MARIAN SCHOOL OF QUEZON CITY
LESSON PLAN
ONLINE LEARNING
GRADE LEVEL:
12
TOPIC:
 Work, Energy and
Energy
Conservation
SUBJECT:
Time Frame:
PHYSICS 1
CONTENT STANDARD:
1 WEEK (WEEK 6)
LESSON OBJECTIVES:
 Calculate the dot or scalar product of vectors
 Determine the work done by a force acting on
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Dot or Scalar Product
Work done by a force
Work-energy relation
Kinetic energy
Power
Conservative and nonconservative
forces
Gravitational potential energy
Elastic potential energy
Equilibria and potential energy
diagrams
Energy Conservation, Work, and
Power Problems




PERFORMANCE STANDARD

Solve, using experimental and
theoretical approaches, multiconcept,
rich-content problems involving
measurement, vectors, motion in 1D
and 2D, Newton’s Laws, Work, Energy,
Center of Mass, momentum, impulse
and collisions



a system
Define work as a scalar or dot product of
force and displacement
Interpret the work done by a force in onedimension as an area under a Force vs.
Position curve
Relate the gravitational potential energy of a
system or object to the configuration of the
system
Relate the elastic potential energy of a
system or object to the configuration of the
system
Explain the properties and the effects of
conservative forces
Use potential energy diagrams to infer force;
stable, unstable, and neutral equilibria; and
turning points
Solve problems involving work, energy, and
power in contexts such as, but not limited to,
bungee jumping, design of roller-coasters,
number of people required to build
structures such as the Great Pyramids and the
rice terraces; power and energy requirements
of human activities such as sleeping vs. sitting
vs. standing, running vs. walking.
VALUING:
 Appreciate the subject matter
RESOURCES AND MATERIALS
 General Physics 1 (Vibal) pp. 123-147
 https://awwapp.com/#
 Powerpoint presentation
VIDEO RECORDING (VC)
5-10 minutes Video Presentation
SCRIPT
CONTENT/NOTES
 Dot Product
 What is energy?
 Very good! Energy is the capacity to do work.
A vector has magnitude (how long it is) and direction
 But in some circumstances not all energy is
Here are two vectors:
available to do work. Because association of
the energy with work, we will begin this lesson They can be multiplied using the "Dot Product"
The Dot Product is written using a central
with a discussion work. Work is intimately
dot:
related to energy and how energy moves from
a·b
one system to another, or changes form.
This means the Dot Product of a and b
 Let’s start with dot product?



How can we calculate a dot product?
What is work?
How can we calculate work done by a force?
We can calculate the Dot Product of two vectors this way:
a · b = |a| × |b| × cos(θ)
Where:
|a| is the magnitude (length) of
vector a
|b| is the magnitude (length) of
vector b
θ is the angle between a and b
So we multiply the length of a times the length of b, then
multiply by the cosine of the angle between a and b
 Work – work done on a system by a constant force
is the product of the component of the force in
the direction of motion times the distance through
which force acts.
W = Fd cos θ
W - work
d - distance
F – force
θ- angle between F and d
1. A person pulls a block 2 m along a horizontal surface by a
constant force F = 20 N. Determine the work done by
force F acting on the block.
Known :
Force (F) = 20 N
Displacement (s) = 2 m
Angle (θ) = 0
Wanted : Work (W)
Solution :
W = F d cos θ = (20)(2)(cos 0) = (20)(2)(1) = 40 Joule
ONLINE PROCESSING (VIDEO CONFERENCING)
PROCESS QUESTIONS/ ESSENTIAL QUESTION
CONTENT/NOTES
Online 1:
Kinetic Energy and Work-Energy Theorem
 What have you learned about the video?
The work-energy theorem states that the work done by all
 What is work energy theorem?
forces acting on a particle equals the change in the
particle’s kinetic energy.
 How can we calculate the net work using
work-energy theorem?
The Work-Energy Theorem
 What is Kinetic energy?
The principle of work and kinetic energy (also known as
 How can we calculate the amount of kinetic
the work-energy theorem) states that the work done by
the sum of all forces acting on a particle equals the change
energy in a body?
in the kinetic energy of the particle. This definition can be
 What is power?
extended to rigid bodies by defining the work of the
 How can we solve problems involving power?
torque and rotational kinetic energy.
The work W done by the net force on a particle equals the
change in the particle’s kinetic energy KE:
W=ΔKE=1/2mvf2−1/2mvi2
where vi and vf are the speeds of the particle before and
after the application of force, and m is the particle’s mass.
Sample Problem:
Suppose that you push on the 30.0-kg package in Figure
3 with a constant force of 120 N through a distance of
0.800 m, and that the opposing friction force averages
5.00 N.
1. Calculate the net work done on the package.
2. Solve the same problem as in part 1, this time by
finding the work done by each force that
contributes to the net force.
Strategy and Concept for Part 1
This is a motion in one dimension problem, because the
downward force (from the weight of the package) and the
normal force have equal magnitude and opposite
direction, so that they cancel in calculating the net force,
while the applied force, friction, and the displacement are
all horizontal. (See Figure 3.) As expected, the net work is
the net force times distance
Solution for Part 1
The net force is the push force minus friction, or Fnet = 120
N – 5.00 N = 115 N. Thus the net work is
Wnet=Fnetd=(115 N)(0.800 m) =9.20 N⋅m=92.0 J
Discussion for Part 1
This value is the net work done on the package. The
person actually does more work than this, because friction
opposes the motion. Friction does negative work and
removes some of the energy the person expends and
converts it to thermal energy. The net work equals the
sum of the work done by each individual force.
Strategy and Concept for Part 2
The forces acting on the package are gravity, the normal
force, the force of friction, and the applied force. The
normal force and force of gravity are each perpendicular
to the displacement, and therefore do no work.
Solution for Part 2
The applied force does work.
Wapp=Fappd(cos0∘)=Fappd =(120 N)(0.800 m) =96.0 J
The friction force and displacement are in opposite
directions, so that θ=180º, and the work done by friction is
Wfr=Ffrd(cos180∘)=Ffrd =−(5.00 N)(0.800 m) =−4.00 J
So the amounts of work done by gravity, by the normal
force, by the applied force, and by friction are,
respectively,
Wgr=0,WN=0,Wapp=96.0 J,Wfr=−4.00. J
The total work done as the sum of the work done by each
force is then seen to be Wtotal = Wgr +WN + Wapp + Wfr =
92.0 J.
Discussion for Part 2
The calculated total work Wtotal as the sum of the work by
each force agrees, as expected, with the work Wnet done
by the net force. The work done by a collection of forces
acting on an object can be calculated by either approach.
Kinetic energy is the energy of motion. An object that has
motion - whether it is vertical or horizontal motion - has
kinetic energy. There are many forms of kinetic energy vibrational (the energy due to vibrational motion),
rotational (the energy due to rotational motion), and
translational (the energy due to motion from one location
to another). To keep matters simple, we will focus upon
translational kinetic energy. The amount of translational
kinetic energy (from here on, the phrase kinetic energy
will refer to translational kinetic energy) that an object has
depends upon two variables: the mass (m) of the object
and the speed (v) of the object. The following equation is
used to represent the kinetic energy (KE) of an object.
KE = 0.5 • m • v2
where m = mass of object
v = speed of object
Sample Problem
1. Determine the kinetic energy of a 625-kg roller coaster
car that is moving with a speed of 18.3 m/s.
KE = 0.5mv2
KE = (0.5) (625 kg) (18.3 m/s)2
KE = 1.05 x105 Joules
Power is the rate at which work is done. It is the
work/time ratio. Mathematically, it is computed using the
following equation.
Power = Work / time
or
Online 2:
 What is the difference between conservative
force and nonconservative forces?
 How can we solve for gravitational potential
energy in a system?
 How can we solve for elastic potential energy
in a system?
 What is conservation of energy?
 How can we solve the conservation of energy
of a system?
P=W/t
Example:
When a car stops, 40000J of work is done by the brakes in
a time of 5s. Calculate the power of the brakes.
P = 40000J/5s
= 8000 W
Conservative force If the work done by a force depends
only on initial and final positions, the force is called a
conservative force. Gravity force is “a” conservative force.
If the work done by a force depends not only on initial and
final positions, but also on the path between them, the
force is called a non-conservative force. Example: Friction
force,Tension, normal force, and force applied by a
person.
Gravitational potential energy is the energy stored in an
object as the result of its vertical position or height. The
energy is stored as the result of the gravitational attraction
of the Earth for the object. The gravitational potential
energy of the massive ball of a demolition machine is
dependent on two variables - the mass of the ball and the
height to which it is raised. There is a direct relation
between gravitational potential energy and the mass of an
object. More massive objects have greater gravitational
potential energy. There is also a direct relation between
gravitational potential energy and the height of an object.
The higher that an object is elevated, the greater the
gravitational potential energy. These relationships are
expressed by the following equation:
PEgrav = mass • g • height
PEgrav = m *• g • h
In the above equation, m represents the mass of the
object, h represents the height of the object
and g represents the gravitational field strength (9.8 N/kg
on Earth) - sometimes referred to as the acceleration of
gravity.
Example
A cart is loaded with a brick and pulled at constant speed
along an inclined plane to the height of a seat-top. If the
mass of the loaded cart is 3.0 kg and the height of the seat
top is 0.45 meters, then what is the potential energy of
the loaded cart at the height of the seat-top?
PE = m*g*h
PE = (3 kg ) * (9.8 m/s/s) * (0.45 m)
PE = 13.2 J
Elastic potential energy is the energy stored in elastic
materials as the result of their stretching or compressing.
Elastic potential energy can be stored in rubber bands,
bungee chords, trampolines, springs, an arrow drawn into
a bow, etc. The amount of elastic potential energy stored
in such a device is related to the amount of stretch of the
device - the more stretch, the more stored energy.
Springs are a special instance of a device that can store
elastic potential energy due to either compression or
stretching. A force is required to compress a spring; the
more compression there is, the more force that is required
to compress it further. For certain springs, the amount of
force is directly proportional to the amount of stretch or
compression (x); the constant of proportionality is known
as the spring constant (k).
Fspring = k • x
Such springs are said to follow Hooke's Law. If a spring is
not stretched or compressed, then there is no elastic
potential energy stored in it. The spring is said to be at
its equilibrium position. The equilibrium position is the
position that the spring naturally assumes when there is
no force applied to it. In terms of potential energy, the
equilibrium position could be called the zero-potential
energy position. There is a special equation for springs
that relates the amount of elastic potential energy to the
amount of stretch (or compression) and the spring
constant. The equation is
PEspring = 0.5 • k • x2
where k = spring constant
x = amount of compression
(relative to equilibrium position)
Example:
The elongation of spring stretched by a force F = 50 N is 10
cm. What is the potential energy of elastic spring if the
elongation of spring is 12 cm.
Known :
Elongation (x) = 10 cm = 0.1 m
Force (F) = 50 N
Wanted : The potential energy of elastic spring
Solution :
Spring constant :
k = F / x = 50 / 0.1 = 500 N/m
The potential energy of elastic spring if the elongation of
spring is 12 cm. :
PE = ½ k x2 = ½ (500)(0.12)2 = (250)(0.0144) = 3.6 Joule.
Conservation of Energy
An object, or a closed system of objects, can have both
kinetic and potential energy. The sum of the kinetic and
potential energy of the object or system is called the total
mechanical energy. If no outside forces act on the system,
then the total mechanical energy is conserved. Energy can
change from kinetic to potential energy, and back, without
reducing the total energy. The sum of the kinetic and
potential energy at an initial time will be equal to the sum
of the kinetic and potential energy at any other time.
Often, a mechanical system is not fully closed. Either the
system can do work on the surroundings (for example, by
heating), or work can be done on the system (for example,
air resistance, or friction). In this case, a term for "other
work" is added to the formula to account for the change in
total mechanical energy. The unit of energy and work is
Joules (J).
K1 = initial kinetic energy (Joules, J)
U1 = initial potential energy (J)
wother = other work, gained or lost to the system (J)
K2 = final kinetic energy (J)
U2 = final potential energy (J)
Example:
An astronaut on the moon picks up a rock, and holds it
out. At that moment, the rock is at rest, and has 5.00 J of
gravitational potential energy. A moment later, the rock
hits the surface of the moon. What was the kinetic energy
of the rock immediately before it hit the surface?
Answer: The stored potential energy of the rock is in the
form of gravitational potential energy. This means that the
potential energy decreases as the rock falls toward the
surface of the moon. At the moment before the rock
reaches the surface, the rock's gravitational potential
energy is zero. Since there is no air on the moon, there is
no air resistance, and so the total mechanical energy is
conserved. The rock is released from rest, and so its initial
kinetic energy is zero. To summarize the known values:
k2 = unknown
Using these values, and the formula for conservation of
energy, the final kinetic energy can be found:
ACTIVITIES/EVALUATION/PERFORMANCE TASK
Formative Test: Online seatwork (Offline 1)
Summative Test: Online Quiz/Problem Solving (Offline 2)
MARIAN SCHOOL OF QUEZON CITY
LESSON PLAN
ONLINE LEARNING
GRADE LEVEL:
12
TOPIC:
 Center of Mass,
Momentum,
Impulse, and
Collisions
SUBJECT:
Time Frame:
PHYSICS 1
CONTENT STANDARD:
 Center of mass
 Momentum
 Impulse
 Impulse-momentum relation
 Law of conservation of momentum
 Collisions
 Center of Mass, Impulse, Momentum,
1 WEEK (WEEK 7)
LESSON OBJECTIVES:
 Differentiate center of mass and geometric
and Collision Problems


center
Relate the motion of center of mass of a
system to the momentum and net external
force acting on the system
Relate the momentum, impulse, force, and
time of contact in a system
VALUING:
 Appreciate the subject matter
PERFORMANCE STANDARD

Solve, using experimental and
theoretical approaches, multiconcept,
rich-content problems involving
measurement, vectors, motion in 1D
and 2D, Newton’s Laws, Work, Energy,
Center of Mass, momentum, impulse
and collisions
RESOURCES AND MATERIALS
 General Physics 1 (Vibal) pp. 153-163
 https://awwapp.com/#
 Powerpoint presentation
VIDEO RECORDING (VC)
5-10 minutes Video Presentation
SCRIPT
CONTENT/NOTES
 Today we will talk about momentum?
 Pictures of bicycle and a semi-truck
 Imagine you are at the bottom of the hill and
 Sample word problems involving momentum
you are faced with the stopping of a running
a. … an electron (m= 9.1 x10-31 kg) moving at 2.18
semi-truck and a running bicycle. Which will
x 106 m/s (as if it were in a Bohr orbit in the H
you choose?
atom).
 You probably choose to stop the bike right?
b. … a 0.45 Caliber bullet (m = 0.162 kg) leaving
 The reason for that is the semi-truck has a
the muzzle of a gun at 860 m/s.
more momentum compare to the bicycle.
c. … a 110-kg professional fullback running across
 Momentum simply means mass in motion.
the line at 9.2 m/s.
 The semi-truck is very massive but also has a
d. … a 360,000-kg passenger plane taxiing down a
large speed which influences momentum
runway at 1.5 m/s
 As well as bike, it has a large speed but its
Answers:
mass is less than the truck therefore its
a.2.0 x 10-24 kg•m/s
momentum is less also.
b. 140 kg•m/s (rounded from 139 kg•m/s)
 This relationship can be expressed through
c. 1.0 x 103 kg•m/s
equation: Momentum = Mass x Velocity
d. 5.4 x 105 kg•m/s
(P=mV)

If you increase the mass and the velocity of
the object, the momentum increases as well.
 In this way, we can see that the truck and a
bicycle has a large momentum. But the truck is
still more because it is far more massive than
the bicycle.
 This also means that an object at rest does not
have momentum because the velocity is 0. In
order for an object to have momentum, it has
to be moving.
 Let me give you some examples.
ONLINE PROCESSING (VIDEO CONFERENCING)
PROCESS QUESTIONS/ ESSENTIAL QUESTION
CONTENT/NOTES
Online 1:
Center of mass – is the average position of the mass of the
 What have you learned about the video?
object
 What is center of mass?
Two point masses 3 kg and 5 kg are at 4 m and 8 m from the
 How can we solve for the
origin on X-axis. Locate the position of center of mass of the
 What is impulse?
 How can we solve for impulse?
two point masses (i) from the origin and (ii) from 3 kg mass.
Solution
Let us take, m1 = 3 kg and m2= 5 kg
(i) To find center of mass from the origin:
The point masses are at positions, x1 = 4 m, x2 = 8 m from
the origin along X axis.
The center of mass xCM can be obtained using equation
5.4.
The center of mass is located 6.5 m from the origin on Xaxis.
(ii) To find the center of mass from 3 kg mass:
The origin is shifted to 3 kg mass along X-axis. The position
of 3 kg point mass is zero (x1 = 0) and the position of 5 kg
point mass is 4 m from the shifted origin (x2 = 4 m).
The center of mass is located 2.5 m from 3 kg point mass,
(and 1.5 m from the 5 kg point mass) on X-axis.
This result shows that the center of mass is located closer to
larger mass.
If the origin is shifted to the center of mass, then the
principle of moments holds good.
m1x1=m2x2; 3x2.5=5x1.5;7.5=7.5
When we compare case (i) with case (ii), the xCM = 2.5m
from 3 kg mass could also be obtained by subtracting 4 m
(the position of 3 kg mass) from 6.5 m, where the center of
mass was located in case (i)
Impulse – is the change in momentum
Δp = FnetΔt
Question: A 50 kg mass is sitting on a frictionless surface.
An unknown constant force pushes the mass for 2 seconds
until the mass reaches a velocity of 3 m/s.
a) What is the initial momentum of the mass?
b) What is the final momentum of the mass?
c) What was the force acting on the mass?
d) What was the impulse acting on the mass?
Part a) What is the initial momentum?
Momentum is mass times velocity. Since the mass is at
rest, the initial velocity is 0 m/s.
momentum = m⋅v = (50 kg)⋅(0 m/s) = 0 kg⋅m/s
Part b) What is the final momentum?
After the force is finished acting on the mass, the velocity
is 3 m/s.
momentum = m⋅v = (50 kg)⋅(3 m/s) = 150 kg⋅m/s
Part c) What was the force acting on the mass?
mv – mv0 = Ft
From parts a and b, we know mv0 = 0 kg⋅m/s and mv =
150 kg⋅m/s.
150 kg⋅m/s – 0 kg⋅m/s = Ft
150 kg⋅m/s = Ft
Since the force was in effect over 2 seconds, t = 2 s.
150 kg⋅m/s = F ⋅ 2s
F = (15 kg⋅m/s) / 2 s
F = 75 kg⋅m/s2
Unit Fact: kg⋅m/s2 can be denoted by the derived SI
unit Newton (symbol N)
F = 75 N
Online 2:
 What is the law of conservation of momentum
states?
 How can we calculate conservation of
momentum?
Part d) What was the impulse acting on the mass?
The impulse is the force multiplied by the time passed. It is
also equal to the change in momentum over the same
time period.
Ft = 75 N ⋅ 2 s
Ft = 150 Ns or 150 kg⋅m/s
The impulse was 150 kg⋅m/s.
Law of conservation of momentum states that:
For two or more bodies in an isolated system acting upon
each other, their total momentum remains constant
unless an external force is applied. Therefore, momentum
can neither be created nor destroyed.
Law of conservation of momentum is an important
consequence of Newton’s third law of motion.
Formula:
Example:
There are cars with masses 4kg and 10kg respectively that
are at rest. Car having the mass 10kg moves towards east
with the velocity of 5m.s-1. Find the velocity of car with
mass 4kg with respect to ground.
Given,
m1 = 4kg
m2 = 10kg
v1 = ?
v2 = 5m.s-1
We know from law of conservation of momentum that,
Pinitial = 0, as the cars are at rest
Pfinal = p1 + p2
Pfinal = m1.v1 + m2.v2
= 4kg.v1 + 10kg.5m.s-1
Pi = Pf
0=4kg.v1+50kg.m.s-1
v1 = 12.5 m.s-1
ACTIVITIES/EVALUATION/PERFORMANCE TASK
Formative Test: Online seatworks and problem solving (Offline 1)
Summative Test: Online Quiz (Offline 2)
MARIAN SCHOOL OF QUEZON CITY
LESSON PLAN
ONLINE LEARNING
GRADE LEVEL:
12
TOPIC:
 Center of Mass,
Momentum,
Impulse, and
Collisions
SUBJECT:
Time Frame:
PHYSICS 1
CONTENT STANDARD:
 Center of mass
 Momentum
 Impulse
 Impulse-momentum relation
 Law of conservation of momentum
 Collisions
 Center of Mass, Impulse, Momentum,
1 WEEK (WEEK 8)
LESSON OBJECTIVES:
 Compare and contrast elastic and inelastic
and Collision Problems
PERFORMANCE STANDARD



collisions
Apply the concept of restitution coefficient in
collisions
Solve problems involving center of mass,
impulse, and momentum in contexts such as,
but not limited to, rocket motion, vehicle
collisions, and ping-pong.
VALUING:
 Appreciate the subject matter
Solve, using experimental and
theoretical approaches, multiconcept,
rich-content problems involving
measurement, vectors, motion in 1D
and 2D, Newton’s Laws, Work, Energy,
Center of Mass, momentum, impulse
and collisions
RESOURCES AND MATERIALS




General Physics 1 (Vibal) pp. 162-172
https://awwapp.com/#
Powerpoint presentation
https://isaacphysics.org/concepts/cp_collisions
VIDEO RECORDING (VC)
5-10 minutes Video Presentation
SCRIPT
CONTENT/NOTES
 Who among of loves playing billiard? How about
 Pictures of different sports that applies
baseball? Or basketball?
collisions
 Did you know that these sports are applications of
 Types of Collision
what we call collision?
- Inelastic collision – momentum is conserved
 What is collision?
- elastic collision - momentum is conserved
 Collisions occur when one object strikes another.
and kinetic energy is conserved.
Problems involving collisions are usually solved
using conservation of
momentum and conservation of energy
 Do you want to know more about collision?
 Then let’s watch this video:
https://www.youtube.com/watch?v=Xe2r6wey26E

Now that you already know about collision and its
type tomorrow we will discuss more of it in
detailed including it computations.
ONLINE PROCESSING (VIDEO CONFERENCING)
PROCESS QUESTIONS/ ESSENTIAL QUESTION
CONTENT/NOTES
Completely Inelastic collision
Online 1:
The easiest collisions to analyse are completely inelastic
 What have you learned about the video?
collisions, where objects stick together after colliding. This
 What is completely in inelastic collision?
is because energy is converted into other forms in the
 How can we solve for the problems involving
collision so we don't have to worry about conserving
completely inelastic collision?
kinetic energy. Also, as we know that the two objects stick
together we have fewer variables to worry about when
conserving momentum.
These collisions can therefore be completely solved just by
considering conservation of momentum.
In
the simple case shown in Figure 1, before the collision,
particle A with mass 𝑚𝖠 is moving towards particle B with
a speed 𝑢𝖠, while particle B with mass 𝑚𝖡 is moving
towards particle A with a speed 𝑢𝖡. This gives a total
momentum (taking to the right as positive)
of 𝑝=𝑚𝖠𝑢𝖠−𝑚𝖡𝑢𝖡.
By conservation of momentum, the momentum after the
collision must be the same as the momentum before the
collision. As both particles are stuck together they move
with a speed 𝑣v and so have a total momentum
of 𝑝′=(𝑚𝖠+𝑚𝖡)𝑣. Equating the total momentum before
and after the collision gives:
Example:
A body of mass 1kg travelling with a
speed 13ms−1 collides with a stationary body of
mass 0.5kg, and they stick together. What speed do the
two bodies have after the collision?
Online 2:
 What is inelastic collision?
Answer = 𝑣=2.00ms−1
Elastic Collisions


What is elastic collision?
How can we solve word problems involving
inelastic and elastic collision?
In elastic
collisions all
of the energy
remains as
kinetic
energy - no
energy is lost to other forms. This means that both kinetic
energy and momentum must be conserved.
In the simple case in Figure 2, before the collision, particle
A with mass 𝑚𝖠 is moving towards particle B with a
speed 𝑢𝖠, while particle B with mass 𝑚𝖡 is moving
towards particle B with a speed 𝑢𝖡. If the collision is
elastic, both momentum and kinetic energy must be
conserved.
Before the collision the particles have a total momentum
(taking to the right as positive) of 𝑝=mAuA−mBuB and a
total kinetic energy of
After the
collision the
particles have a total momentum (taking to the right as
positive) of 𝑝′=−mAvA+mBvB and a total kinetic energy of
Equating momentum and kinetic energies gives:
These two equations can be re-arranged to find any two of
the
variables 𝑚𝖠mA, 𝑚𝖡mB, 𝑢𝖠uA, 𝑢𝖡uB, 𝑣𝖠vA or 𝑣𝖡vB in
terms of the other four. As an example, we will re-arrange
these to find the final velocities 𝑣𝖠vA and 𝑣𝖡vB for the
case when particle B starts out stationary (so 𝑢𝖡=0uB=0)
and both particles have the same mass 𝑚m.
From the conservation of momentum
equation, 𝑣𝖡=𝑚𝑢𝖠+𝑚𝑣𝖠𝑚/m =𝑢𝖠+vA. This can be
substituted into the kinetic energy equation. This gives:
This has two
solutions.
One solution gives 𝑣𝖠=−𝑢𝖠, but since 𝑣𝖠 was drawn in the
negative direction then this actually means that the final
velocity of particle A is 𝑢 in the positive direction - the
same as its initial velocity! This solution is if there is no
collision, and is therefore useless.
The other solution gives 𝑣𝖠=0, and so 𝑣𝖡=𝑢𝖠. In this case
all of the kinetic energy and momentum has been
transferred to particle B.
Example:
Two particles, A and B, of
masses 𝑚𝖠=𝑚and 𝑚𝖡=2𝑚collide head-on with speeds
of 𝑢𝖠=2𝑢 and 𝑢𝖡=−𝑢. Find the speeds 𝑣𝖠vA and 𝑣𝖡vB of
both particles after the collision.
Answer: 𝑣𝖠=−2𝑢vA=−2u and 𝑣𝖡=𝑢
Inelastic Collision
In inelastic collisions some kinetic energy is converted to
another form. In fully inelastic collisions the maximum
possible kinetic energy is lost and the objects stick
together. However in many inelastic collisions this is not
the case - only some kinetic energy is lost and this must be
taken into consideration in the calculation of the
movement of the objects after the collision.
In an inelastic collision:
Kinetic Energy before collision = Kinetic Energy after
collision + Energy converted into other forms
We can use this along with the conservation of
momentum, which is always conserved, in order to work
out the motion of objects after the collision.
Figure 3 shows an inelastic collision between two
particles, both of mass 𝑚m, in which Δ𝐾=𝑋JΔK=XJ of
sound and heat are produced by the collision. The particle
motion involved in the sound and heat has net zero
momentum.
Example:
A body of mass 1kg moving with a speed 13ms-1 collided
inelastically with a stationary body of mass 2kg. After the
collision, the body of mass 1kg was stationary. How much
energy was converted to sound and heat in this collision?
Answer: 2.25J of energy was converted to sound and heat.
ACTIVITIES/EVALUATION/PERFORMANCE TASK
Formative Test: Online seatworks and problem solving (Offline 1)
Summative Test: Online Quiz (Offline 2)