Elephant has more inertia than skaterat

m < M

Moving

Not Moving

Elephant has no momentum at rest

What about momentum?

Moving

Elephant got momentum!

Momentum is inertia in motion

\mathrm{Momentum} = \mathrm{mass} \times \mathrm{velocity}

\mathrm{Momentum} = \mathrm{mass} \times \mathrm{velocity}

p = mv

p = mv

A container ship has lots of momentum even when moving slow because it has a lot of mass

Checkpoint

Which has more momentum, a 3000 kg elephant moving at 1 km/h or a 0.3 kg mice moving at 10000 km/h?

Both have the same momentum of 3000 kg km/h

p = mv

p = mv

Impulse is change in momentum

\mathrm{Impulse} = \mathrm{change \ in \ momentum}

\mathrm{Impulse} = \mathrm{change \ in \ momentum}

\mathrm{Force} \times \mathrm{time} = \Delta(mv)

\mathrm{Force} \times \mathrm{time} = \Delta(mv)

Ft = \Delta(mv)

Ft = \Delta(mv)

Impulse-momentum relationship

\mathrm{Impulse} = \mathrm{Force} \times \mathrm{time}

\mathrm{Impulse} = \mathrm{Force} \times \mathrm{time}

Ft = \Delta(mv)

Ft = \Delta(mv)

Impulse-Momentum Relationship

Ft = \Delta(mv)

Ft = \Delta(mv)

Large t, small F

Large F, small t

Impulse-Momentum Relationship

Dynamic ropes reduce the force by increasing the time to stop

Checkpoint

If a dynamic rope is able to increase the duration of a rock climber's fall by four times, how much would the force of impact change?

The force is reduced by four

Ft = \Delta(mv)

Ft = \Delta(mv)

Bouncing increases impulse

Change in momentum of teddy bear vs. bouncing ball

Bouncing increases impulse

Conservation of momentum

In the absence of an external force, the momentum of a system remains unchanged

p_\mathrm{cannon} + p_\mathrm{ball} = 0

p_\mathrm{cannon} + p_\mathrm{ball} = 0

p_\mathrm{cannon} + p_\mathrm{ball} = 0

p_\mathrm{cannon} + p_\mathrm{ball} = 0

Initial momentum

Final momentum

=

=

Conservation of momentum

m(v)

m(v)

Initial momentum

Final momentum

Big Fish

Small Fish

5m(\frac{v}{5})

5m(\frac{v}{5})

4m(0)

4m(0)

Small Fish

Big Fish

0

0

+

+

+

+

=

=

?

Elastic Collisions

Inelastic Collisions

Energy and momentum are conserved

Momentum is conserved but energy is not

Momentum is always conserved regardless of the type of collision!

vs.

Objects rebound without lasting deformation or heat generation

Objects get tangled and/or generate heat

Collisions

Net momentum before collision = Net momentum after collision

Checkpoint

What is the velocity of the blue truck after the collision?

5 m/s

See Solution

\sum{p_\mathrm{before}} = \sum{p_\mathrm{after}}

\sum{p_\mathrm{before}} = \sum{p_\mathrm{after}}

Checkpoint

What is the velocity of grandma and grandson after the collision?

4 m/s

See Solution

\sum{p_\mathrm{before}} = \sum{p_\mathrm{after}}

\sum{p_\mathrm{before}} = \sum{p_\mathrm{after}}

Momentum IS A VECTOR!

The sum of the momenta vectors is the same before and after the collision/explosion

Energy

Energy may be the most familiar concept in science, yet it is one of the most difficult to define

Here we are going to focus on Mechanical Energy: Work, Kinetic Energy and Potential Energy

Work

Work is proportional to force and distance

Work

Work is proportional to force and distance

\mathrm{Work} = \mathrm{Force} \times \mathrm{distance}

\mathrm{Work} = \mathrm{Force} \times \mathrm{distance}

W = F d

W = F d

Work

W = F d

W = F d

The wall is not moving, so d = 0 and thus W = 0

Is he doing work on the wall?

Work

W = Fx d

Fx

Fy

Fy is not doing any work since the block doesn't move on the y direction

Is the Earth doing work on the Moon?

No because the force is not in the direction of motion

Checkpoint 1

Units of Work

W = F d

W = F d

\mathrm{Work \ Units} = \mathrm{Force \ Units} \times \mathrm{Distance \ Units}

\mathrm{Work \ Units} = \mathrm{Force \ Units} \times \mathrm{Distance \ Units}

= N \times m = kg \frac{m}{s^2} \times m = kg \frac{m^2}{s^2} = 1 \ \mathrm{Joule}

= N \times m = kg \frac{m}{s^2} \times m = kg \frac{m^2}{s^2} = 1 \ \mathrm{Joule}

Work has units of Energy

Work is Energy in transfer!

Checkpoint 2

How much work is needed to lift a bag of groceries that weights 300 N to a height of 2 m?

600 Joules

W = F d

W = F d

Power

\mathrm{Power} = \frac{\mathrm{Work \ done}}{\mathrm{Time \ interval}} = \frac{W}{t}

\mathrm{Power} = \frac{\mathrm{Work \ done}}{\mathrm{Time \ interval}} = \frac{W}{t}

\mathrm{Power \ Units} = \frac{\mathrm{Work \ Units}}{\mathrm{Time \ Units}} = \frac{J}{s} = \mathrm{watt}

\mathrm{Power \ Units} = \frac{\mathrm{Work \ Units}}{\mathrm{Time \ Units}} = \frac{J}{s} = \mathrm{watt}

1 horsepower = 746 watts

Power is the rate at which work is done.

Power is the rate at which energy is transferred!

Checkpoint 3

To lift a bag of groceries up the stairs in half the time you need ________ as much power?

Twice

\mathrm{Power} = \frac{W}{t}

\mathrm{Power} = \frac{W}{t}

With twice as much power you can do either the same work in half the time or twice as much work in the same time

\mathrm{Work} = \Delta{\mathrm{E}}

\mathrm{Work} = \Delta{\mathrm{E}}

Work-Energy theorem

The Floor is doing work on the bicycle, thus the bicycle's Energy must change (it decreases as it stops moving).

The Energy due to the bicycle's motion turns into heat

Kinetic Energy

The energy due to an object moving

Kinetic Energy

\mathrm{KE} = \frac{1}{2} \mathrm{mass} \times \mathrm{velocity^2}

\mathrm{KE} = \frac{1}{2} \mathrm{mass} \times \mathrm{velocity^2}

\mathrm{KE} = \frac{1}{2} m v^2

\mathrm{KE} = \frac{1}{2} m v^2

Potential Energy

The energy that is stored and held in readiness due to the position of an object (the arrow in this case)

Potential Energy -> Kinetic Energy

The Potential Energy of the arrow is converted to Kinetic Energy

Potential Energy

The energy that is stored and held in readiness due to the position of an object (the car in this case)

Work --> Potential Energy

Potential Energy --> Kinetic Energy

Gravitational Potential Energy

The potential energy of the ball is independent of how the ball got to its position, in all cases

PE = mg (3 meters)

Gravitational Potential Energy = weight x height

\mathrm{PE} = mg \ h

\mathrm{PE} = mg \ h

h = 3 meters

The energy associated to the gravitational force and the work it does on objects

Checkpoint 4

Ignoring friction, which path requires the most work? (hint: remember that work is transfer of energy)

They all require the same amount of work =

\mathrm{PE} = mg \ h

\mathrm{PE} = mg \ h

Mechanical Energy

E = \frac{1}{2} m v^2 + mg h

E = \frac{1}{2} m v^2 + mg h

\mathrm{Mechanical \ Energy} = \mathrm{Kinetic \ Energy} + \mathrm{Potential \ Energy}

\mathrm{Mechanical \ Energy} = \mathrm{Kinetic \ Energy} + \mathrm{Potential \ Energy}

E = \mathrm{KE} + \mathrm{PE}

E = \mathrm{KE} + \mathrm{PE}

When the Potential Energy is gravitational:

Conservation of Energy

Energy cannot be created or destroyed: it may be transformed from one form into another, but the total amount of energy never changes

Conservation of Energy

E = \frac{1}{2} m v^2 + mg h = \mathrm{Constant}

E = \frac{1}{2} m v^2 + mg h = \mathrm{Constant}

E = \mathrm{KE} + \mathrm{PE} = \mathrm{Constant}

E = \mathrm{KE} + \mathrm{PE} = \mathrm{Constant}

When the Potential Energy is gravitational

Momentum and Energy