momentum - coblejphysics.files.wordpress.com file28/04/2009 · p = mv • momentum will always be...

MOMENTUM is

•

“inertia in motion”•

Momentum is a vector quantity equal to the product of an object’s mass and velocity. Momentum is represented with the variable ‘p’

and its SI unit of measure is a

kilogram·meter/second (kg·m/s).

p = mv

p = mv

•

momentum will always be in the same direction as the velocity

•

the momentum of that object will be directly proportional to the velocity

•

momentum will remain constant unless acted upon by an external force.

p = mv

p

m vx

Change in Momentum

Δp = p2

– p1

= mv2

– mv1

= m(v2

– v1

)

Δp = m Δv

ΔMomentum = Impulse•

Newton’s second law of motion

F = ma = mΔv/ ΔtOr

FΔt = mΔv•

The left side of the equation is a quantity called impulse.

•

The impulse is the product of the average force and the time interval, measured in Newton·seconds (N·s), which are equal to (kg·m/s)

The impulse placed on an object is ALWAYS equal to the Δp

experienced by the object.

A Summer Driving Experience

orThe tale of an unlucky Bug!

(refer to handout provided by your teacher)

p = mv

•

momentum will always be in the same direction as the velocity

•

the momentum of that object will be directly proportional to the velocity

•

momentum will remain constant unless acted upon by an external force.

ΔMomentum = Impulse•

Newton’s second law of motion

F = ma = mΔv/ ΔtOr

FΔt = mΔv•

The left side of the equation is a quantity called impulse.

•

The impulse is the product of the average force and the time interval, measured in Newton·seconds (N·s), which are equal to (kg·m/s)

The impulse placed on an object is ALWAYS equal to the Δp

experienced by the object.

SOLVING Δ

momentum/impulse problems

Ask yourself this Question

Does the problem mention Δp, impulse, time, or force?

If yes it’s a FΔt = mΔv. If no continue to evaluate.

Equation

F Δt = m Δv

Or

FAT –

MAV

Impulse Change in Momentum an example problem

Tiger Woods hits a 0.050 kg golf ball, giving ita speed of 75 m/s. What impulse does heimpart on the ball?

Tiger Woods hits a 0.050 kg golf ball, giving ita speed of 75 m/s. What impulse does heimpart on the ball?


Does the problem mention Δp, impulse, time, or force? If yes it’s a FΔt = mΔv.

If no continue to evaluate.

Impulse Change in Momentum

Tiger Woods hits a 0.050 kg golf ball, givingit a speed of 75 m/s. What impulse does heimpart on the ball?

FΔt = mΔvimpulse

∆p


Tiger Wood hits a 0.050 kg golf ball, giving ita speed of 75 m/s. What impulse does heimpart on the ball?

FΔt = mΔvm = 0.050 kgv1 = 0v2 = 75 m/s


Tiger Woods hits a 0.050 kg golf ball, givingIt a speed of 75 m/s. What impulse does heimpart on the ball?m = 0.050 kgv1 = 0v2 = 75 m/s

FΔt

= 0.050kgx(75m/s-0)


FΔt

= 0.050kgx(75m/s-0)Since the impulse impressed on an object

ALWAYS equals the change in momentum it

experiencesImpulse = 3.8 kgxm/s

or (Nxs)

Another example

Wayne hits a stationary 0.12kg hockey puck with a force that lasts 0.010 s and makes the puck shoot across the ice at 20.0 m/s. What force was applied to the puck?

Another exampleWayne hits a stationary 0.12kg hockey puck with

a force that lasts 0.010 s and makes the puck shoot across the ice at 20.0 m/s. What force was applied to the puck?


Does the problem mention Δp, impulse, time, or force?If yes it’s a FΔt = mΔv.




FΔt = mΔvm = 0.12 kgv1 = 0v2 = 20.0 m/st1 = 0t2 = 0.010 s



m = 0.12 kg FΔt = mΔvv1 = 0 F = (mΔv)/Δtv2 = 20.0 m/s

F = 0.12kgx(20.0m/s-0)/(0.010 s –

0)

t1 = 0t2 = 0.010 s

Another example

FΔt = mΔv

F = 0.12kgx(20.0m/s-0)/(0.010 s –

0)F = 240 N

Still another exampleA 0.60 kg tennis ball traveling at 10 m/s

is

returned. It leaves the racket with a speed of36 m/s

in the opposite direction. What is the

Δp

of the ball? If the ball is in contact with theracket 0.020 s what is the force impressed on the

ball by the racket?

Ask yourself this Question Does the problem mention Δp, impulse, time, or force?

If yes it’s a FΔt = mΔv.



is



Δp


ball by the racket?FΔt = mΔv

m = 0.060 kgv1 = -10 m/sv2 = 36 m/st1 = 0t2 = 0.020 s

∆p


is



Δp


ball by the racket?m = 0.60 kgv1 = -10 m/sv2 = 36 m/st1 = 0t2 = 0.020 s

∆p = m∆v= 0.60kgx(36-(-10))m/s= 2.8 kgxm/s


is



Δp


ball by the racket?m = 0.60 kgv1 = -10 m/sv2 = 36 m/st1 = 0t2 = 0.020 s

F∆t = m∆vF = (m∆v)/∆t= 0.60kgx(36-(-10))m/s/(0.020s –

0)

F = 140 N

Your Teacher will now give you a problem packet and assign Δp = Impulse

problems on which you will work.

ΔMomentum = Impulse Problems

LAW OF CONSERVATION OF MOMENTUM

The momentum of a system before an event or incident is equal to the momentum of the system after the event or incident if the system is closed.A closed system is a system that doesn’t have objects entering or leaving, and an isolated system is one without external forces acting on it.

Types of eventsInelastic:

Objects are tangled or stuck

together after the event. Objects become deformed and heat & sound are generated during the event.

Elastic:

Objects are apart after the event. Objects are not deformed and no heat or sound are generated during the event.

SOLVING conservation of momentum problems

Remember these 5 steps1) Does the problem mention Δp, impulse, time, or force?

If yes it’s a FΔt = mΔv

If no continue to # 2.2) Identify the objects.3) Identify the incident or event. 4) Are the objects together or apart AFTER the event? 5) If the objects are together AFTER the event this is an

inelastic event. -

If the objects are apart AFTER the event this is an elastic event.

Inelastic EventsIn a closed system when the objects are together

AFTER the event the event is inelastic.

Momentum of the system before the event = Momentum of the system after the event

pb

= pa

m1

v1 + m2

v2 = (m1 + m2

)va

Inelastic EventsIn a closed system when the objects are together

AFTER the event the event is inelastic.

m1 –

mass of the first objectm2

-

mass of the second objectv1

-

velocity of the first object before the eventv2

-

velocity of the second object before the eventva

-

velocity of the combined objects after the event

Family fun at Happy Wheels

orAdventures

of an ancient Roller Derby Queen

Inelastic EventsGranny (aka Roller Ruth) whizzes around the rink at3m/s and confronts Ambrose scared motionless infront of her. Thinking quickly she picks up Ambrose andcontinues on. What is their velocity after averting this neardisaster?

1) Does the problem mention Δp, impulse, time, or force? If yes it’s a FΔt = mΔv

If no continue to # 2.2) Identify the objects.3) Identify the incident or event. 4) Are the objects together or apart AFTER the event? 5) If the objects are together AFTER the event this is an inelastic event. -



pb

= pam1

v1 + m2

v2 = (m1 + m2

)vam1

= 80kg m2

= 40kg v1

= 3m/s V2 = 0 va

- ?


pb

= pam1

v1 + m2

v2 = (m1 + m2

)vam1

= 80kg 80kgx3m/s + 40 kgx0 = (80+40)kgxvam2

= 40kg 240kgxm/s = 120kgxva

v1

= 3m/s 120kg 120 kgv2

= 3m/s (240kgxm/s)/120kg = vava

-

? 2m/s = va

Inelastic EventsA truck with a mass of 2000kg moving at15m/s hits a1000kg car moving in theopposite direction at 10m/s. Whatis the direction and speed of the resultingcombined wreckage?1) Does the problem mention Δp, impulse, time, or force?




Inelastic EventsA truck with a mass of 2000kg moving at 15m/s hits a1000kg car moving in the opposite direction at 10m/s. Whatis the direction and speed of the resulting combinedwreckage?

pb

= pam1

v1 + m2

v2 = (m1 + m2

)vam1

= 2000kg m2

= 1000kg v1

= 15m/s V2 = -10ms va

- ?

Inelastic EventsA truck with a mass of 2000kg moving at 15m/s hits a1000kg car moving in the opposite direction at 10m/s. Whatis the direction and speed of the resulting combinedwreckage?

pb

= pa

m1

v1 + m2

v2 = (m1 + m2

)va

2000kgx15m/s + 1000 kgx-10m/s = (2000+1000)kgxva20000kgxm/s

= 3000kgxva

3000 kg 3000 kg(20000kgxm/s)/3000kg = va

6.7m/s = va

m1

= 2000kgm2

= 1000kg v1

= 15m/s V2 = -10m/s

va

- ?

Your Teacher will now choose problems fromthe packet and assign

inelastic Eventproblems on which you will work.

Conservation of Momentum Inelastic Event Problems

Elastic EventsIn a closed system when the objects are

apart AFTER the event, the event is elastic.pb

= pa

m1

v1 + m2

v2 = m1

v'1 + m2

v'2

Elastic EventsIn a closed system when the objects are apart

AFTER the event, the event is elastic.

m1 –

mass of the first objectm2

-

mass of the second objectv1

-

velocity of the first object before the eventv2

-

velocity of the second object before the eventv'1

-

velocity of the first object after the eventv'2

-

velocity of the second object after the event

Elastic EventsIn a closed system when the objects are apart AFTER the

event, the event is elastic.A person docks a canoe. The person, with mass 50 kg steps toward

thedock at 4 m/s. What is the resultant velocity of the canoe (100 kg).

1) Does the problem mention Δp, impulse, time, or force?







pb

= pa

m1

v1 + m2

v2 = m1

v'1 + m2

v'2m1 –

50 kgm2

–

100 kgv1

- 0v2

- 0v'1

- 4 m/sv'2

- ?




pb

= pa

m1

v1 + m2

v2 = m1

v'1 + m2

v'2m1 –

50 kgm2

–

100 kgv1

- 0v2

- 0v'1

- 4 m/sv'2

- ?

50 kgx0 + 100 kgx0 = 50 kg x 4m/s + 100 kgxv'20 = 200 kgxm/s

+

100 kgxv'2-200 kgxm/s

= 100 kgxv'2(-200 kgxm/s)/100kg = v'2

-2 m/s

= v'2


event, the event is elastic.A bullet of mass 0.03 kg is fired from a gun with a muzzle velocity of500m/s. The gun recoils with a velocity of 5 m/s. What is the mass of thegun?






event, the event is elastic.A bullet of mass 0.03 kg is fired from a gun with a muzzle velocity of500m/s. The gun recoils with a velocity of 5 m/s. What is the mass of thegun?

pb

= pa

m1

v1 + m2

v2 = m1

v'1 + m2

v'2m1 0.03 kgm2

?v1

0v2

0v'1

500 m/sv'2

- 5 m/s


event the event is elastic.A bullet of mass 0.03 kg is fired from a gun with a muzzle velocity of500m/s. The gun recoils with a velocity of 5 m/s. What is the mass of thegun?

pb

= pa

m1

v1 + m2

v2 = m1

v'1 + m2

v'2m1 0.03 kgm2

?v1

0v2

0v'1

500 m/sv'2

- 5 m/s

0.03 kgx0 + m2

x0 = 0.03 kg x 500m/s + m2

x -5m/s

0 = 15 kgxm/s

+

m2

x -5m/s

-15 kgxm/s

= m2

x -5m/s

(-15 kgxm/s)/ -5m/s = m2

3 kg = m2


event the event is elastic.A 2 kg cart traveling right at 30 m/s

collides and bounces off a 1 kg carttraveling at 10 m/s. After the collision the velocity of the first cart was18m/s. What was the velocity of the second cart?






event the event is elastic.A 2 kg cart traveling right at 30 m/s

collides and bounces off a 1 kg carttraveling at 10 m/s. After the collision the velocity of the first cart was18m/s. What was the velocity of the second cart?

pb

= pa

m1

v1 + m2

v2 = m1

v'1 + m2

v'2m1 – 2 kgm2

– 1 kgv1

- 30m/sv2

- 10v'1

-

18 m/sv'2

- ?


event, the event is elastic.A 2 kg cart traveling right at 30 m/s

collides and bounces off a 1 kg carttraveling at 10 m/s. After the collison

the velocity of the first cart was18m/s. What was the velocity of the second cart?

pb

= pa

m1

v1 + m2

v2 = m1

v'1 + m2

v'2m1 – 2 kgm2

– 1 kgv1

- 30 m/sv2

- 10 m/sv'1

-

18 m/sv'2

- ?

2 kg x 30m/s + 1 kg x 10m/s = 2 kg x 18m/s + 1 kgxv'260 kgxm/s

+ 10 kgxm/s

= 36 kgxm/s

+

1 kgxv'270 kgxm/s

-

36 kgxm/s

= 1 kgxv'2(34 kgxm/s)/1kg = v'2

34 m/s

= v'2

Your Teacher will now choose problems fromthe packet and assign

Elastic Eventproblems on which you will work.

Conservation of Momentum Elastic Event Problems

momentum - coblejphysics.files.wordpress.com file28/04/2009 · p = mv • momentum will always be...

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