. Ignoring air resistance, determine the apple’s gravitational potential energy, U, kinetic energy, K, and total mechanical energy, E, when its height above the ground is each of the following: 4.0 m, 2.0 m, and 0.0 m. Take ground level to be the point of zero potential energy.

Pendulums and Energy

Conservation

� Energy goes back and forth between

K and U.

� At highest point, all energy is U.

� As it drops, U goes to K.

� At the bottom , energy is all K.

h

Pendulum Energy

½mvmax2 = mghFor minimum and maximum points of swing

K1 + U1 = K2 + U2For any points 1 and 2.

Sample problemWhat is the speed of the pendulum bob at point B if it is

released from rest at point A?

1.5 m

A

B

40o

Springs and Energy

Conservation

� Transforms energy back and forth between K and U.

� When fully stretched or extended, all energy is U.

� When passing through equilibrium, all its energy is K.

� At other points in its cycle, the energy is a mixture of U and K.

Monday,

November 3, 2008

Energy conservation

Announcements

� Conservation of Energy lab

in Dr. Bertrand’s room LC 210 this week

� Turn in HW

Chapter 7 (33, 35, 37)

Spring Energy

m

m-x

m

x

0

½kxmax2 = ½mvmax2For maximum and minimum displacements from equilibrium

K1 + U1 = K2 + U2 = E For any two points 1 and 2

All U

All U

All K

Spring Simulation

Spring Physics Simulation

Sample problem (#8.18)A 1.60 kg block slides with a speed of 0.950 m/s on a frictionless, horizontal surface until it encounters a spring with a force constant of 902 N/m. The block comes to rest after compressing the spring 4.00 cm. Find the spring potential energy, U, the kinetic energy of the block, K, and the total mechanical energy of the system, E, for the following compressions: 0 cm, 2.00 cm, 4.00 cm.

Pendulum lab

� Figure out how to demonstrate conservation of energy with a pendulum using the equipment provided.

� The photogates must be set up in “gate” mode this time.

� The width of the pendulum bob is an important number. To

get it accurately, use the caliper.

� Repeat for at least THREE different release heights.

Pendulum lab - Theory

� Calculate the velocity of the bob at the bottom of the swing using the photogate.

� Compare velocity to the theoretical v given by conservation of mechanical energy for a set height of release

h

Pendulum Lab write-up

In your handwritten report in your lab book, you should have the following

1. A simple sketch of your apparatus.

2. A clearly labeled data table. This table should include raw data plus calculated potential energy and kinetic energy values for comparison purposes. Measure multiple trials for at least three different heights

3. A discussion regarding how well your results support the Law of Conservation of Mechanical Energy. If your results do not support this law very well, what might be the explanation for the discrepancy?

Sample Data Table

3

2

1

Kmax

(J)

v (m/s)

t (s)d (m)∆U (J)Ugf (J)Ugi (J)Xf(m)

Xi (m)

Trial

Law of Conservation of

Energy

� E = U + K + Eint= Constant

� Eint is thermal energy.

� ∆U + ∆K + ∆ Eint = 0

� Mechanical energy may be converted

to and from heat.

Work done by non-

conservative forces

� Wnet = Wc + Wnc

� Net work is done by conservative and non-conservative forces

� Wc = -∆U • Potential energy is related to conservative forces only!

� Wnet = ∆K• Kinetic energy is related to net force (work-energy theorem)

� ∆K = -∆U + Wnc

• From substitution

� Wnc = ∆U + ∆K = ∆E

� Nonconservative forces change mechanical energy. If nonconservative work is negative, as it often is, the mechanical energy of the system will drop.

Solution (#8.22)

Wnc = ∆U + ∆K

= Uf – Ui + Kf – Ki

= mghf – mghi + ½ mvf2 – ½ m vi

2

= m[g(hf –hi) + ½ (vf2 –vi

2)]

= 72[(9.8)(0 - 1.75) + ½ (8.22 – 1.32)]

= 1125 J

Tuesday,

November 4, 2008

Non-conservative forces and the Law of Conservation of Energy

Announcements

� Lab downstairs in Dr. Bertrand’s

room.

� HW turned in at the back

Sample problem

If 60.0 J of work are required to stretch a spring from a 2.00 cm

elongation to a 5.00 cm elongation, how much additional work is

needed to stretch it from a 5.00 cm elongation to a 8.00 cm

elongation?

Sample problem (#8.29)

A 1.75-kg rock is released from rest at the surface of a pond 1.00 m deep. As the rock falls, a constant upward force of 4.10 N is exerted on it by water resistance. Calculate the nonconservative work, Wnc, done by the water resistance on the rock, the gravitational potential energy of the system, U, the kinetic energy of the rock, K, and the total mechanical energy of the system, E, for the following depths below the water’s surface: d = 0.00 m, d = 0.500 m, d = 1.00 m. Let potential energy be zero at the bottom of the pond.

Solution (#8.29) – for 0.00 m

Wnc = F∆r = 0

E = U + K

= mgh + 0 = mgh

= (1.75 kg)(9.8 m/s2)(1.00 m) = 17.15 J

Therefore

Wnc = 0 (the rock hasn’t moved yet)

E = 17.15 J(will be reduced by the drag force of water)

U = 17.15 J (maximum value)

K = 0 (minimum value)

Solution (#8.29) – for 0.50 m

Wnc = F∆r cos θ= (4.10 N)(0.50 m)cos(180°) = -2.05 J = ∆E

E = 17.15 J – ∆E = 17.15 J - 2.05 J = 15.1 J

E = U + K = mgh + K

15.1 J = (1.75 kg)(9.8 m/s2)(0.50 m) + K

15.1 J = 8.6 J + K

K = 15.1 – 8.6 J = 6.5 J

Therefore

Wnc = -2.05 J

E = 15.1 J(reduced by the drag force of water)

U = 8.6 J (determined by height)

K = 6.5 J (reduced by the drag force of water)

Solution (#8.29) – for 1.00 m

Wnc = F∆r = (4.10 N)(1.00 m) cosθ= -4.10 J = ∆E

E = 17.15 J – ∆E = 17.15 J - 4.10 J = 13.05 J

E = U + K = 0 + K = K

13.05 J = K

Therefore

Wnc = -4.10 J

E = 13.05 J (reduced by the drag force of water)

U = 0 (lowest point in problem)

K = 13.05 J (maximum value)

Wednesday,

November 5, 2008

Energy Review

Announcements

� Thursday we will be UPSTAIRS, not in

the lab classrooms.

� Turn in HW

Practice problem (not in packet)

A skeleton runner in the Winter Olympics drops 104 m in elevation from top to bottom of the run.

a) In the absence of nonconservative forces, what would the speed of a rider be at the end of the track. Assume initial velocity of zero.

b) In reality, the best riders reach the bottom at a speed of 35.8 m/s (80 mph). How much work is done on an 86.0 kg rider and skeleton by nonconservative forces?

Shown below is a graph of velocity versus time for an object that moves along a straight, horizontal line under the, perhaps intermittent, action of a single force exerted by an external agent.

Rank the intervals shown on the graph, from greatest to least, on the basis of the work done on the object by the external agent.

5 10 15 20 25 30 35

-10

-8

-6

-

4

-2

0

2

4

6

8

10

Velocity (m/s)

Time (s)

AB C

D

E F

G

A B C

3 m 5 m 4 m

0.600 kg

0.750 kg

0.400 kg

D E F

4 m 3.5 m 3 m

0.750 kg

0.400 kg0.500 kg

Sample Problem (not in packet)

� Two masses are suspended by a string over each side of a pulley (Atwood’s machine). They are initially at rest

at the same height, after they are released, the large

mass m2 falls through a height h and hits the floor, and the small mass m1 rises through a height h.

� Find the speed of the masses just before m2 lands, giving your answer in terms of m1, m2, g, and h. Assume

the ropes and pulley have negligible mass and that friction can be ignored.

Thursday,

November 6, 2008

Special Speaker

Nick Antonas

Announcements

Friday, November 7, 2008

Linear Momentum

Announcements

� Tuesday and Wednesday lab groups

have lab write-ups due TODAY

� Thursday lab group has write-up due

MONDAY

� Next week’s lab:

conservation of momentum

� Turn in HW: Ch. 8 (23, 25, 28)

Which do you think has more

momentum?

Momentum

Momentum is a measure of how hard it is to stop or turn a moving object.

� What characteristics of an object would make it hard to stop or turn?

� Let’s watch Mad Scientist Guy on Momentum and Newton’s Third Law of Motion!

Calculating Momentum

� For one particle

p = mv

Note that momentum is a vector with the same

direction as the velocity!

� For a system of multiple particles

p = Σpi --- add up the vectors

� The unit of momentum is…

kg m/s or Ns

Sample Problem� Calculate the momentum of a 65-kg sprinter

running east at 10 m/s.

Sample Problem� Calculate the momentum of a system composed of a 65-kg

sprinter running east at 10 m/s and a 75-kg sprinter running north

at 9.5 m/s.

Change in momentum

� Like any change, change in

momentum is calculated by looking at

final and initial momentums.

� ∆∆∆∆p = pf – pi

� ∆∆∆∆p: change in momentum

� pf: final momentum

� pi: initial momentum

Momentum change

demonstration� Using only a meter stick, find the momentum

change of each ball when it strikes the desk from a height of exactly one meter.

� Which ball, Bouncy or Lazy, has the greatest change in momentum?

Wording

dilemma

� In which case is the magnitude of the momentum change

greatest?

� In which case is the

change in the magnitude of the

momentum greatest?

Shown below is a graph of velocity versus time for an object that moves along a straight, horizontal line under the, perhaps intermittent, action of a single force exerted by an external agent.

Rank the intervals shown on the graph, from greatest to least, on the basis of the work done on the object by the external agent.

5 10 15 20 25 30 35

-10

-8

-6

-4

-2

0

2

4

6

8

10

Velocity (m/s)

Time (s)

AB C

D

EF

G

Monday,

November 10, 2008

Impulse

Announcements

� Lab in Dr. Bertrand’s room (LC 207)

all week

� Set Packet FR Problem #2 out for

stamp check.

Impulse (J)

� Impulse is the product of an external force

and time, which results in a change in

momentum of a particle or system.

� J = F t and J = ∆P

� Therefore Ft = ∆P� Units: N s or kg m/s (same as momentum)

Impulsive Forces

� Usually high magnitude, short duration.

� Suppose the ball hits the bat at 90 mph and leaves the bat at 90 mph, what is the magnitude of the momentum change?

� What is the change in the magnitude of the momentum?

Impulse (J) on a graph

F(N)

t (ms)0 1 2 3 40

1000

2000

3000

area under curve

Sample Problem

� Suppose a 1.5-kg brick is dropped on a glass

table top from a height of 20 cm.

a) What is the magnitude and direction of the

impulse necessary to stop the brick?

b) If the table top doesn’t shatter, and stops the

brick in 0.01 s, what is the average force it

exerts on the brick?

c) What is the average force that the brick exerts

on the table top during this period?

Sample Problem

� This force acts on a 1.2 kg object moving at 120.0 m/s. The

direction of the force is aligned with the velocity. What is the new

velocity of the object?

0.20 0.40 0.60 0.80 t(s)

F(N)

1,000

2,000

Tuesday, November 11, 2008

Law of Conservation of Momentum

Announcements

� Lab: Conservation of Momentum

� HW due: Ch. 9 (15,17,18)

Impulsive Forces in Driving

Law of Conservation of

Momentum

� If the resultant external force on a

system is zero, then the vector sum of

the momentums of the objects will

remain constant.

� ΣPbefore = ΣPafter

Sample problem� A 75-kg man sits in the back of a 120-kg canoe that is at rest

in a still pond. If the man begins to move forward in the

canoe at 0.50 m/s relative to the shore, what happens to the

canoe?

External versus internal

forces

� External forces: forces coming from outside the system of particles whose momentum is being considered.

� External forces change the momentum of the

system.

� Internal forces: forces arising from interaction of particles within a system.

� Internal forces cannot change momentum of

the system.

An external force in golf

� The club head exerts an external impulsive force on the ball and changes its momentum.

� The acceleration of the ball is greater because its mass is smaller.

The System

An internal force in pool

� The forces the balls exert on each other are internal and do

not change the momentum of the

system.� Since the balls have equal

masses, the magnitude of

their accelerations is equal.

The System

Explosions

� When an object separates suddenly, as in an explosion, all forces are internal.

� Momentum is therefore conserved in an explosion.

� There is also an increase in kinetic energy

in an explosion. This comes from a potential energy decrease due to chemical

combustion.

Recoil

� Guns and cannons “recoil” when fired.

� This means the gun or cannon must move

backward as it propels the projectile forward.

� The recoil is the result of action-reaction force pairs, and is entirely due to internal forces. As the

gases from the gunpowder explosion expand, they push the projectile forwards and the gun or

cannon backwards.

Wednesday,

November 12, 2008

Inelastic Collisions

Announcements

� Labs today

Collisions

� When two moving objects make contact with each other, they undergo a collision.

� Conservation of momentum is used to analyze all collisions.

� Newton’s Third Law is also useful. It tells us

that the force exerted by body A on body B in a collision is equal and opposite to the

force exerted on body B by body A.

Collisions

� During a collision, external forces are ignored.

� The time frame of the collision is very short.

� The forces are impulsive forces (high force, short duration).

Collision Types

� Elastic collisions

� Also called “hard” collisions

� No deformation occurs, no kinetic energy lost

� Inelastic collisions

� Deformation occurs, kinetic energy is lost

� Perfectly Inelastic (stick together)

� Objects stick together and become one

object

� Deformation occurs, kinetic energy is lost

(Perfectly) Inelastic

Collisions

� Simplest type of collisions.

� After the collision, there is only one velocity, since there is only one object.

�

Sample Problem� An 80-kg roller skating

grandma collides inelastically with a 40-kg kid. What is their velocity after the collision?

� How much kinetic energy is lost?

Sample Problem � A fish moving at 2 m/s swallows a stationary fish which is 1/3 its mass. What is the velocity of the big fish after dinner?

Sample problem

� A car with a mass of 950 kg and a speed of 16 m/s to the east approaches an intersection. A 1300-kg minivan traveling north at 21 m/s approaches the same intersection. The vehicles collide and stick together. What is the resulting velocity of the vehicles after the collision?

Conservation of Momentum

Thursday,

November 13, 2008

Elastic Collisions

Announcements

Sample problem� Suppose a 5.0-kg projectile launcher shoots a

209 gram projectile at 350 m/s. What is the recoil velocity of the projectile launcher?

Sample Problem� An exploding object breaks into three fragments. A 2.0 kg fragment

travels north at 200 m/s. A 4.0 kg fragment travels east at 100 m/s.

The third fragment has mass 3.0 kg. What is the magnitude and

direction of its velocity?

Elastic Collision

� In elastic collisions, there is no deformation of colliding objects, and no change in kinetic energy of the system. Therefore, two basic

equations must hold for all elastic collisions

� Σpb = Σpa (momentum conservation)

� ΣKb = ΣKa (kinetic energy conservation)

Sample Problem� A 500-g cart moving at 2.0 m/s on an air track elastically strikes a

1,000-g cart at rest. What are the resulting velocities of the two

carts?

Sample Problem

� Suppose three equally strong, equally massive astronauts decide to play a game as follows: The first astronaut throws the

second astronaut towards the third astronaut and the game begins. Describe

the motion of the astronauts as the game proceeds. Assume each toss results from

the same-sized "push." How long will the game last?

2D-Collisions

� Momentum in the x-direction is conserved.

� ΣPx (before) = ΣPx (after)

� Momentum in the y-direction is conserved.

� ΣPy (before) = ΣPy (after)

� Treat x and y coordinates independently.

� Ignore x when calculating y

� Ignore y when calculating x

� Let’s look at a simulation:

� http://surendranath.tripod.com/Applets.html

Sample problem� Calculate velocity of 8-kg ball after the collision.

3 m/s

2 kg 8 kg

0 m/s

Before

2 m/s

2 kg

8 kgv

After

50o

x

y

x

y

Friday, November 14, 2008

Review of Momentum

Monday,

November 17, 2008

Energy and Momentum Review

Tuesday,

November 18, 2008

Energy and Momentum EXAM