AQA GCSE Physics 2-3
Work, Energy & Momentum
GCSE Physics pages 146 to 159
July 2010
AQA GCSE Specification
WORK & ENERGY
12.3 What happens to the movement energy when
things speed up or slow down?
Using skills, knowledge and understanding of how
science works:
• to discuss the transformation of kinetic energy to other
forms of energy in particular situations.
Skills, knowledge and understanding of how science
works set in the context of:
• When a force causes a body to move through a
distance, energy is transferred and work is done.
• Work done = energy transferred.
• The amount of work done, force and distance are
related by the equation:
work done = force applied × distance moved in direction
of force
• Work done against frictional forces is mainly
transformed into heat.
• Elastic potential is the energy stored in an object when
work is done on the object to change its shape.
• The kinetic energy of a body depends on its mass and
its speed.
HT Calculate the kinetic energy of a body using the
equation:
kinetic energy = ½ × mass × speed2
MOMENTUM
12.4 What is momentum?
Using skills, knowledge and understanding of how
science works:
• to use the conservation of momentum (in one
dimension) to calculate the mass, velocity or
momentum of a body involved in a collision or
explosion
• to use the ideas of momentum to explain safety
features.
Skills, knowledge and understanding of how science
works set in the context of:
• Momentum, mass and velocity are related by the
equation:
momentum = mass × velocity
• Momentum has both magnitude and direction.
• When a force acts on a body that is moving, or able to
move, a change in momentum occurs.
• Momentum is conserved in any collision/explosion
provided no external forces act on the
colliding/exploding bodies.
HT Force, change in momentum and time taken for the
change are related by the equation:
force = change in momentum / time taken for the change
Work
When a force causes a body to move through a distance,
energy is transferred and work is done.
Work done = energy transferred.
Both work and energy are measured in joules (J).
Work and friction
Work done against frictional
forces is mainly transformed into
heat.
Rubbing hands together causes
them to become warm.
Brakes pads become hot if they
are applied for too long. In this
case some of the car’s energy
may also be transferred to sound
in the form of a ‘squeal’
The work equation
The amount of work done, force and distance are
related by the equation:
work done = force applied × distance moved
in the direction
of the force
Work is measured in joules (J)
Force is measured in newtons (N)
Distance is measured in metres (m)
also:
force = work done ÷ distance moved
and:
distance = work done ÷ force
work
force
distance
Question 1
Calculate the work done when a force of
5 newtons moves through a distance of
3 metres.
work = force x distance
= 5N x 3m
work = 15 joules
Question 2
Calculate the work done when a force of
6 newtons moves through a distance of
40 centimetres.
work = force x distance
= 6 N x 40 cm
= 6 N x 0.40 m
work = 2.4 joules
Question 3
Calculate the value of the force required to
do 600 joules of work over a distance of 50
metres.
work = force x distance
becomes:
force = work done ÷ distance
= 600 J ÷ 50 m
force = 12 newtons
Question 4
Calculate the distance moved by a force of
8 newtons when it does 72 joules of work.
work = force x distance
becomes:
distance = work done ÷ force
= 72 J ÷ 8 N
distance moved = 9 metres
Question 5
Calculate the work done by
a child of weight 300N who
climbs up a set of stairs
consisting of 12 steps each
of height 20cm.
work = force x distance
The child must exert an
upward force equal to its
own weight.
Therefore: force = 300N
This force is exerted
upwards and so the
distance must also be
measured upwards.
= (12 x 20cm)
= 2.4m
therefore:
work = 300 N x 2.4 m
work = 720 J
Question 6
Calculate the work done by a person of mass 80kg who
climbs up a set of stairs consisting of 25 steps each of
height 10cm.
work = force x distance
the person must exert an upward force equal their weight
the person’s weight = (80kg x 10N/kg) = 800N
the distance moved upwards equals (10 x 25cm) = 2.5m
work = 800 N x 2.5 m
work = 2000 J
Complete
Answers
work
force
distance
150 J
50 N
3m
800 J
40 N
20 m
500 J
250 N
2m
80 kJ
4000 N
2m
2 MJ
3.03
400 N
5 km
Choose appropriate words to fill in the gaps below:
force
Work is done when a _______
moves through a distance.
energy transferred is also equal to the work
The amount of _______
heat
done. When a car brakes energy is transformed to ______.
equal to the force _________
multiplied by the distance
Work done is ______
moved in the __________
of the force. The work done is
direction
measured in ______
joules if the force is measured in newtons and
distance in metres.
the _________
WORD SELECTION:
energy direction force equal multiplied distance heat joules
Energy and work
Notes questions from pages 146 & 147
1. What is meant by ‘work’?
2. Copy both of the equations for work on page
146 along with the units used.
3. Copy and answer questions (a) and (b) on
pages 146 and 147.
4. Explain two ways in which the force of friction
causes energy to be transformed into heat.
5. Copy the Key Points on page 147.
6. Answer the summary questions on page 147.
Energy and work
ANSWERS
In text questions:
(a) To the surroundings
as heat energy and
sound energy
(b) 300 J
Summary questions:
1. (a) 96 J
(b) 96 J
2. (a) (i) 90 J
(ii) 4500 J
(b) 0.60 m
Potential energy
Elastic potential energy is
the energy stored in an
object when work is done on
an object to change its
shape.
An elastic object regains its
shape after being stretched
or squashed.
Elastic potential energy is
stored in the bow string when
it is pulled by the archer.
Gravitational potential
energy is the energy stored
in an object when work is
done in moving the object
upwards.
The potential energy stored
is equal to the weight of the
object multiplied by the
height lifted.
The weightlifter stores
gravitational potential energy
when he lifts the weights.
Kinetic energy
Kinetic energy is the energy possessed by a
body because of its speed and mass.
kinetic energy = ½ x mass x (speed)2
kinetic energy is measured in joules (J)
mass is measured in kilograms (kg)
speed is measured in metres per second (m/s)
Question 1
Calculate the kinetic energy of a car of mass
1000kg moving at 5 m/s.
kinetic energy = ½ x mass x (speed)2
kinetic energy = ½ x 1000kg x (5m/s)2
kinetic energy = ½ x 1000 x 25
kinetic energy = 500 x 25
kinetic energy = 12 500 joules
Question 2
Calculate the kinetic energy of a child of mass
60kg moving at 3 m/s.
kinetic energy =
k.e. = ½ x 60kg
k.e. = ½ x 60 x
k.e. = 30 x 9
kinetic energy =
½ x mass x (speed)2
x (3m/s)2
9
270 J
Question 3
Calculate the kinetic energy of a apple of mass
200g moving at 12m/s.
kinetic energy = ½ x mass x (speed)2
k.e. = ½ x 200g x (12m/s)2
k.e. = ½ x 0.200kg x 144
k.e. = 0.100 x 144
kinetic energy = 14.4 J
Question 4
Calculate the mass of a train if its kinetic energy is
2MJ when it is travelling at 4m/s.
kinetic energy = ½ x mass x (speed)2
2MJ = ½ x mass x (4m/s)2
2 000 000J = ½ x mass x 16
2 000 000 = 8 x mass
2 000 000 ÷ 8 = mass
mass = 250 000 kg
Question 5
Calculate the speed of a car of mass 1200kg if its kinetic
energy is 15 000J.
kinetic energy = ½ x mass x (speed)2
15 000J = ½ x 1200kg x (speed)2
15 000 = 600 x (speed)2
15 000 ÷ 600 = (speed)2
25 = (speed)2
speed = 25
speed = 5 m/s
Question 6
Calculate the speed of a ball of mass 400g if its kinetic
energy is 20J.
kinetic energy = ½ x mass x (speed)2
20J = ½ x 400g x (speed)2
20 = ½ x 0.400kg x (speed)2
20 = 0.200 x (speed)2
20 ÷ 0.200 = (speed)2
100 = (speed)2
speed = 100
speed = 10 m/s
Complete
Answers
kinetic energy
mass
speed
8J
4 kg
2 m/s
27 J
6 kg
3 m/s
1000 J
80 kg
5 m/s
6.4 kJ
200 kg
8 m/s
3.2 J
3.03g
400
4 m/s
Choose appropriate words to fill in the gaps below:
potential energy is the energy stored when an object
Elastic ________
squashed This energy is released when the
is stretched or ________.
returns to its original shape.
object ________
Kinetic energy is the energy possessed by an object due to its
speed and mass. If the mass of an object is ________
doubled its
_______
kinetic energy doubles. If the speed is doubled the kinetic
four
energy will increase by ______
times.
stretched
When a __________
elastic band is released elastic potential
kinetic
energy is converted into _________
energy.
WORD SELECTION:
returns speed four kinetic potential squashed doubled stretched
Kinetic energy
Notes questions from pages 148 & 149
1.
2.
3.
4.
5.
6.
7.
8.
How can gravitational potential energy be calculated? (see the
practical on page 148)
Copy the equation for kinetic energy at the top of page 149 along
with the units used.
Repeat the calculation below the kinetic energy equation but this
time with a mass of 400 kg moving at a speed of 8 m/s.
What does ‘elastic’ mean?
What is elastic potential energy?
Copy and answer question (b) on page 149.
Copy the Key Points on page 149.
Answer the summary questions on page 149.
Kinetic energy
ANSWERS
In text question:
(b) Heat energy
transferred to the
surroundings, the
foot and the shoe;
also sound energy.
Summary questions:
1. (a) (i) Chemical energy from the loader is
transferred into elastic potential energy of the
catapult and some is wasted as heat energy.
(ii) Elastic potential energy in the catapult is
transformed into kinetic energy of the object
and the rubber band and heat energy (plus a
little sound energy).
(b) (i) 10 J
(ii) 10 J
2. (a) 3800 N
(b) Friction due to the brakes transforms it
from kinetic energy of the car to heat energy
in the brakes.
(c) 800 kg
Momentum
momentum = mass x velocity
mass is measured in kilograms (kg)
velocity is measured in metres per second (m/s)
momentum is measured in:
kilogram metres per second (kg m/s)
Momentum has both
magnitude and
direction.
Its direction is the same
as the velocity.
The greater the mass of a
rugby player the greater is
his momentum
Question 1
Calculate the momentum of a rugby player,
mass 120kg moving at 3m/s.
momentum = mass x velocity
= 120kg x 3m/s
momentum = 360 kg m/s
Question 2
Calculate the mass of a car that when
moving at 25m/s has a momentum of
20 000 kg m/s.
momentum = mass x velocity
becomes: mass = momentum ÷ velocity
= 20000 kg m/s ÷ 25 m/s
mass = 800 kg
Complete
Answers
momentum
mass
velocity
150 kg m/s
50 kg
3 m/s
160 kg m/s
8 kg
20 m/s
1500 kg m/s
250 kg
6 m/s
4 kg m/s
500 g
8 m/s
3 kg m/s
kgkg
6
50 cm/s
Momentum conservation
Momentum is conserved in any collision or
explosion provided no external forces act
on the colliding or exploding bodies.
The initial momentum of the yellow car has been
conserved and transferred to the red car
Question 1
A truck of mass 0.5kg
moving at 1.2m/s collides
and remains attached to
another, initially stationary
truck of mass 1.5kg.
Calculate the velocity of
the trucks after the
collision.
total momentum before collision
momentum = mass x velocity
0.5 kg truck: = 0.5 kg x 1.2 m/s = 0.6 kg m/s
1.5 kg truck: = 1.5 kg x 0 m/s = 0 kg m/s
total initial momentum = 0.6 kg m/s
Momentum is conserved in the collision
so total momentum after collision = 0.6 kg m/s
total momentum = total mass x velocity
0.6 kg m/s = 2.0 kg x velocity
0.6 ÷ 2.0 = velocity
velocity = 0.3 m/s
Question 2
A train wagon of mass 800 kg moving at 4 m/s
collides and remains attached to another wagon
of mass 1200 kg that is moving in the same
direction at 2 m/s. Calculate the velocity of the
wagons after the collision.
total momentum before collision
momentum = mass x velocity
800 kg wagon: = 800 kg x 4 m/s = 3200 kg m/s
1200 kg truck: = 1200 kg x 2 m/s = 2400 kg m/s
total initial momentum = 5600 kg m/s
Momentum is conserved in the collision
so total momentum after collision = 5600 kg m/s
total momentum = total mass x velocity
5600 kg m/s = 2000 kg x velocity
5600 ÷ 2000 = velocity
velocity = 2.8 m/s
Choose appropriate words to fill in the gaps below:
mass multiplied
The momentum of an object is equal to its ______
direction
by its velocity. Momentum has _________,
the same as the
velocity, and is measured in kilogram _______
metres per second.
forces act
In any interaction of bodies, where no external _______
momentum is conserved.
on the bodies, __________
In snooker, a head-on collision of a white ball with a red ball
same initial
can result in the red ball moving off with the ______
velocity of the white ball. This is an example of momentum
conservation
____________.
WORD SELECTION:
direction forces same conservation
metres momentum mass
Momentum
Notes questions from pages 214 & 215
1.
2.
3.
4.
5.
6.
7.
8.
What is ‘momentum’?
Copy out the equation at the top of page 214. State the
units for each quantity in the equation.
Copy and answer question (a) on page 214.
Under a heading “Conservation of momentum” copy
out the statement in bold at the bottom of page 214.
Copy out the worked example on page 215.
Copy and answer question (b) on page 215.
Copy the Key Points on page 215.
Answer the summary questions on page 215.
Momentum
ANSWERS
In text questions:
(a) 240 kg m/s
(b) 0.48 m/s
Summary questions:
1. (a) mass, velocity
(b) momentum, force
2. (a) 5000 kg m/s
(b) velocity
= momentum / mass
= 5000 / 2500
= 2.0 m/s
Head-on collisions
In this case bodies are moving in opposite directions.
Momentum has direction.
One direction is treated as positive, the other as negative.
In calculations the velocity of one of the colliding bodies
must be entered as a NEGATIVE number.
DIRECTION OF MOTION
NEGATIVE
POSITIVE
+ ve
velocity
- ve
velocity
Question 1
A car of mass 1000 kg moving at 20 m/s makes a
head-on collision with a lorry of mass 2000 kg
moving at 16 m/s. Calculate their common
velocity after the collision if they remain attached
to each other.
lorry, mass 2000kg
car, mass 1000kg
20 m/s
16 m/s
DIRECTION OF MOTION
NEGATIVE
POSITIVE
total momentum before collision
momentum = mass x velocity
car: = 1000 kg x +20 m/s = +20000 kg m/s
lorry: = 2000 kg x -16 m/s = -32000 kg m/s
total initial momentum = -12000 kg m/s
Momentum is conserved in the collision
so total momentum after collision = -12000 kg m/s
total momentum = total mass x velocity
-12000 kg m/s = 3000 kg x velocity
-12000 ÷ 3000 = velocity
common velocity = - 4 m/s
The lorry/car combination will move in the negative
direction (to the left in this case) with a common
velocity of 4 m/s.
Question 2
A car of mass 1000 kg moving at 30 m/s makes a
head-on collision with a lorry of mass 2000 kg
moving at 15 m/s. Calculate their common
velocity after the collision if they remain attached
to each other.
lorry, mass 2000kg
car, mass 1000kg
30 m/s
15 m/s
DIRECTION OF MOTION
NEGATIVE
POSITIVE
total momentum before collision
momentum = mass x velocity
car: = 1000 kg x +30 m/s = +30000 kg m/s
lorry: = 2000 kg x -15 m/s = -30000 kg m/s
total initial momentum = 0 kg m/s
Momentum is conserved in the collision
so total momentum after collision = 0 kg m/s
The lorry/car combination will not move after
the collision.
Explosions
Before an explosion the total momentum is zero.
As momentum is conserved, the total momentum
afterwards must also be zero.
This means that the different parts of the exploding
body must move off in different directions.
Question 1
An artillery gun of mass 1500kg fires a shell of
mass 20kg at a velocity of 150m/s. Calculate the
recoil velocity of the gun.
artillery gun, mass
1500kg
recoil
shell,
mass
20kg
150 m/s
The total momentum before and after the explosion
is ZERO
momentum = mass x velocity
shell: = 20 kg x +150 m/s = +3000 kg m/s
This must cancel the momentum of the gun.
Therefore the gun’s momentum must be -3000 kg m/s
gun: = 1500 kg x recoil velocity = -3000 kg m/s
recoil velocity = - 3000 ÷ 1500
= - 2m/s
The gun will recoil (move to the left)
with a velocity of 2 m/s.
Question 2
A girl of mass 60kg throws a boy, mass 90kg out
off a swimming pool at a velocity of 2m/s. What is
the girl’s recoil velocity?
boy, mass 90kgboy, mass 90kg
girl,
girl,mass
mass60kg
60kg
2 m/s
2 m/s
recoil
recoil
DIRECTION OF MOTION
NEGATIVE
POSITIVE
The total momentum before and after throwing the
boy is ZERO
momentum = mass x velocity
boy: = 90 kg x +2 m/s = +180 kg m/s
This must cancel the momentum of the girl.
Therefore the girl’s momentum must be -180 kg m/s
gun: = 60 kg x recoil velocity = -180 kg m/s
recoil velocity = - 180 ÷ 60
= - 3m/s
The girl will recoil (move to the left)
with a velocity of 3 m/s.
More on collisions and explosions
Notes questions from pages 152 & 153
1. Apart from size what other property does
momentum have?
2. Copy and answer question (a) on page 152.
3. Explain how conservation of momentum
applies in an explosion.
4. Why do guns recoil?
5. Copy and answer question (b) on page 153.
6. Copy the Key Points on page 153.
7. Answer the summary questions on page 153.
More on collisions and explosions
ANSWERS
In text questions:
(a) The boat and the
person who jumps
off move away with
equal and opposite
amounts of
momentum.
(b) 25 m/s
Summary questions:
1. (a) momentum
(b) velocity
(c) force
2. (a) 60 kg m/s
(b) 1.5 m/s
Force and momentum
A force will cause the velocity of an object to
change and therefore also its momentum.
The greater the force the faster the momentum
changes.
force =
change in momentum
time taken for the change
force is measured in newtons (N)
change in momentum is measured in:
kilogram metres per second (kg m/s)
time is measured in seconds (s)
Equation proof
acceleration = velocity change ÷ time taken
multiplying both sides of this equation by ‘mass’
gives:
(mass x acceleration) = (mass x velocity) change ÷ time
but:
(mass x acceleration) = force
and:
(mass x velocity) = momentum
therefore:
force = momentum change ÷ time taken
Question 1
Calculate the force required to change the
momentum of a car by 24000 kgm/s over a
6 second period.
force = momentum change ÷ time taken
= 24000 kgm/s ÷ 6 s
force = 4000N
Question 2
Calculate the time taken for a force of
6000N to cause the momentum of truck to
change by 42000 kgm/s.
force = momentum change ÷ time taken
becomes:
time taken = momentum change ÷ force
= 42000 kgm/s ÷ 6000 N
force = 7 seconds
Complete
Answers
force
time taken
200 N
momentum
change
8000 kgm/s
25 N
500 kgm/s
20 s
500 N
3000 kgm/s
6s
800 N
8000 kgm/s
10 s
4N
480 kgm/s
2 minutes
40 s
Car safety features
Crumple zones, air bags and a collapsible steering wheel
are designed to increase the time taken for a driver or
passenger to change momentum to zero during a crash.
The equation: force = momentum change ÷ time taken
shows that if the time taken is increased for the same
momentum change the force exerted is decreased so is
the injury to the driver or passenger.
Playground flooring question
ANSWER:
The picture shows rubber
tiles used for playground
When a child falls to the floor its
flooring. Explain how these
can reduce injury to children. momentum changes from a high
value to zero.
The rubber flooring tiles increase
the time taken for this change.
force = change in momentum ÷
time taken for the change
Therefore the force on the child is
reduced and so is the potential
injury.
Choose appropriate words to fill in the gaps below:
momentum
The force exerted on an object is equal to the __________
time
change caused divided by the ______
taken for the change.
crash
An airbag activates during a car _______.
The inflated
increases the time taken for a driver’s or
airbag _________
velocity to fall to zero. The time taken for their
passenger’s ________
momentum to fall to ______zero
is also increased. Therefore the
_______ exertedforce
on the driver or passenger is __________
decreased
injury
and
so is the potential ________ caused.
WORD SELECTION:
time velocity zero momentum force
decreased injury increases crash
Changing momentum
Notes questions from pages 154 & 155
1. What is the purpose of a car’s crumple zones?
2. Copy the key points on page 155.
3. Copy and answer questions (a), (b) and (c) on
pages 154 and 155.
4. Copy the Key Points on page 155.
5. Answer the summary questions on page 155.
Changing momentum
ANSWERS
In text questions:
(a) If a child falls off a
swing, the rubber mat
reduces the impact
force by increasing the
impact time when the
child hits the ground.
(b) The force is bigger
(c) 1800 N
Summary questions:
1. (a) stays the same
(b) increases
(c) decreases
2. (a) 24 000 kg m/s
(b) (i) 2000 N
(ii) 800 N
Virtual Physics Laboratory Simulations
NOTE: Links work only in school
Roller Coaster.exe - Energy considerations
Online Simulations
Work (GCSE) - Powerpoint
presentation by KT
Kinetic Energy (GCSE) Powerpoint presentation by KT
Gravitational Potential Energy
(GCSE) - Powerpoint presentation
by KT
Energy Skate Park - Colorado Learn about conservation of
energy with a skater dude! Build
tracks, ramps and jumps for the
skater and view the kinetic
energy, potential energy and
friction as he moves. You can
also take the skater to different
planets or even space!
Rollercoaster Demo Funderstanding
Energy conservation with falling
particles - NTNU
Ball rolling up a slope - NTNU
Pulley System - Fendt
BBC AQA GCSE Bitesize
Revision:
Work, force and distance
Potential and kinetic energy
Kinetic energy equation
Forces for safety
Notes questions from pages 156 & 157
1. Answer questions 1 and 2 on pages 156
and 157.
Forces for safety
ANSWERS
1. The air bag increases the time taken to stop
the person it acts on. This reduces the force of
the impact. Also, the force is spread out across
the chest by the air bag so its effect is lessened
again.
2. (a) 26 100 kg m/s
(b) 34.8 m/s
(c) Yes.
How Science Works
ANSWERS
(a)
(b)
(c)
(d)
(e)
(f)
(g)
No. The upright position is slightly greater although there is no
significant difference between the three sets of readings.
The upright position propels the ball further. This prediction is even
stronger if the 2nd go/front measurement is considered to be an
anomaly.
No. The measurements have a wide range within each set. There
is even overlap of results.
Position of the releae point.
Categoric.
By measuring the angle of the spoon to the upright.
More information can be obtained. A graph can also be drawn and
a pattern discerned (or not).