CHAPTER 7CHAPTER 7MOMENTUM MOMENTUM

AND COLLISIONSAND COLLISIONS

Momentum is the product of Momentum is the product of the mass and velocity of a the mass and velocity of a body. Momentum is a vector body. Momentum is a vector quantity that has the same quantity that has the same direction as the velocity of direction as the velocity of the object. the object.

MOMENTUMMOMENTUM

MOMENTUMMOMENTUM

m = mass (kg)

v = velocity (m/s)

p = momentum (kg•m/s)

p = mv

PROBLEM 1PROBLEM 1

m = 2250 kg

v = 25 m/s

p = mv = 2250(25)

= 56250 kg•m/s East

A 2250 kg pickup truck has a velocity A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the of 25 m/s to the east. What is the momentum of the truck?momentum of the truck?

Impulse is the product of the Impulse is the product of the average force and the time average force and the time interval during which the force interval during which the force is exerted.is exerted.

IMPULSE-MOMENTUM IMPULSE-MOMENTUM THEOREMTHEOREM

F = ma

ΔvF = m

ΔtF×Δt = m×Δv

Impulse = change in momentum

Applications of Impulse Momentum Applications of Impulse Momentum TheoremTheorem

Used to determine stopping distances Used to determine stopping distances and safe following distances of cars and and safe following distances of cars and trucks.trucks.

Used in designing safety equipment that Used in designing safety equipment that reduces force exerted on the object reduces force exerted on the object during collisions. Example-nets and giant during collisions. Example-nets and giant air mattresses for fire-fighters.air mattresses for fire-fighters.

Also used in sports equipment and Also used in sports equipment and games.games.

A baseball of mass 0.14 kg is moving at 35 m/s.

A) Find the momentum of the baseball.

B) Find the velocity of a bowling ball, mass 7.26 kg, if its momentum is the same as the baseball.

PROBLEM 3PROBLEM 3

m = 0.14 kg

v = 35 m/s

p = mv

= 0.14(35)

= 4.9 kg•m/s

4.9 /

7.26 kg

4.9

7.260.67 /

p kg m s

m

pvm

m s

The brakes exert a 640 N force on a car weighing 15689 N and moving at 20 m/s. The car finally stops. A)What is the car’s mass? B)What is the car’s initial momentum? C)What is the change in the car’s momentum?D) How long does the braking force act on the car to bring it to a halt?

PROBLEM 6PROBLEM 6

640

15689

?

20 /

0 /i

f

F N

W N

t

v m s

v m s

?

?

?

?

t

p

p

m

i

15689

15689

9.8

1600.92

W N

Wm

g

kg

20 /

1600.92

1600.92 20

32018.4 /

i

i i

v m s

m kg

p mv

kg m s

i

f

f i

p 32018 kg m/s

p 0 kg m/s

Δp p p

32018.4 kg m/s

32018.4 kg m/s

640 N50 s

F t p

t

CONSERVATION OF CONSERVATION OF MOMENTUMMOMENTUM

States that the momentum of any closed, isolated system does not change.

finalinitial pp

Isolated System has no net force acting on it.Closed System is a system in which objects neither enter nor leave .

Glider A of mass 0.355 kg moves along a frictionless air track with a velocity of 0.095 m/s. It collides with a glider B of mass 0.710 kg moving in the same direction at a speed of 0.045 m/s. After collision glider A continues in the same direction with a velocity of 0.035 m/s. What is the velocity of glider B after collision?

PROBLEM 11PROBLEM 11

A

Ai

Af

Glider A

m = 0.355 kg

v = 0.095 m/s

v = 0.035 m/s

B

Bi

Bf

Glider B

m = 0.710 kg

v = 0.045 m/s

v = ? m/s

initial final

Ai Bi Af Bf

A Ai B Bi A Af B Bf

Conservation of momentum means:

p = p

p + p = p + p

m v + m v = m v + m v

A Ai B Bi A Af B Bf

Bf

Bf

Bf

Bf

Bf

m v + m v = m v + m v

0.355(0.095) + (0.710)(0.045) = 0.355(0.035) + (0.710)v

0.033725 + 0.03195 = 0.012425 + (0.710)v

0.065675 - 0.012425 = (0.710)v

0.05325 = (0.710)v

v = 0.075 m/s

Internal forces are forces between objects within a system.

Example: If we consider a single car as our system, forces are exerted on objects within the car during a collision (i.e. a crash dummy)

INTERNAL FORCESINTERNAL FORCES

INTERNAL FORCESINTERNAL FORCES

External force is a force exerted by an object outside the system.

Example: Our single car from the previous example is still considered our system. If the car collides with a tree, then the force the tree exerts on the car is an external force.

EXTERNAL FORCESEXTERNAL FORCES

EXTERNAL FORCESEXTERNAL FORCES

Two campers dock a canoe. One camper steps on the dock. This camper has a mass of 80 kg and moves forward at 4 m/s. With what speed and direction do the canoe and the other camper move if their combined mass is 110 kg?

PROBLEM 17PROBLEM 17

a

ai

af

Camper

m = 80 kg

v = 0 m/s

v = 4 m/sb

bi

bf

Camper and Canoe

m = 110 kg

v = 0 m/s

v = ? m/s

ai bi af bf

a af b bf

a afbf

b

bf

p + p = p + p

0 = m v + m v

-m vv =

m

v = -2.9m/s

Types of CollisionsTypes of Collisions

ELASTIC – objects are apart ELASTIC – objects are apart after the collisionafter the collision

INELASTIC – objects “stick” INELASTIC – objects “stick” together after the collisiontogether after the collision

COLLISIONSCOLLISIONS

INELASTICINELASTICMomentum is Momentum is

conservedconservedSome KE is Some KE is

changed into changed into other formsother forms

TOTALLY TOTALLY ELASTICELASTIC

Momentum is Momentum is conserved.conserved.

KE is KE is conserved.conserved.

ELASTIC COLLISIONSELASTIC COLLISIONS

http://www.youtube.com/watch?NR=1&v=SBesU12g58Ihttp://www.youtube.com/watch?NR=1&v=SBesU12g58I

INELASTIC COLLISIONSINELASTIC COLLISIONS

http://www.youtube.com/watch?v=wFoPawE0LxAhttp://www.youtube.com/watch?v=wFoPawE0LxA

SKATER/MEDICINE BALL SKATER/MEDICINE BALL EXAMPLEEXAMPLE

A 15 kg medicine ball is thrown at a A 15 kg medicine ball is thrown at a velocity of 20 km/hr to a 60 kg velocity of 20 km/hr to a 60 kg skater who is at rest on ice. The skater who is at rest on ice. The skater catches the ball and slides skater catches the ball and slides with the ball across the ice. with the ball across the ice. Determine the velocity of the Determine the velocity of the skater and the ball after the skater and the ball after the collision. collision.

The collision occurs between a skater and a medicine ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the skater to gain momentum. After the collision, the ball and the skater travel with the same velocity ("v") across the ice.

Before Before Collision Collision

After After Collision Collision

Skater Skater 0 0 60 * v 60 * v

Medicine ball Medicine ball 300 300 15 * v 15 * v

Total Total 300 300 300 300

before afterp = p

60v 15v 300

75v 300

v 4 km/hr

SKATER/MEDICINE BALL EXSKATER/MEDICINE BALL EX

GRANNY/RALPH EXAMPLEGRANNY/RALPH EXAMPLE

GRANNY/RALPH EXAMPLEGRANNY/RALPH EXAMPLEBefore Collision Before Collision After Collision After Collision

Granny Granny 80 * 6 = 480 80 * 6 = 480 80 * v 80 * v

Ralph Ralph 0 0 40 * v 40 * v

Total Total 480 480 480 480

before afterp = p

80v 40v 480

120v 480

v 4m/s

CAR/TRUCK EXAMPLECAR/TRUCK EXAMPLE

CAR/TRUCK EXAMPLECAR/TRUCK EXAMPLE

Before Collision Before Collision After Collision After Collision

Truck Truck 3000 * 10 = 30 000 3000 * 10 = 30 000 3000 * v 3000 * v

Car Car 0 0 1000 * 15 = 15 000 1000 * 15 = 15 000

Total Total 30 000 30 000 30 000 30 000

before afterp = p

30000 = 3000v +15000

15000 = 3000v

v = 5m/s

Elastic

Elastic

Elastic

Above is a representation of  3 bullets with equal mass running into 3 blocks of wood with equal mass. The first bullet passes through the block and maintains much of its original momentum   As a result, very little momentum gets transferred to the block. The second bullet, expands as it enters the block of wood which prevents it from passing all the way through it.  As a result, most of its momentum gets transferred to the block.  (This is an inelastic collision.) The third bullet (a rubber bullet) bounces off the block transferring all of it's own momentum and then borrowing some more from the block.  This has the most momentum transferred to the block.  (This is an elastic collision.)

Totally (Perfectly) Elastic Collision KE is conserved: no sound or heat produced Example: sub atomic particles (electrons), attracting fields http://www.youtube.com/watch?v=x6n-QgjM4Ss

(Partially) Elastic Collision some KE is lost as heat or sound Is a broad range containing most collisionsExample: billiard balls http://www.youtube.com/watch?v=CgDDiDa3Kzk

http://www.youtube.com/watch?v=wFoPawE0LxA

KE and CollisionsKE and Collisions

Completely Inelastic Collision KE is conserved: no sound or heat produced Example: sub atomic particles (electrons), repelling fields http://www.youtube.com/watch?v=NN_wwbx6Bew

Inelastic Collision more KE is lost as heat or sound the colliding objects stick togetherExample: Coupling railroad cars http://www.youtube.com/watch?

v=qzV8ovAobfE

KE and CollisionsKE and Collisions

PROBLEM 20PROBLEM 20

A 2575 kg van runs into the A 2575 kg van runs into the back of a 825 kg compact back of a 825 kg compact car at rest. They move off car at rest. They move off together at 8.5 m/s. together at 8.5 m/s. Assuming no friction with Assuming no friction with the ground, find the initial the ground, find the initial speed of the van.speed of the van.

smv

smv

kgm

af

ai

a

/.

/?

58

2575

smv

v

kgm

bf

bi

b

/.58

0

825

smv

v

v

v

vmvmvmvm

ai

ai

ai

ai

bfbafabibaia

/.

..

).().()(

211

289002575

570125218872575

5882558257508252575

PROBLEM 21PROBLEM 21

A 5g bullet is fired with a A 5g bullet is fired with a velocity of 100 m/s toward a velocity of 100 m/s toward a 10 kg stationary solid block 10 kg stationary solid block resting on a frictionless resting on a frictionless surface. What is the change surface. What is the change in momentum of the bullet if in momentum of the bullet if it becomes embedded in the it becomes embedded in the block?block?

xv

smv

kgm

af

ai

a

/

.

100

005

xv

v

kgm

bf

bi

b

0

10

smx

x

xx

vmvmvmvm bfbafabibaia

/.

..

)()(.)()(.

05

005105

10005010100005

skgm

vv aiaf

/.

)(..

)(m

Bullet of Momentum in Change

a

50

10005005