Momentum and Collisions - schools.misd.orgschools.misd.org/page/download/11856/0/Momentum PPT Student.pdf · in its state of motion. ... • Which of Newton’s Laws is this associated

Momentum and Collisions

Resource

http://www.physicsclassroom.com/

Class/momentum/momtoc.html

Define Inertia

• The property of any body to resist changes

in its state of motion.

• The measure of Inertia is:

• Mass (Kg)

• Which of Newton’s Laws is this associated

with?

• First Law

Why is a bullet that is thrown

not as dangerous as a bullet

that is fired from a rifle?

Think about this:

Which does more damage in striking a tree,

an F-150 (Ford Truck) 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?

• Which requires a greater stopping force?

Why?

Sooooo

• There is a relationship between

• Mass

• Velocity

• Force

Momentum is inertia “mass” in

motion• Momentum combines the first and second

law of 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.

the product ofmass and velocity

of an object

momentum = mass x velocity

ρ = mvWhere

p is momentum in kgm/s m is mass in kgv is velocity in m/s

SI units are kilogram x meters per second (kgm/s)

•Vector (direction matching that of the velocity)

Is momentum scalar or vector?

• impetus was the quality of an object that

was moving independent of an observed

force. Impetus comes from the Latin in- +

petere to go to, seek -- from Greek

petesthai to fly, piptein to fall, pteron wing.

Also, push and pull derive from the Latin

pellere.

Example 1

• An ostrich with a mass of 146kg is running to

the right with a velocity of 17m/s. Find the

momentum of the ostrich.

(146 kg)(17 m/s) = 2482 kg * m/s

Example 2

• What velocity would a 5.5g bullet have, if

its momentum was the same as the

ostrich in the previous problem?

v = p/m

5.5g = 0.0055 kg

v = 2482 kg * m/s/ 0.0055 kg

v = 451,273 m/s

3. A car has a momentum of 20 000 kg m/s . What would be the car's new momentum if ...

A. its velocity were doubled. B. its velocity were tripled. C. its mass were doubled (by adding more passengers and a greater load) D. both its velocity were doubled and its mass were doubled.

3. A car has a momentum of 20 000 kg m/s . What would be the car's new momentum if ...

A. its velocity were doubled. doubledB. its velocity were tripled. tripledC. its mass were doubled (by adding more passengers and a greater load) doubledD. both its velocity were doubled and its mass were doubled. quadrupled

• How can you change momentum of an object?

• Change the velocity. (Mass could change, but

then you are changing the object)

• 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.

The quantity of force applied during a time

interval is called Impulse

• Impulse is a change in momentum

• Impulse = FΔt

• Where F = Force and t = time

• What unit is impulse measured in?

• Ns (SI Unit)

• FYI: The symbol of Impulse is an I or a J.

We typically just write out Impulse

In your head…..

• Calculate the impulse when an average

force of 10N is exerted upon a cart for 2.5

seconds.

• 25N*s

• As impulse increases what happens to

momentum?

• It increases

• What happens to momentum if the impulse

decreases?

• It decreases

Impulse = change in momentumaka the Impulse–Momentum Theorem

FΔt = m(Vf-Vi)Ns = kgm/s

How can these be equal? This is and FYI

FΔt = mΔv

Impulse = change in momentum

FΔt = m(Vf-Vi)

So what does this mean?In simple terms, a __________ acting for a long time can produce the same change in momentum as a large force acting for a __________.

Small force

Short time

FΔt = mΔv

Increasing Momentum by

Increasing VelocityApplies to an object.

Therefore mass is usually constant.

If you increase momentum you get a greater

change in velocity.

Why would you want to increase

momentum? List some examples

NOTE: the time refers to how long the force is acting on the object

Example 4

• A hockey puck has a mass

of 0.12 kg and is at rest. A

hockey player makes a shot,

exerting a constant force of

30.0 N on the puck for 0.1 s.

With what speed does it

head toward the goal?

Example 4

• m = .12 kg

• vi = 0m/s

• t = 0.1 s

• F = 30N

• F∆t = m∆v

• Solve for ∆v (vf – vi)

• ∆v = F∆t/m

• ∆v = (30 N)(0.1sec)/0. 12kg

• ∆v = 25 m/s

Example 5

• A hockey puck has a mass

of 0.12 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 5

• F∆t = m∆v

• Solve for ∆v (vf – vi)

• ∆v = F∆t/m

• ∆v = (30 N)(0.16sec)/0. 12kg

• ∆v = 40 m/s

Example 6

• A hockey puck has a mass

of 0.12 kg and is at rest. A

hockey player makes a shot,

exerting a constant force of

36.0 N on the puck for 0.1 s.

With what speed does it

head toward the goal?

Example 6

• F∆t = m∆v

• Solve for ∆v (vf – vi)

• ∆v = F∆t/m

• ∆v = (36 N)(0.1 sec)/0. 12kg

• ∆v = 30 m/s

Example 7

• A hockey puck has a mass

of 0.12 kg and is at rest. A

hockey player makes a shot,

exerting a constant force of

36.0 N on the puck for 0.16

s. With what speed does it

head toward the goal?

Example 7

• F∆t = m∆v

• Solve for ∆v (vf – vi)

• ∆v = F∆t/m

• ∆v = (36 N)(0.16 sec)/0. 12kg

• ∆v = 48 m/s

The player should use

more force and follow

through!!!!

Apply the greatest force for as

long as possible

We just examined ways to

Increase Momentum by causing

an increase in velocity

Can you think of an example where mass

would change and therefore increase

momentum?

Infiniti

Decreasing Velocity

• You are driving at 50 mph and lose control of your car. You can hit a wall or a haystack. Which do you choose? Why?

• How is the momentum different?

• It isn’t

• Why ?

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

Decreasing Velocity

• So what is different?

• Remember impulse is Force x time

• Hitting the haystack increases the time, thus decreasing the Force

• The change in momentum is the same regardless!

• Actually affecting Force or Time

Example 8

• A 1400kg car is travelling eastward at a

velocity of 15m/s, when it veers off the

road and collides with a pole and is

brought to rest in 0.30s. How much force

is exerted on the car during the collision?

Example 8

• F∆t = m∆v

• Solve for F

• F = (m∆v)/ ∆t

• F = [(1400 kg)(0-15 m/s)]/0.3 s

• F = -70000 N

• Why is F negative?

• It is a stopping force

Example 9

• A 1400kg car is travelling eastward at a

velocity of 15m/s, when it veers off the road

and instead of colliding with a pole it collides

with a barrier of sand and is brought to rest in

0.70s. How much force is exerted on the car

during the collision?

Example 9

• F∆t = m∆v

• Solve for F

• F = (m∆v)/ ∆t

• F = [(1400 kg)(0-15 m/s)]/0.7 s

• F = -30000 N

• How does this force compare to previous example?

• Smaller force: hopefully less damage to the car!

Can you think of other

examples where you would

want to decrease force?

Effect of Collision Time Upon the

Force….or why a boxer “rides”

the punch

Spreading impulse out over a longer time means that the

force will be less; either way, the change in momentum of

the boxing glove, fist, and arm will be the same.

What about vertical situations?

• Think of an object falling: its vi is 0 m/s, it

accelerates and reaches a vf max just

before impact. We are considering the

time frame of the stopping force.

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 PART II

and Multiple Objects

• This section we will observe more than

one object, and how they interact.

Law of Conservation of

Momentum• The momentum of any closed, isolated

system does not change.

• individual parts of the system 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.

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.

• 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 :

Solve for v2’

m1v1 + m2 v2 = m1v1’ + m2v2’

[m1v1 + m2 v2 - m1v1’ ] /m2 = v2’

There are 2 main types of

collisions• Inelastic collisions: two objects stick or join after the

collision. They each have the same velocities after the event.

• Elastic collision: two objects "bounce" apart when they collide. They each have different velocities after the event. They may or may not go in the same direction

• Explosions: special type. One object splits into multiple objects after explosion. Momentum before is zero. Sum of momentum after is zero.

The animation below portrays the inelastic collision between a 1000-kg car and a 3000-kg truck. The before- and after-collision velocities and momentum are shown in the data tables.

The animation below portrays the elastic collision between a 3000-kg truck and a 1000-kg car. The before- and after-collision velocities and momentum are shown in the data tables.

When objects collide and

“STICK”

• Two objects collide and continue jointly in

same direction

• This is called Inelastic

• Train hits Car - Train crashes into Car -

Train Car Accident Shocking video –

YouTube

Example A

Example A:

• A grandma is roller skating at 6 m/s and collides with a little boy who is stationary. If grandma has a mass of 80 kg and the boy has a mass of 40 kg, what is their velocity after impact?

• m1v1 + m2v2 = m1v1’ + m2v2’

• Modify formula to show they stick and that you are solving for final velocity

Example A

Grandma is m1 Child is m2

m1v1 + m2v2 = (m1 + m2)v3

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

(80 kg)(6 m/s) + (40 kg)(0 m/s)/(80 kg + 40 kg) = v3

v3 = 4 m/s

A grandma is roller skating at 6 m/s and collides with a little boy

who is stationary. If grandma has a mass of 80 kg and the boy

has a mass of 40 kg, what is their velocity after impact?

When objects go different

directions or “BOUNCE”This is called an elastic collision

Exercise Ball Fails Compilation! – YouTube

Determine the formula: Indicate Opposing Directions

In these ex I am assuming v1 starts right, v2starts left)

Then Rearrange Formula to Solve for v2’

EX B

• 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 velocity of the blue

stone after the collision.

• Determine the formula: Red in m1, blue is m2

• Indicate Directions

• Then Rearrange Formula to Solve for v ’

m1v1 + m2v2 = m1v1’ + m2v2’

[m1v1 + m2v2 – m2v2’]/m1 = v1’

(17 kg x -5 m/s) + 0 – (17 kg x 1.25 m/s) /20

kg = -5.31 m/s

Explosions

• One object explodes (seperates) into two

objects moving in opposite direction

• Canon Recoil

• Recoil Music Video - YouTube

Ex C 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?

• Determine the formula: Astronaut is m1, tank is m2

• Then Rearrange Formula to Solve for v1’

Ex C

m1v1 + m2v2 = m1v1’ + m2v2’

0 = m1v1’ + m2v2’

– m2v2’/m1 = v1’

– (10 kg x 12 m/s)/63 kg = -1.90 m/s

More Practice

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 B

• 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?

Sample B

(5kg)(2.4m/s) + (6kg)(4.6m/s) = (6kg)(4.1m/s) + (5kg)(? m/s)

(? m/s) = (39.6 kg*m/s – 24.6 kg*m/s)/(5kg)

3m/s