Momentum

Notes (HRW p196)

Momentum 

Definition

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The measure of how difficult it is to stop a moving object Mass in Action!

Formula

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p = mass * velocity, or p = mv

Units

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kg.m/s

Momentum is a vector! Mass in Action!

Example 

Q: What is the momentum of a 1000 kg Civic traveling at 30 m/s?

• p = mass x velocity = m x v • p = 1000 x 30 = 30,000 kg.m/s

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Q: What is the momentum of a 100,000 kg locomotive traveling at 30 m/s?

• P = mass x velocity • P = 100,000 x 30 = 3,000,000 kg.m/s

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Q: What is the momentum of a 40,000 ton (40,000,000 kg) oil tanker traveling at 5 m/s?

• P = m * v = 40,000,000 * 5 • P = 200,000,000 kg.m/s Mass in Action!

Impulse 

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Definition

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The product of the force (F) acting on an object and the duration of the force (t)

Formula

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Impulse = F * t

Units

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Newtons.seconds (N.s)

Examples

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Striking a golf/foot/base ball Seatbelts (how do they work?)

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Auto safety developments since 50’s

Mass in Action!

Impulse Example

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Example: Wall exerts a force of 10,000 N on the van. The contact time is 0.01 s. What is the impulse? Solve: Impulse = F * t = 10,000 * 0.01 • Impulse = 100 N-s Mass in Action!

Ex. Impulse = Momentum Change

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Impulse = change in momentum (final – initial) Impulse = 0 – mv Ft = -mv (momentum is a vector!) F = mv/t (force felt is the momentum/duration of the applied force) Mass in Action!

Impulse and Momentum Change 

According to Newton’s 2nd Law

• The application of a net force on an object will • • • • • •

result in the object accelerating (aka changing velocity) F = ma = m(Δv/t) = m(vf – vo)/t, but Ft= mvf – mvo = pf – po thus Ft (or impulse) = change in momentum Ft = Δmv = mΔv F = mΔv/t F = m(vf – v0)/t (Newton’s 2nd law in momentum terms)

Mass in Action!

Example – Madness!

Mass in Action!

Impulse Example 

A 1000 kg Civic is traveling at 30 m/s and accelerates to 40 m/s in 10 seconds.

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What is the momentum of the car before accelerating?

• po = m*v = 1000 * 30 = 30,000 kg.m/s

What is the momentum of the car after accelerating for 10 seconds?

• pf = m*v = 1000*40 = 40,000 kg.m/s

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What is the change in momentum?

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What is the impulse?

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What is the net force that causes the change?

• Δp = pf – po = 40,000 – 30,000 = 10,000 kg.m/s • J = Change in momentum = 10,000 kg.m/s • F = change in momentum/time = 10,000/10 = 1,000 N

Mass in Action!

F*t = change in momentum = mΔv

Impulse Example 

A 1000 kg Civic is traveling at 30 m/s and hits a lamp post.

• What is the momentum of the car while moving?

• po = m*v = 1000*30 = 30,000 kg.m/s

• What is the momentum of the car after hitting the post?

• pf = m*v = 1000*0 = 0 kg.m/s

• What is the change in

momentum?

• Δp = pf – po = 0 – 30,000 = -30,000 kg.m/s

Mass in Action!

Impulses and Contact Time How does momentum of the vehicle relate to the impulse in the 2 scenarios below?

Force is spread over a longer duration

Force is spread over a shorter duration! Mass in Action!

Summary: Momentum - Impulse 

Interrelated

• Momentum • Impulse • Change in momentum IMPULSE Force * time F*t

Mass in Action!

Change in momentum F*t = m(vf – vo)

Momentum Δp = mvf - mvo

Summary

Mass in Action!

Practice 

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A stationary 0.12 kg hockey puck is hit with a force that lasts for 0.01 seconds and makes the puck move at 20 m/s. With what force was the puck hit? Impulse = Change in momentum

• Ft = mΔv = m(vf – vo) • F = mΔv/t • F = 0.12 (20 – 0)/0.01 = 240 N

Mass in Action!

Practice 

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If a 5kg object experiences a 10 N force for 0.1 second, what is the change in momentum of the object? Solve

• Change in momentum = impulse • Impulse = F*t • Change in momentum = F*t = 10*0.1 = 1N-s

Mass in Action!

Practice 

A 50 kg driver of a sports car is traveling at 35 m/s when she hits a large deer. She strikes the air bag/seatbelt combination that brings her body to a stop in 0.5 seconds.

• What average force does the bag/belt exert on her?

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Solve: m=50, vo=35, vf=0, t=0.5, F=?

• F = m*(vf-vo)/t • F = 50*(-35)/0.5 = -3500 N Mass in Action!

Practice (continued) 

What if the driver in the previous example was not wearing a seatbelt and there were no airbags, and the windshield stopped her head in 0.002 s.

• What is the average force on her head?

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Solve

• F = 50*-35/0.002 = -875,000 N!!!

Mass in Action!

Conservation of Momentum Sum of the momenta of all elements before the event = Sum of the momenta of all elements after the event

Momentum Before firing = p(rifle) + p(bullet) = 0 Momentum After firing = p(rifle) + p(bullet) = 0 After firing, the opposite momenta cancel – direction is important in vector arithmetic! Mass in Action!

Conservation of Momentum – Collisions (aka events) 

2 basic types of collisions for analysis

• Elastic

• Bodies collide and bounce apart – no energy loss • Bowling ball/pin • Pool

• Inelastic

• Bodies collide and stick together – some energy transformation into heat • Auto rear-ender • Picking up an object (combining)

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Real world – a bit of both! Mass in Action!

Momentum Table Before vs After BEFORE

Object 1

Object 2

Total Momentum

Mass (m)

m1

m2

Velocity (v)

v1

v2

Momentum

m1v1

m2v2

m1v1 + m2v2

AFTER

Object 1

Object 2

Total Momentum

Mass (m)

m1

m2

Velocity (v)

v1

v2

Momentum

m1v1

m2v2

m1v1 + m2v2

Equate the total momentum (before) with the total momentum (after) to solve!

Mass in Action!

Practice using table 

A 1000 kg Honda @ 30 m/s collides (head-on glancing blow) with a 2000 kg Camry @ 20 m/s. If the Honda continues @ 20 m/s,

• what is the speed of the Camry after the impact?

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Solve: Honda, Camry, collision

• Total momentum (before) = Total momentum (after) • phonda + pcamry = phonda + pcamry • 1000*30 + 2000*(-20) = 1000*20 + 2000*v • 30,000 – 40,000 = 20,000 + 2000v • v = -15 m/s (negative direction) Mass in Action!

Practice using table 

A 200 kg astronaut leaves the Shuttle for a space walk and while stationary his tether breaks. To return to the craft he throws a 2 kg hammer @ 3m/s directly away from the shuttle.

• What is his returning speed?

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Solve: astronaut, hammer, throw

• pastro + phammer (before) = pastro + phammer (after) • 200*0 + 2*0 = 200*v + 2*3 • v = -200/6 = -0.03 m/s Mass in Action!

Practice using table 

A 500 kg bumper car @ 5 m/s runs into the back of a similar car @ 2 m/s. If the 2nd car bounces forward at 4 m/s,

• What is the speed of the 1st car?

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Solve: bumper car #1, #2, collision

• (p1 + p2)before = (p1 + p2)after • 500*5 + 500*2 = 500*v + 500*4 • 2500 + 1000 = 500v + 2000 • v = 3 m/s Mass in Action!