MOMENTUM is inertia in motion Momentum is a vector quantity - - PowerPoint PPT Presentation

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 v x

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/ Δt Or 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.

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• r

The 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/ Δt Or 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 it a speed of 75 m/s. What impulse does he impart on the ball?

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

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.

Impulse Change in Momentum

Tiger Woods hits a 0.050 kg golf ball, giving it a speed of 75 m/s. What impulse does he impart on the ball? FΔt = mΔv

impulse ∆p

Impulse Change in Momentum

Tiger Wood hits a 0.050 kg golf ball, giving it a speed of 75 m/s. What impulse does he impart on the ball? FΔt = mΔv m = 0.050 kg v1 = 0 v2 = 75 m/s

Impulse Change in Momentum

Tiger Woods hits a 0.050 kg golf ball, giving It a speed of 75 m/s. What impulse does he impart on the ball? m = 0.050 kg v1 = 0 v2 = 75 m/s FΔt = 0.050kgx(75m/s-0)

Impulse Change in Momentum

FΔt = 0.050kgx(75m/s-0) Since the impulse impressed on an object ALWAYS equals the change in momentum it experiences Impulse = 3.8 kgxm/s

• r (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 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?

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.

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?

FΔt = mΔv

m = 0.12 kg v1 = 0 v2 = 20.0 m/s t1 = 0 t2 = 0.010 s

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?

m = 0.12 kg FΔt = mΔv v1 = 0 F = (mΔv)/Δt v2 = 20.0 m/s

F = 0.12kgx(20.0m/s-0)/(0.010 s – 0)

t1 = 0 t2 = 0.010 s

Another example

FΔt = mΔv F = 0.12kgx(20.0m/s-0)/(0.010 s – 0) F = 240 N

Still another example

A 0.60 kg tennis ball traveling at 10 m/s is

• returned. It leaves the racket with a speed of

36 m/s in the opposite direction. What is the Δp

• f the ball? If the ball is in contact with the

racket 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. If no continue to evaluate.

Still another example

A 0.60 kg tennis ball traveling at 10 m/s is

• returned. It leaves the racket with a speed of

36 m/s in the opposite direction. What is the Δp

• f the ball? If the ball is in contact with the

racket 0.020 s what is the force impressed on the ball by the racket? FΔt = mΔv m = 0.060 kg v1 = -10 m/s v2 = 36 m/s t1 = 0 t2 = 0.020 s ∆p

Still another example

A 0.60 kg tennis ball traveling at 10 m/s is

• returned. It leaves the racket with a speed of

36 m/s in the opposite direction. What is the Δp

• f the ball? If the ball is in contact with the

racket 0.020 s what is the force impressed on the ball by the racket? m = 0.60 kg v1 = -10 m/s v2 = 36 m/s t1 = 0 t2 = 0.020 s

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

Still another example

A 0.60 kg tennis ball traveling at 10 m/s is

• returned. It leaves the racket with a speed of

36 m/s in the opposite direction. What is the Δp

• f the ball? If the ball is in contact with the

racket 0.020 s what is the force impressed on the ball by the racket? m = 0.60 kg v1 = -10 m/s v2 = 36 m/s t1 = 0 t2 = 0.020 s

F∆t = m∆v F = (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

• r 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

• bjects entering or leaving, and an isolated system

is one without external forces acting on it.

Types of events

Inelastic: 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 steps

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

Inelastic Events

In 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 Events

In a closed system when the objects are together AFTER the event the event is inelastic. m1 – mass of the first object m2

• mass of the second object

v1

• velocity of the first object before the event

v2

• velocity of the second object before the event

va

• velocity of the combined objects after the event

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Adventures

• f an ancient

Roller Derby Queen

Inelastic Events

Granny (aka Roller Ruth) whizzes around the rink at 3m/s and confronts Ambrose scared motionless in front of her. Thinking quickly she picks up Ambrose and continues on. What is their velocity after averting this near disaster?

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

Inelastic Events

Granny (aka Roller Ruth) whizzes around the rink at 3m/s and confronts Ambrose scared motionless in front of her. Thinking quickly she picks up Ambrose and continues on. What is their velocity after averting this near disaster? pb = pa m1 v1 + m2 v2 = (m1 + m2 )va m1 = 80kg m2 = 40kg v1 = 3m/s V2 = 0 va

• ?

Inelastic Events

Granny (aka Roller Ruth) whizzes around the rink at 3m/s and confronts Ambrose scared motionless in front of her. Thinking quickly she picks up Ambrose and continues on. What is their velocity after averting this near disaster? pb = pa m1 v1 + m2 v2 = (m1 + m2 )va m1 = 80kg 80kgx3m/s + 40 kgx0 = (80+40)kgxva m2 = 40kg 240kgxm/s = 120kgxva v1 = 3m/s 120kg 120 kg v2 = 3m/s (240kgxm/s)/120kg = va va

• ? 2m/s = va

Inelastic Events

A truck with a mass of 2000kg moving at 15m/s hits a1000kg car moving in the

• pposite direction at 10m/s. What

is the direction and speed of the resulting combined wreckage?

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

Inelastic Events

A truck with a mass of 2000kg moving at 15m/s hits a 1000kg car moving in the opposite direction at 10m/s. What is the direction and speed of the resulting combined wreckage? pb = pa m1 v1 + m2 v2 = (m1 + m2 )va m1 = 2000kg m2 = 1000kg v1 = 15m/s V2 = -10ms va

• ?

Inelastic Events

A truck with a mass of 2000kg moving at 15m/s hits a 1000kg car moving in the opposite direction at 10m/s. What is the direction and speed of the resulting combined wreckage? pb = pa m1 v1 + m2 v2 = (m1 + m2 )va

2000kgx15m/s + 1000 kgx-10m/s = (2000+1000)kgxva 20000kgxm/s = 3000kgxva 3000 kg 3000 kg (20000kgxm/s)/3000kg = va 6.7m/s = va m1 = 2000kg m2 = 1000kg v1 = 15m/s V2 = -10m/s va

• ?

Your Teacher will now choose problems from the packet and assign inelastic Event problems on which you will work. Conservation of Momentum Inelastic Event Problems

Elastic Events

In 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 Events

In a closed system when the objects are apart AFTER the event, the event is elastic. m1 – mass of the first object m2

• mass of the second object

v1

• velocity of the first object before the event

v2

• velocity of the second object before the event

v'1

• velocity of the first object after the event

v'2

• velocity of the second object after the event

Elastic Events

In 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 the dock at 4 m/s. What is the resultant velocity of the canoe (100 kg). 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. - If the objects are apart AFTER the event this is an elastic event.

Elastic Events

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

pb = pa m1 v1 + m2 v2 = m1 v'1 + m2 v'2

m1 – 50 kg m2 – 100 kg v1

v2

v'1

• 4 m/s

v'2

• ?

Elastic Events

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

pb = pa m1 v1 + m2 v2 = m1 v'1 + m2 v'2

m1 – 50 kg m2 – 100 kg v1

v2

v'1

• 4 m/s

v'2

• ?

50 kgx0 + 100 kgx0 = 50 kg x 4m/s + 100 kgxv'2 0 = 200 kgxm/s + 100 kgxv'2

• 200 kgxm/s

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

• 2 m/s

= v'2

Elastic Events

In a closed system when the objects are apart AFTER the event, the event is elastic.

A bullet of mass 0.03 kg is fired from a gun with a muzzle velocity of 500m/s. The gun recoils with a velocity of 5 m/s. What is the mass of the gun? 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. - If the objects are apart AFTER the event this is an elastic event.

Elastic Events

In a closed system when the objects are apart AFTER the event, the event is elastic.

A bullet of mass 0.03 kg is fired from a gun with a muzzle velocity of 500m/s. The gun recoils with a velocity of 5 m/s. What is the mass of the gun?

pb = pa m1 v1 + m2 v2 = m1 v'1 + m2 v'2

m1 0.03 kg m2 ? v1 v2 v'1 500 m/s v'2

• 5 m/s

Elastic Events

In a closed system when the objects are apart AFTER the event the event is elastic.

A bullet of mass 0.03 kg is fired from a gun with a muzzle velocity of 500m/s. The gun recoils with a velocity of 5 m/s. What is the mass of the gun?

pb = pa m1 v1 + m2 v2 = m1 v'1 + m2 v'2

m1 0.03 kg m2 ? v1 v2 v'1 500 m/s v'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

Elastic Events

In a closed system when the objects are apart AFTER the event the event is elastic.

A 2 kg cart traveling right at 30 m/s collides and bounces off a 1 kg cart traveling at 10 m/s. After the collision the velocity of the first cart was 18m/s. What was the velocity of the second cart? 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. - If the objects are apart AFTER the event this is an elastic event.

Elastic Events

In a closed system when the objects are apart AFTER the event the event is elastic.

A 2 kg cart traveling right at 30 m/s collides and bounces off a 1 kg cart traveling at 10 m/s. After the collision the velocity of the first cart was 18m/s. What was the velocity of the second cart?

pb = pa m1 v1 + m2 v2 = m1 v'1 + m2 v'2

m1 – 2 kg m2 – 1 kg v1

• 30m/s

v2

• 10

v'1

• 18 m/s

v'2

• ?

Elastic Events

In a closed system when the objects are apart AFTER the event, the event is elastic.

A 2 kg cart traveling right at 30 m/s collides and bounces off a 1 kg cart traveling at 10 m/s. After the collison the velocity of the first cart was 18m/s. What was the velocity of the second cart?

pb = pa m1 v1 + m2 v2 = m1 v'1 + m2 v'2

m1 – 2 kg m2 – 1 kg v1

• 30 m/s

v2

• 10 m/s

v'1

• 18 m/s

v'2

• ?

2 kg x 30m/s + 1 kg x 10m/s = 2 kg x 18m/s + 1 kgxv'2 60 kgxm/s + 10 kgxm/s = 36 kgxm/s + 1 kgxv'2 70 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 from the packet and assign Elastic Event problems on which you will work. Conservation of Momentum Elastic Event Problems