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Chapter 7; Chapter 8
KD5 CONSERVATION OF
MOMENTUM AND ENERGY
Review: Momentum
 Momentum - Mass in motion
 momentum = mass X velocity
 Vector quantity (magnitude and direction)
Review: Momentum
 What are the 2 ways to
increase momentum?
 If the velocity of an object
doubles, what happens to
the momentum?
 Which has more momentum
a car moving down a hill or a
roller skate moving with the
same speed?
 Which has more momentum
a skateboard moving at 2
m/s or a truck at a red
Increase mass OR
increase velocity
Momentum Doubles
Car due to more mass
with the same velocities
Skateboard due to the
Truck has no velocity
Thus no momentum
Review: Momentum
 Calculate the momentum of a
90 kg football player running
at 6 m/s.
540 kg m/s
Law of Conservation
 Physical properties in an isolated
system don’t change
 Just transferred from 1 object to another
 Examples:
 Law of Conservation of Momentum
 Law of Conservation of Energy
 Law of Conservation of Mass
Law of Conservation of
Momentum
 Example: 2 cars collide
 The momentum of the 2 vehicles before the
collision is equal to the total momentum of
the 2 vehicles after the collision
 The momentums of each car might change but
the total momentum remains the same
Law of Conservation of
Momentum
 The total momentum of any closed,
isolated system does not change
 Momentum is NEVER lost!
 Just transferred
Law of Conservation of
Momentum
 p = p or
 m1v1 + m2v2 = m1v1 + m2v2
Example
 A ball with a mass of 0.10 kg moves to
the right with a speed of 2.0 m/s. It
hits a .040 kg ball which is standing
still. After the balls hit, the 2nd ball
moves to the right with a speed of 0.80
m/s. What is the velocity of the 1st
ball?
Answer
 A ball with a mass of 0.10 kg moves to the right
with a speed of 2.0 m/s. It hits a .040 kg ball
which is standing still. After the balls hit, the
2nd ball moves to the right with a speed of 0.80
m/s. What is the velocity of the 1st ball?
m(ball 1) v(ball 1) + m(ball 2) v(ball 2) = m(ball 1) v(ball 1) + m(ball 2) v(ball 2)
.1kg (2.0 m/s) + .04kg (0 m/s) = .1 kg (v) + .04 (.8 m/s)
.2 + 0 = .1 (v) + .032
.168 = .1 (v)
1.68 = v
Collisions
 p (before a collision or explosion) = p
(after a collision or explosion)
2 Types of Collisions
 1. Elastic Collision-Objects hit and
bounce off
 Both momentum and kinetic energy are
conserved
 Ex) Pool balls collide
2 Types of Collisions
 2. Inelastic Collision-Objects hit and
stick together
 Only momentum is conserved
 Ex) 2 cars collide and stick together
Energy
 Energy-The ability to do work
 Forms of energy:
 Electrical
 Chemical
 Solar
 Thermal
 Mechanical
Mechanical Energy
 1. Kinetic energy
 2. Gravitational Potential Energy
 3. Elastic Potential Energy
Mechanical Energy: Kinetic
 Kinetic Energy-Energy of motion
 KE = ½ mv²
 m=mass (kg)
 v=velocity (m/s)
 KE=kinetic energy (J)
Mechanical Energy: Gravitational
 Gravitational Potential Energy-Stored
energy of position
 GPE = m g h
 m=mass (kg)
 g=9.8 m/s²
 h=height (m)
 GPE=gravitational potential energy (J)
Mechanical Energy: Elastic
 Elastic Potential Energy-Stored energy
of stretch or compression
 EPE = ½ k x²
 k=elastic constant (N/m)
 x=distance stretched/compressed from
rest (m)
 EPE=elastic potential energy (J)
Elastic Potential Energy
Law of Conservation of Energy
 The total energy of a system remains
constant
 Energy is neither created or destroyed
(only transferred to another form)
 Total Energy Before = Total Energy After
 KE + GPE + EPE = KE´ + GPE´ + EPE´
“The Process” to Solve Energy
Problems
 1. Sketch a picture of the problem
 2. Identify the types of mechanical
energy present at the important stages
of the problem
 3. Set the before energies equal to the
after
 4. Substitute
 5. Solve and Label
Example:
 Bill throws a 0.1 kg ball straight up with
a speed of 7.5 m/s. How high did the
ball go?
Remember
 Bill throws a 0.1 kg ball straight up with
a speed of 7.5 m/s. How high did the
ball go?
KE + GPE + EPE = KE´ + GPE´ + EPE´
½ mv2 + mgh + 1/2 kx2 = ½ mv2 + mgh + 1/2 kx2
Only velocity to start with (only KE)
= Only height to end with (only GPE)
½ m v2 = mgh
½ (.1 kg) (7.5 m/s)2 = .1kg (9.8 m/s2) v2