Download Date: SPH4U Energy and Momentum Lesson 1

Date: ________________________
WORK
Work (W): the product of the magnitude of an object’s displacement and the component of the
applied force in the direction of the displacement. Work is a scalar.
𝑊 = 𝐹∆𝑑 cos 𝜃
𝑊 - Work done by the force in Joules (1 J = 1 N m)
𝐹 - the magnitude of the force applied in Newtons
∆𝑑 - the magnitude of the object’s displacement
𝜃 - the angle between the force and the displacement
Note that the component of force in the direction of motion is 𝐹𝑐𝑜𝑠𝜃. Recall that if θ = 0 then cosθ
= 1, if θ = 90˚ then cosθ = 0 and if θ = 180˚ then cosθ = -1.
Example 1: A wheelbarrow is being pushed with a force of 75 N at an angle of 40.0˚ to the
horizontal. Calculate the work done after the wheelbarrow has been pushed for 75 m.
SPH4U Energy and Momentum
Lesson 1 - Work
Page 1 of 3
Date: ________________________
Example 2: A hockey player stops by pushing her blades into the ice with a constant force of 105
N. The angle between the blades and the direction of motion is 135˚. If the player stops in 0.75
m, calculate the work done by the ice on the hockey player.
Example 3: A ball on a string is twirled in a horizontal circle. Determine the work the string
tension does on the ball over a complete revolution.
SPH4U Energy and Momentum
Lesson 1 - Work
Page 2 of 3
Date: ________________________
Example 4: A skid is being pulled across a rough surface with a constant force of 175 N at an
angle of 62˚ with the direction of motion. A frictional force of 82 N also acts on the skid. If the
skid is pulled for 45 m, calculate a) the work done by the pulling force, b) the work done by
friction and c) the total work done on the skid.
HOMEWORK – Nelson Physics 12
p. 166 practice #1,2,4 p. 167 practice #1,3 p. 168 practice #1,
p. 169 practice #1,2 p. 170 questions #1,2,4,5 Read pages 171 to 176 to prepare for Lesson 2
SPH4U Energy and Momentum
Lesson 1 - Work
Page 3 of 3
Date: ________________________
WORK-KINETIC ENERGY THEOREM
Kinetic Energy (Ek): the energy an object has because of its motion. Kinetic
energy is a scalar.
1
𝐸𝑘 = 𝑚𝑣 2
2
𝐸𝑘 - energy an object has due to its mass and speed in J
𝑚 - the mass of the object in kg
𝑣 - the speed of the object in m/s
Example 1: A car with a mass of 2155 kg is travelling at 22.5 m/s. Calculate the
car’s kinetic energy.
Work-Energy Theorem: the total work done on an object by an external force
equals the change in kinetic energy of the object. This assumes no other forms of
energy transformations, and that the applied force is constant.
1
1
𝑚𝑣𝑓2 − 𝑚𝑣𝑖2
2
2
= 𝐸𝑘𝑓 − 𝐸𝑘𝑖
𝑊=
𝑊 = ∆𝐸𝑘
SPH4U Energy and Momentum
Lesson 2 - Work -Kinetic Energy Theorem
Page 1 of 3
Date: ________________________
Example 2: A car with a mass of 2135 kg is travelling at 22.5 m/s when it starts
to speed up. If 2.95 x 105J of work are done on the car, what is the car’s final
speed?
SPH4U Energy and Momentum
Lesson 2 - Work -Kinetic Energy Theorem
Page 2 of 3
Date: ________________________
Example 3: A curler wants to release a 17 kg stone that will travel 31 m before
stopping. If the coefficient of kinetic friction between the ice and the stone is
0.012, calculate the initial speed at which the curler must release the stone.
HOMEWORK – Nelson Physics 12
p. 172 practice #1,2 p. 175 practice #1,2,3 p. 176 question #1, 3, 7, 10
Read pages 177 to 181 to prepare for Lesson 3.
SPH4U Energy and Momentum
Lesson 2 - Work -Kinetic Energy Theorem
Page 3 of 3
Date: ________________________
GRAVITATIONAL POTENTIAL ENERGY
Potential Energy: the stored energy an object has that can be converted into another form of
energy. For example, gasoline has chemical potential energy, an object on the edge of a cliff has
gravitational potential energy and a stretched rubber band has elastic potential energy.
Mechanical Energy: the sum of an object’s kinetic and mechanical potential energies.
Gravitational Potential Energy (Eg): the stored energy an object has because of its position and
the applied gravitational force.
When lifting an object, the work can be calculated by:
𝑊 = 𝑚𝑔∆𝑦
𝑊 - work to lift the object in J
𝑚 - the mass of the object in kg
𝑔 - the acceleration due to gravity in m/s2
∆𝑦 – the vertical displacement of the object in m
The work done when lifting an object is transformed into gravitational potential energy (GPE) of
the object. Change in gravitational potential energy can be calculated by:
∆𝐸𝑔 = 𝑚𝑔∆𝑦
∆𝐸𝑔 - the change in GPE of an object relative to a reference
position, in J
𝑚 - the mass of the object in kg
𝑔 - the acceleration due to gravity in m/s2
∆𝑦 – the vertical displacement of the object in m, relative to
the reference position
Note: the equation is valid only for situations where the change in height does not result in a
significant change in the value of g.
SPH4U Energy and Momentum
Lesson 3 – Gravitational Potential Energy
Page 1 of 2
Date: ________________________
Example 1: A box of books is lifted 0.75 m from the ground to the back of a truck, increasing the
gravitational potential energy by 88 J. Calculate the mass of the box of books.
Example 2: A weightlifter raises a 75 kg barbell from his shoulders to a position 0.45 m above his
shoulders. a) How much work was done? b) What is the gravitational potential energy of the
barbell relative to his shoulders?
HOMEWORK – Nelson Physics 12
p. 180 practice #1,2 p. 181 question #1, 3, 4, 7
Read pages 184 to 191 to prepare for Lesson 4
SPH4U Energy and Momentum
Lesson 3 – Gravitational Potential Energy
Page 2 of 2
Date: ________________________
THE LAW OF CONSERVATION OF ENERGY
The Law of Conservation of Energy: Energy is neither created nor destroyed in. It can only
change form.
Conservation of energy is one of the fundamental principles in Physics. The total initial energy of a
closed system must be equal to the total final energy of a closed system.
Closed or isolated system: a system that cannot interact or exchange energy with external
systems
𝐸𝑇𝑖 = 𝐸𝑇𝑓
𝐸𝑇𝑖
- initial total energy in J
𝐸𝑇𝑓
- final total energy in J
If a closed system only contains gravitational potential energy and kinetic energy, then:
𝐸𝑔𝑖 + 𝐸𝑘𝑖 = 𝐸𝑔𝑓 + 𝐸𝑘𝑓
𝐸𝑔𝑖
𝐸𝑘𝑖
𝐸𝑔𝑓
𝐸𝑘𝑓
- initial gravitational potential energy in J
- initial kinetic energy in J
- final gravitational potential energy in J
- final kinetic energy in J
Open system: a system that can interact with other external systems and exchange various forms
of energy including thermal, elastic, electrical, chemical and biochemical, light and sound.
Biochemical energy: a type of chemical potential energy stored in the cells and other basic
structures of living organisms
SPH4U Energy and Momentum
Lesson 4 – The Law of Conservation of Energy
Page 1 of 3
Date: ________________________
Example 1: A 52 kg skier slides to the top of a 40.0 m high hill at a speed of 2.00 m/s, and
continues down the frictionless surface. a) What is the skier’s total mechanical energy? b) What
will the skier’s speed be at 25.0 m?
SPH4U Energy and Momentum
Lesson 4 – The Law of Conservation of Energy
Page 2 of 3
Date: ________________________
Power: the rate of work done by a force over time, or the rate at which the energy of an open
system changes.
𝑃=
𝑊
𝑡
𝑃
- power in Watts (W), 1 W = 1 J/s
𝑊
- total work done in J
𝑡
- time to complete the work in s
Example 2: A 62 kg woman climbs 22 flights of stairs in 30.0 minutes. Each flight is 4.0 m high. a)
What is her power output? b) Where does she get the energy from?
HOMEWORK – Nelson Physics 12
p. 187 practice #1,2 p. 190 practice #1,3 p. 191 questions #1,3,4,5,6
Read pages 192 to 200 to prepare for Lesson 5
SPH4U Energy and Momentum
Lesson 4 – The Law of Conservation of Energy
Page 3 of 3
Date: ________________________
ELASTIC POTENTIAL ENERGY
Spring Force – A spring that is stretched or compressed from its rest position has stored energy
called Elastic Potential Energy, due to the force of the compressed or stretched spring.
(Diagram from Nelson Physics 12, 2012)
Hooke’s Law – The force exerted by a spring is directly proportional to the displacement of the
spring.
𝐹⃗𝑥 = −𝑘∆𝑥⃗
𝐹⃗𝑥 – force exerted by a stretched or compressed spring, measured in N
𝑘 – spring constant that corresponds to the stiffness of the spring, in N/m.
Stiffer springs have
higher spring constants.
∆𝑥⃗ – the displacement of the spring from the rest position in m.
Notice – the direction of force is opposite to the direction of the displacement
Ideal spring – any spring that obeys Hooke’s Law; it does not experience any friction (internal or
external).
SPH4U Energy and Momentum
Lesson 5 – Elastic Potential Energy and Simple Harmonic Motion
Page 1 of 4
Date: ________________________
Example 1: A 0.250 g mass hangs from a spring with a spring constant of 32.0 N/m. If the mass is
lifted so that the spring is compressed by 0.150 m from the rest position, calculate the
acceleration of the mass when it is released.
Elastic Potential Energy – the potential energy due to the stretching or compressing of an elastic
material. The elastic potential energy in Joules of a compressed or stretched spring can be
calculated by:
1
𝐸𝑒 = 𝑘(∆𝑥)2
2
SPH4U Energy and Momentum
Lesson 5 – Elastic Potential Energy and Simple Harmonic Motion
Page 2 of 4
Date: ________________________
Example 2: A spring-loaded toy is compressed, so that it will eventually pop up from the table. If
a force of 35 N compresses the spring by 0.012 m, calculate the elastic potential energy of the toy.
Periodic Motion – When a mass attached to a stretched or compressed spring is released, it will
oscillate back and forth. This is a type of periodic motion. If this happens on a frictionless
surface, the force experienced by the mass will be maximized at peak amplitude (displacement of
the spring) and decrease as the spring approaches its rest position. Hence the acceleration of the
mass will be proportional to the displacement from the rest position. This is a type of simple
harmonic motion.
Simple Harmonic Motion (SHM) – periodic motion in which the acceleration of the moving object is
proportional to its displacement.
To calculate the period (T) for the SHM of a mass on a spring:
𝑚
𝑘
𝑇 = 2𝜋√
T – period in s
SPH4U Energy and Momentum
Lesson 5 – Elastic Potential Energy and Simple Harmonic Motion
Page 3 of 4
Date: ________________________
Of course, frequency of any periodic motion is always:
𝑓=
1
𝑇
f – frequency in Hz (cycles per second)
Example 3: A mass-spring system undergoes SHM. The elastic potential energy at maximum
stretch is 8.25 J, the mass is 0.30 kg and the spring constant is
228 N/m. Calculate the frequency and amplitude of oscillation.
HOMEWORK – Nelson Physics 12
p. 195 practice #1,2 p. 196 practice #1,2 p. 199 practice #1, p. 200 questions #1,4,7
Read pages 201 to 208 to prepare for Lesson 6
SPH4U Energy and Momentum
Lesson 5 – Elastic Potential Energy and Simple Harmonic Motion
Page 4 of 4
Date: ________________________
SPRINGS AND CONSERVATION OF ENERGY
Elastic potential energy is just another form of mechanical energy. In systems such as safety
systems, the law of conservation of energy is upheld. Springs or other elastic materials “absorb”
energy, so that it is not transferred to humans or valuable equipment.
Example 1: A 2.0 kg block is on a frictionless ramp with a height of 1.5 m, to the right of a spring
that has a spring constant of 4900 N/m. If the block is released
a) what is the maximum compression of the spring? b) calculate the speed of the block when the
spring has been compressed by 2.0 cm.
SPH4U Energy and Momentum
Lesson 7 – Springs and Conservation of Energy
Page 1 of 3
Date: ________________________
Example 2: A block with a mass of 3.5 kg is held against a compressed spring. The spring is
compressed by 16 cm from the equilibrium position and has a spring constant of 220 N/m. After
the block is released it travels along a frictionless surface and then up a frictionless ramp that has
an angle of inclination of 27°.
a)
Determine the stored elastic potential energy stored in the spring before the block is
released.
b) Calculate the speed of the block on the horizontal surface.
c) Determine how far along the ramp the block will travel before it stops.
SPH4U Energy and Momentum
Lesson 7 – Springs and Conservation of Energy
Page 2 of 3
Date: ________________________
Perpetual Motion Machine – a machine that can operate forever without restarting or refuelling.
Such a machine is impossible to make, because due to friction, energy in a system is always lost,
often as heat.
Damped Harmonic Motion – Periodic motion that is changed by friction. Typically amplitude
decreases gradually over time. If the amplitude decreases rapidly, it is called overdamped. If the
amplitude decreases very slowly, it is called underdamped. If the amplitude decrease time falls
between an overdamped and underdamped case, it is called critically damped.
HOMEWORK – Nelson Physics 12
p. 205 practice #1, 2,3, 4 p. 208 questions #1, 3, 5,9
Read pages 210 to 211 and prepare for Investigation 4.5.1
SPH4U Energy and Momentum
Lesson 7 – Springs and Conservation of Energy
Page 3 of 3
Date: ________________________
MOMENTUM AND IMPULSE
Linear momentum (𝑝⃗) – a quantity that describes the motion of an object travelling in a straight
line as the product of its mass and velocity. Momentum is a vector.
𝑝⃗ = 𝑚𝑣⃗
𝑝⃗ – linear momentum, kg∙m/s
𝑚 – mass, kg
𝑣⃗ – velocity, m/s
Note: the term momentum will refer to linear momentum in class notes and the textbook.
Example 1: Calculate the momentum of a 4.10 x 105 kg jet travelling north at 658 km/h.
Impulse ( 𝐹⃗ ∆𝑡 ) – the product of force and time that acts on an object to produce a change in
momentum.
𝐹⃗ ∆𝑡 = ∆𝑝⃗
𝐹⃗ ∆𝑡 – impulse, N∙s
𝐹⃗ – applied force, N
∆𝑡 – time that force is applied, s
∆𝑝⃗ – change in momentum after force has been applied, N∙s
SPH4U Energy and Momentum
Lesson 10 – Momentum and Impulse
Page 1 of 2
Date: ________________________
Example 2: A 0.060 kg tennis ball is hit horizontally by a tennis racket. The ball’s final velocity is
43 m/s [forward]. Calculate the impulse applied by the racket. If the racket was in contact with
the ball for 0.021 s, what average force was applied to the ball?
Force – time Graphs – The area under a force time graph is equal to the impulse.
HOMEWORK – Nelson Physics 12
p. 223 practice # 1,2, p. 226 practice # 1,2 p. 227 questions #2,3,5,7,10,12
Read pages 228 to 232 to prepare for Lesson 10
SPH4U Energy and Momentum
Lesson 10 – Momentum and Impulse
Page 2 of 2
Date: ________________________
CONSERVATION OF MOMENTUM IN ONE DIMENSION
The Law of Conservation of Momentum– When two or more objects collide, the collision does not
change the total momentum of the two objects. Whatever momentum is lost by one object in the
collision is gained by the other. The total momentum of the system is conserved.
𝑚1 𝑣⃗𝑖1 + 𝑚2 𝑣⃗𝑖2 = 𝑚1 𝑣⃗𝑓1 + 𝑚2 𝑣⃗𝑓2
Notes:

The law of conservation also applies to complex systems with multiple objects. For example,
when the cue ball hits the other balls in the break shot in a game of pool, the total
momentum of all the balls will equal the initial momentum of the cue ball.

Momentum is also conserved in the case of an explosion.
Example 1: Skater 1, a 72 kg man, is moving at 3.1 m/s [forward] when he comes up behind Skater
2 and grabs her waist. If Skater 2 is moving at 2.2 m/s [forward] when this happens, and she has a
mass of 47 kg, what is the final velocity of the two skaters?
SPH4U Energy and Momentum
Lesson 11 – Conservation of Momentum
Page 1 of 2
Date: ________________________
Example 2: An asteroid is headed towards Earth. NASA experts must blow it into two equal parts
that will break apart from one another. The original asteroid has a mass of 1.8 x 109 kg. To safely
miss Earth, the explosion must cause each half to travel a minimum of 8.0 x 10 6 m at a right angle
away from Earth within 24 h. Assume this is a one dimensional problem.
a) Calculate the momentum of each part after the explosion. b) Determine the impulse delivered
to each part during the explosion.
HOMEWORK – Nelson Physics 12
p. 231 practice # 1,2, p. 232 questions #1,3,4,5,6,8
Read pages 233 to 239 to prepare for Lesson 11
SPH4U Energy and Momentum
Lesson 11 – Conservation of Momentum
Page 2 of 2
Date: ________________________
ELASTIC AND INELASTIC COLLISIONS
Elastic Collision – A collision in which both momentum and kinetic energy are conserved. Perfectly
elastic collisions are idealized collisions in which external forces are minimized to the point where
both momentum and kinetic energy are perfectly conserved.
Inelastic Collision – A collision in which only momentum is conserved. Kinetic energy is not
conserved.
Perfectly Inelastic Collision - A collision in which two colliding objects stick together after a
collision, therefore having the same final velocity.
In today’s lesson we will only be considering collisions in one dimension.
Example 1: A 0.500 kg block of plastic slides across the ice at 3.15 m/s [forward]. It collides with
a stationary 3.50 kg block of plastic. Assume that the collision is perfectly elastic and find the
final velocity of both of the blocks of wood.
SPH4U Energy and Momentum
Lesson 12 – Collisions
Page 1 of 3
Date: ________________________
SPH4U Energy and Momentum
Lesson 12 – Collisions
Page 2 of 3
Date: ________________________
Example 2: A car with a mass of 1825 kg is moving forward at 14.7 m/s when it is struck by a car
with a mass of 2252 kg that is moving forward at 22.6 m/s. Assume that this is a perfectly
inelastic collision and calculate the final velocity of the cars.
HOMEWORK – Nelson Physics 12
p. 236 practice # 1,2, p. 238 practice # 1,2 p. 239 questions #1,2,4,6,8
Read pages 241 to 244 to prepare for Lesson 12
SPH4U Energy and Momentum
Lesson 12 – Collisions
Page 3 of 3
Date: ________________________
HEAD ON ELASTIC COLLISIONS
Perfectly Elastic Head-On Collisions in One Dimension – These occur when two objects approach
from opposite directions. After the collision both momentum and kinetic energy is conserved. By
manipulating the momentum and energy equations (see Nelson Physics 12 pages 240 – 241) we can
find the final velocities using the following equations:
𝑣⃗𝑓1 = (
𝑚1 − 𝑚2
2𝑚2
) 𝑣⃗𝑖1 + (
) 𝑣⃗
𝑚1 + 𝑚2
𝑚1 + 𝑚2 𝑖2
𝑣⃗𝑓2 = (
𝑚2 − 𝑚1
2𝑚1
) 𝑣⃗𝑖2 + (
) 𝑣⃗
𝑚1 + 𝑚2
𝑚1 + 𝑚2 𝑖1
Example 1: A car with a mass of 1820 kg moving at 3.55 m/s [W] strikes a stationary car that has
a mass of 2110 kg. Find the find the final velocity of each car after the collision.
SPH4U Energy and Momentum
Lesson 13 – Head on Elastic Collisions
Page 1 of 3
Date: ________________________
Example 2: A car with a mass of 1820 kg moving at 3.55 m/s [W] strikes a car moving at 1.78 m/s
[E] that has a mass of 2110 kg. Find the find the final velocity of each car after the collision.
SPH4U Energy and Momentum
Lesson 13 – Head on Elastic Collisions
Page 2 of 3
Date: ________________________
Special Cases
Case 1: OBJECTS HAVE THE SAME MASS
If the two colliding objects have the same mass, then the final velocity equations can be
manipulated to discover that 𝑣⃗𝑓1 = 𝑣⃗𝑖2 and 𝑣⃗𝑓2 = 𝑣⃗𝑖1 . (In English we would say that the two objects
switched velocities).
Case 2: A LIGHTER OBJECT COLLIDES WITH A MUCH HEAVIER, STATIONARY OBJECT
If a moving object collides with a much heavier stationary object, then the equations can be
manipulated to discover the 𝑣⃗𝑓1 = −𝑣⃗𝑖1 and 𝑣⃗𝑓2 = 0. (In other words the lighter object bounces off
the heavier object with a velocity equal and opposite in direction that it hit at.)
HOMEWORK – Nelson Physics 12
p. 243 practice # 1,2, p. 248 questions #1,2,3,4
Read pages 244 to 248 to prepare for Lesson 13
SPH4U Energy and Momentum
Lesson 13 – Head on Elastic Collisions
Page 3 of 3
Date: ________________________
HEAD ON ELASTIC COLLISIONS WITH CONSERVATION OF MECHANICAL
ENERGY
When two objects collide and one or both have compressible bumpers attached, the time of contact
between the two objects is increased (compared to a collision without bumpers). This occurs
because during the collision, the bumper compresses and then springs back to its original position.
Some of the energy of the collision is temporarily stored as elastic potential energy. When the
bumper springs back the energy is converted to kinetic energy. By applying the law of conservation
of energy to this scenario, we get:
1
1
1
1
1
2
2
2
2
𝑚1 𝑣𝑖1
+ 𝑚2 𝑣𝑖2
= 𝑚1 𝑣𝑓1
+ 𝑚2 𝑣𝑓2
+ 𝑘𝑥 2
2
2
2
2
2
In the graph below, (Nelson Physics 12, p. 244) we can see that as the bumper compresses, Ee
increases and total Ek of the two objects decrease. The total mechanical energy remains constant.
As the compression decreases, the Ee decrease, and the total Ek increases. We see that the total
mechanical energy remains constant through the collision.
Note: When two objects collide and the spring is at maximum compression they will have the same
velocity!
SPH4U Energy and Momentum
Lesson 14 – Head on Collisions with Cushioning
Page 1 of 3
Date: ________________________
Example 1: Cart 1, a 1.75 kg dynamics cart with a spring attached (spring constant of 7.5 x 10 4
N/m) is moving at 3.0 m/s [left] when it collides head on with a 2.5 kg dynamics cart (Cart 2). Cart
2 has no spring and is moving at 1.5 m/s [right] when the carts collide. Assume this is an elastic
collision. a) What is the compression of the spring when cart 2 is moving at 1.0 m/s [right]? b)
Calculate the maximum compression of the spring.
SPH4U Energy and Momentum
Lesson 14 – Head on Collisions with Cushioning
Page 2 of 3
Date: ________________________
HOMEWORK – Nelson Physics 12
p. 247 practice # 1,2 p. 248 question 5
SPH4U Energy and Momentum
Lesson 14 – Head on Collisions with Cushioning
Page 3 of 3
Date: ________________________
COLLISIONS IN TWO DIMENSIONS: GLANCING COLLISIONS
Glancing collision: a collision in which the first object, after an impact with the second, travels at
an angle to the direction it was originally travelling. For example, in pool, the cue ball strikes
another ball, and after the collision, the angle of travel of the cue ball has changed.
In glancing
collisions the x and y components of force and momentum must be considered separately.
The diagram below illustrates a cue ball striking another ball at an angle, causing both balls to
change direction (Nelson Physics 12, p.250)
Example 1: A hockey puck of mass 0.16 kg, sliding on a nearly frictionless surface of ice with a
velocity of 2.0 m/s [E] strikes a second puck at rest with a mass of 0.17 kg. The first puck has a
velocity of 1.5 m/s [E 31˚ N
] after the collision. Determine the velocity of the second puck
after the collision.
SPH4U Energy and Momentum
Lesson 15 – Collisions in 2 Dimensions
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Date: ________________________
SPH4U Energy and Momentum
Lesson 15 – Collisions in 2 Dimensions
Page 2 of 3
Date: ________________________
Example 2: An automobile collides with a truck at an intersection. The car, of mass 1.4 x 103 kg, is
travelling at 8.9 m/s [S]; the truck has a mass of 2.6 x 10 4 kg and is travelling at 13.3 m/s [E]. The
collision is perfectly inelastic. Determine their velocity just after the collision.
HOMEWORK – Nelson Physics 12
p. 252 practice # 1,2 p. 253 question 1, 5, 7
SPH4U Energy and Momentum
Lesson 15 – Collisions in 2 Dimensions
Page 3 of 3