Download Chapter 9 - AS-A2

Along the flat and up the hill
Question 240W: Warm-up Exercise
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Think carefully, and calculate a little.
Two speeds, one principle
Travelling slowly, at about 5 miles an hour, the retarding forces acting on a cyclist are about 5 N.
1. What propulsive force must the cyclist provide to travel at constant speed?
2. How much energy does the cyclist have to provide to cover 5 m?
At a higher speed, nearly 10 mph the retarding forces are much larger, nearly 8 N.
3. Write down the propulsive force required at this speed.
4. Calculate the energy the cyclist must provide to cover 5 m at this constant speed.
Climbing a hill at a steady speed of 5 mph the cyclist finds herself perspiring rather more than on the
flat. The mass of the cycle plus cyclist is 80 kg. The ergometer pedals fitted report the energy
supplied to the bicycle as 185 J to cover a 5 m stretch of road.
5. How much energy is used to lift the cyclist uphill?
6. How much height does she gain in travelling along the 5 m stretch of road?
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7. How much energy would she have to provide to cover this same stretch of road at 10 mph?
8. What is the retarding force acting on her, when travelling uphill at 10 mph?
9. Draw a vector diagram for the retarding forces (due to drag and gravity) acting on her at 10 mph.
10. Calculate the retarding force due to the slope as a fraction of her weight. Comment on the
advantages of using the slope of the hill rather than lifting bike and rider vertically by the same
distance.
Calculated steps
Question 20S: Short Answer
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Here you work on predicting positions, calculating where and when people can be found.
A lost group
The questions ask you to consider the plight of a Duke of Edinburgh assessor, faced with a group,
location currently unknown. Here is the map of the area in which they are working. They are expected
to move at 4 km h–1 over the fairly flat terrain.
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check point
WILTSHIRE
10 km
1. The group have correctly dropped in at the checkpoint. The assessor has their route card which
says they will travel for 45 minutes north-east. Draw a displacement vector to show where he
should look for them. Label this vector a.
2. Now assume they have headed off north-west. Where might they be? Draw in a vector to show
this displacement (from the checkpoint). Label it b.
3. Assume the group is at the place shown by vector b. Mark in the displacement vector, c, that
shows what they must do to get from where they are to where they should be.
After a fruitless search, the assessor gives up guessing where they might have gone wrong,
assuming only that they travel in a straight line.
4. How much area must he search after 1 hour? after 2 hours, after 3 hours? Is there a general
pattern, relating the time before looking to the difficulty of finding the group?
The moral is to have lots of checkpoints, otherwise predicting the future gets unreliable!
5. Compare the area to be searched by the assessor after one hour if the group were travelling at
3 km h–1, 4 km h–1, 5 km h–1. Is there a general pattern, relating the speed of travel to the difficulty
of finding the group?
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6. Is there a comparable moral for speeds?
Coping with graphs
Question 30S: Short Answer
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Slopes and tangents
Graphs give us a visual impression of how different quantities relate to each other, when we have
learned to read them. They might not tell more than an equation does, but they make relationships
easier to take in and remember. In this activity you are asked to relate two graphs showing different
features of the same event: an object accelerating uniformly in a straight line.
What to do
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20
10
0
0
2
4
6
time / s
8
10
12
120
100
80
60
40
20
0
0
2
4
6
time / s
8
10
12
These graphs are speed–time and distance–time graphs for an accelerated object.
1. What feature of which graph shows that the acceleration of the object is constant?
2. Use and explain two ways for calculating the value of the acceleration from the data in the first
graph.
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Use the second graph to find:
3. the speed of the object at 4 s;
4. the speed of the object at 10 s;
5. the average speed of the object.
6. Use your answers to questions 3 and 4 to estimate the acceleration of the object.
7. Do these values correspond with the values you can obtain from graph 1? If not, suggest a
reason.
Things to remember
1. The speed–time graph of an object moving in a straight line with constant acceleration is a
straight line.
2. How to use a linear speed–time graph to calculate an acceleration by
using total change in speed divided by time taken for speed to change
using the slope of the graph.
3. How to estimate speed at a given time by drawing a tangent to a distance–time graph.
4. How to use two speed values taken at the beginning and end of a time interval to calculate the
acceleration in the time interval.
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Calculating accelerated steps
Question 50S: Short Answer
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Here you work on predicting positions, calculating where objects that are accelerated can be found.
A funicular problem – stepwise
A funicular railway has an antiquated braking system that stops the car 2.0 s after being triggered by
a trackside bollard. The owners wish to know where to put the rescue platform for the passengers to
step out onto. You are to calculate this using the following steps. The stepwise calculation is needed
as the brakes fade with time.
Time
interval / s
Speed at
start of
interval /
m s–1
Acceleration
during
interval /
m s–2
0–0.5
5.00
–4.00
0.5–1.0
–3.00
1.0–1.5
–2.00
1.5–2.0
–1.00
Speed at
end of
interval /
m s–1
Average
speed
during
interval /
m s–1
Distance
covered
during this
interval / m
Where must the platform go?
Uniform acceleration
Question 60S: Short Answer
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You will be able to tackle these questions when you have mastered the kinematic equations of
accelerated motion. You need to have the definitions of the basic ideas, displacement, velocity and
acceleration, very clear in your mind before starting work on these problems.
Using simple kinematics
A light aircraft has a take-off speed of 60 m s–1. Its engine can produce an average acceleration of 4
m s–2.
1. How long does the aircraft take to reach take-off speed?
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2. What is the minimum length of runway the plane needs?
3. Write a few sentences considering what difference it makes if the aircraft takes off into a
head-wind blowing at a speed of 10 m s–1.
The following data were produced for a car:
Time / s
Speed / m s–1
0
0.00
1
3.20
2
6.40
3
9.40
4
12.5
5
15.4
6
18.0
7
20.0
8
20.8
9
20.8
4. Draw a speed–time graph for these data.
5. Describe the motion of the car during the first 9 s.
6. Suggest reasons why the car reached a steady speed.
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7. Use the graph to estimate how far the car travelled in the first 5 s and then in the first 9 s.
8. Use the graph to estimate the car's acceleration at a time of 6 s.
Braking distance and the Highway Code
Question 70S: Short Answer
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Instructions
You will need to use the kinematic equations and some algebra to tackle this set of questions.
Only a fool breaks the 2-second rule!
These are braking distance data from the Highway Code. You could set up a spreadsheet to help with
the repetitive calculations that follow.
Speed /
km h–1
Speed /
m s–1
Thinking
distance /
m
Braking
distance /
m
Stopping
distance /
m
50
13.9
9.7
14.7
24.4
65
18.1
12.6
24.9
37.5
80
22.2
15.6
37.7
53.3
95
26.4
18.5
53.1
71.6
110
30.6
21.4
71.2
92.6
1. Show how the speed is converted from km h–1 to m s–1.
2. The thinking distance data imply that the driver has a constant reaction time. Use these data to
work out this reaction time.
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3. The police believe that a reaction time of 1 s is realistic for an alert driver. A driver who is tired
may have a reaction time of 1.3 s. Calculate the total stopping distance for cars being driven by
drivers with these reaction times, at 110 km h–1.
A police car was driven over the road at an accident scene as part of an accident investigation. The
brakes were applied and the car decelerated from 50 km h–1 (the legal speed limit in a built-up area)
to rest. In the process a skid 13 m in length was produced. Study the graph.
relationship between speed2 and braking distance
1000
900
800
700
600
500
400
300
200
100
0
10
20
30
40
50
60
70
braking distance / m
80
90
100
110
good road surface
highway code
poor road surface
Speed /
km h–1
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Speed /
m s–1
Speed2 /
m2 s–2
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Highway
Code data
for braking
Braking
distance on
a good road
Braking
distance on
a poor road
120
Speed /
km h–1
Speed /
m s–1
Speed2 /
m2 s–2
Highway
Code data
for braking
distance / m
Braking
distance on
a good road
surface / m
Braking
distance on
a poor road
surface / m
50
13.9
193
14.7
6.1
24.6
65
18.1
326
24.9
10.4
41.5
80
22.2
494
37.7
15.7
62.9
95
26.4
696
53.1
22.2
88.7
110
30.6
934
71.2
29.7
119.0
Source
4. Use the data from the police reconstruction above to decide which road conditions model applied
at the time of the police experiment.
5. Why was v2 plotted against braking distance in the graph? Use the kinematic equations to help
you with your answer.
Throwing a ball
Question 80S: Short Answer
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Skills
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Instructions
You will need to have studied the kinematic equations and be confident with vector components to
tackle these questions.
Cricket!
This question is about the path followed by an object moving freely under gravity.
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A cricketer lobs a cricket ball from close to the ground at an angle of 45 above the horizontal. Here
we show the variation of height, h, with horizontal distance covered, x. Assume that the ground is
horizontal, and that air resistance is negligible.
15
10
5
0
0
10
20
30
x/m
40
50
1. The horizontal component of the ball's velocity is 15 m s–1. Use the graph to calculate the time for
which the ball is in the air.
2. Write down the ball's vertical component of velocity at the moment of release. Justify your
answer.
3. Use the graph to calculate the time the ball takes to reach its maximum height. Explain your
answer.
4. The maximum height reached is 11.5 m. Use this value to calculate the deceleration of the ball
during this time.
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A slide
Question 90S: Short Answer
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A child is playing with a nearly frictionless car down a smooth flexible track. She arranges the
identical length of track in two ways.
In both arrangements the track runs from the same place on the edge of a table to floor level.
1. She thinks that the car runs off the track at very nearly the same speed in both arrangements
when started from rest at the top of the track (the point nearest the table). Is she right or wrong?
Give a reason for your answer.
2. She thinks the car takes longer to run down the top arrangement than down the bottom
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arrangement, when started from rest at the top of the track. Is she right or wrong? Again, justify
your answer.
3. Suppose she now lets a car of twice the mass of the previous one run from rest down the top
arrangement. Will this car, which has twice as great a mass take a shorter, the same, or a longer
time to run down the track? Give a reason for your answer.
Accurate archery
Question 110S: Short Answer
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Here you predict the positions of objects that are accelerated.
Challenges in target archery
One archery competition requires archers to fire a total of 90 arrows for a maximum possible score of
900. The targets are 1.22 m in diameter at 60, 50 and 40 metres distance. You can model the motion
of the arrow to find out what problems the archer faces.
The arrow leaves the bow at 60 m s–1 and travels at almost this speed horizontally for the whole of its
flight.
The arrow, of course, falls because of the acceleration due to gravity. You can find its position at any
moment by working out how far it has moved horizontally and how far it has fallen vertically.
1. The archer shoots the arrow horizontally at the 40 m target, How far does it drop over this range?
2. How would the archer make allowance for this fall? Think carefully before committing yourself.
3. Now try to calculate the fall at 50 m and 60 m.
4. 50 years ago the release speed of an arrow was about 30 m s–1. What effect would this have on
the vertical distance the arrow fell? Calculate the drop for a range of 60 m to check your answer.
We have ignored the effect of air resistance in these calculations. How could you take account of it?
You can explore it most easily by making a computer model of the motion.
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Lifting a car
Question 150S: Short Answer
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Instructions
This is a short answer question requiring use of F = m a and force as a vector about using a
helicopter to transport a heavy load. Take g = 9.8 N kg–1.
Accelerated lifting
A vehicle is suspended beneath a helicopter on a steel cable.
50 m
steel cable
ground
1. The helicopter lifts the vehicle of mass 1500 kg at a vertical acceleration of 3.0 m s–2. Show that
the tension in the cable is 1.9  104 N.
Moving forwards at a constant speed
At a later time, the helicopter is moving forward in level flight at a constant velocity of 50 m s–1. The
cable carrying the vehicle now hangs at a steady angle of 15 to the vertical as shown in the diagram
below.
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15
vertical
2. Draw labelled arrows on the diagram to show and name the forces acting on the vehicle.
3. Calculate the tension in the cable in this case.
Newton's second law
Question 160S: Short Answer
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Simple calculations and a little thinking
An 80 kg skier has a force of 200 N exerted on him down the slope.
1. Calculate his acceleration down the slope.
2. Is the slope less than or more than 45? Explain your answer.
An ice hockey player has a sudden impact force of 2000 N exerted on him due to unexpected
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collision with the wall. The mass of the player is 100 kg.
3. Find his acceleration.
4. Compare this with the acceleration when he free falls.
Coming out of a dive 75 kg astronauts in training experience an acceleration of 40 m s–2.
5. Calculate the force acting on them.
6. Compare this with the force which normally acts on them when stationary on Earth.
7. Why is it important that they are seated and strapped in before the dive ends?
A 50 g tennis ball may be accelerated at 1000 m s–2 to reach a service speed of 130 mph.
8. Calculate the force required to accelerate the ball.
9. Is your answer reasonable? Comment.
When a force of 200 N is exerted on an asteroid it accelerates at 0.002 m s–2.
10. Find the mass of the asteroid.
A skateboarder
Question 220S: Short Answer
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This question is about a skateboarder coasting up a ramp. Assume that the skateboarder behaves as
a rigid vertical object during the motion. Write the answers in the spaces provided.
A first ramp
A 60 kg skateboarder approaches a ramp.
1.6 m
x
His speed and horizontal displacement, x, are measured at three places as he coasts up the ramp.
These results are used to plot the graph of kinetic energy against x.
300
200
100
0
0.0
0.5
1.0
1.5
x/m
Draw a straight line through the points on the graph and deduce, explaining your
reasoning:
1. The kinetic energy of the skateboarder before mounting the ramp.
2. Whether the skateboard will reach the top of the ramp.
A second run at the same ramp
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Suppose, with a different approach speed, the skateboarder just comes to rest at the top of the ramp.
3. On the axes above add a second graph, as accurately as you can, of this motion. Label this
graph B.
4. Hence show that the initial speed required for the 60 kg skateboarder to stop at the top of the
ramp is about 3 m s–1.
A new ramp
1.6 m
The skateboarder approaches a curved ramp with the same kinetic energy as in the second run.
Both ramps have the same height.
5. On the same axes sketch a third graph to show how kinetic energy may change with x for this
new ramp.
Power and cars
Question 230S: Short Answer
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Teaching Notes | Key Terms
You will need to do some careful thinking and prior research to answer this question. You use some
easily available data for a particular car, and then use your knowledge of physics to build a
mathematical model of the car's behaviour. At the end you should appreciate some of the difficulties
facing car designers.
Data
Record the data for your chosen car:
Power, P =
kW
Mass, m =
kg
Top speed, v =
Useful conversion factors:
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m s–1
1 horsepower =
746 W
1 mph =
0.44 m s–1
1 km h–1 =
0.28 m s–1
Thinking:
At the top speed:
Force distance covered each second = Energy transformed each
second.
That is
d E
F  .
t
t
Or
F  v  P.
You can use this to estimate the driving force Fdrive generated by the motor and transmission.
Use:
F
P
v
to find the thrust.
Thrust, F =
N
More thinking:
The frictional forces on the car increase very rapidly as the speed
increases. In fact Fdrag = k v2 where v is the speed in m s–1 and k is the
drag constant.
Since you know both F and v you should be able to work out k.
Use:
k
Fdrag
v2
to find the drag constant.
Drag constant, k =
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N s2 m–2
Final thoughts:
Now you are in a position to work out the maximum acceleration of the
car at any speed.
Firstly find the forces:
Fnet  Fdrive  Fdrag .
Next find the acceleration:
F
a  net .
m
Now complete the table and use it to plot a graph of acceleration against speed for the car.
Speed / m s–1
Fdrag / N
Fdrive / N
Fnet / N
a / m s–2
5
10
15
20
25
30
35
40
45
50
Now plot a graph of acceleration against speed.
Write a sentence to say why the acceleration gets less as the speed increases.
The bow as a spring
Question 250S: Short Answer
Teaching Notes | Key Terms | Answers | Key Skills
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Bows first appeared between 30 000 and 50 000 years ago in several cultures almost simultaneously.
They represented man's first attempt at machines in which energy is stored before being released.
You transfer energy to the bow during the draw, then the bow transfers this energy to the arrow when
it is released.
With some arrows, release speeds have increased over the last 50 years from about 30 m s–1 to
about 60 m s–1. This represents a considerable increase in the kinetic energy transferred to the
arrows.
Since the energy is transferred from the bow to the arrow, the bow must have stored energy. This in
turn means that more work (force  distance) must have been done by the archer in stretching the
bow. The distance is limited by the size of the archer, so the main improvement has been to increase
the force available. Recreational archers have become stronger! (But military archers were another
matter: one of the sixteenth-century bows found in the wreck of the Mary Rose would have taken
800 N to draw it fully, and the average starting speed of the heavy 60 g war arrows of the time is
estimated at 45–55 m s–1.)
Bows behave rather like springs. For example, a practice bow is quoted as having a draw force of
82 N at 0.7 m. From these figures you can work out its spring constant.
You can also find the energy stored in the bow.
1. If the work done on the bow is equal to the energy carried by the arrow, what kinetic energy does
the arrow carry from a practice bow?
2. A 28 inch (0.7 m) practice arrow has a mass of 26 g. Calculate its release speed from the kinetic
energy calculated in question 1.
3. Calculate the spring constant of the bow.
Rolling up and down slopes
Question 260S: Short Answer
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These questions ask you to think carefully about energy transfer, amongst other things.
Energy graphs
These two graphs represents the motion of an object rolling up a slope and then back down again.
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potential energy
200
180
160
140
120
100
80
60
40
20
0
0.0
0.5
1.0
1.5
2.0
2.5
time / s
3.0
3.5
4.0
4.5
5.0
1. For one graph there was significant friction. Label this graph.
2. The object had a mass of 5 kg. Estimate the height it rose up the slope:
when there was significant friction
when there was little friction.
Working out with a cycle
Question 280S: Short Answer
Teaching Notes | Key Terms | Hints | Answers
You will need to resolve forces and think about energy conservation in this question.
Uphill or down – the same principles apply
A particularly macho mountain biker sets out to prove something. He attacks a 20% hill:
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Quick Help
1. Draw the forces acting on the cyclist, whilst in motion.
The mass of the cyclist plus bicycle is 100 kg.
2. Calculate the size of the retarding force due to gravity, acting along the slope.
3. Calculate the energy he must supply to move 200 m up this slope.
The cyclist covers this 200 m of road, whilst travelling up the hill, in 75 s. Previous tests show that the
retarding force at this speed is 15 N.
4. Calculate the energy he must also supply, just to cover any 200 m at this speed.
5. Find his power output.
On another training session he is seen freewheeling down a hill at a steady speed of 15 m s–1.
6. Use the graph to find the force acting at this steady speed.
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70
60
50
40
30
20
10
0
2
4
6
8
10
12
–1
speed / m s
7. Calculate the steepness of the hill.
Displacement–time graphs
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16
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Question 10M: Multiple Choice
Quick Help
Teaching Notes | Key Terms | Answers | Key Skills
Instructions
You need to be very clear about the connection between velocity and displacement to solve this
problem. Observing collisions and motion on an air track will help you.
Bouncing on an air track
The graphs A to E, show how the quantity plotted on the y-axis varies with time.
B
A
0
0
time
0
0
0
0
time
0
time
E
D
0
C
time
0
0
time
1. Which graph correctly shows the variation with time of the displacement of a vehicle moving
along an air track without friction, between stiff elastic buffers, measured from one end of the
track?
2. Suggest which graph best shows how the height of a bouncing ball varied with time. Why is the
graph not exactly right?
Acceleration–time graphs
Question 40M: Multiple Choice
Teaching Notes | Key Terms | Answers | Key Skills
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Instructions
To solve this question you need to visualise the motion of air-track vehicles. You will also need to
think about how the acceleration changes during collisions of such a vehicle with a barrier.
Acceleration during collision
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The following graphs show how the acceleration of an object varies with time.
A
0
B
time
0
time
0
time
E
D
0
C
time
0
time
An air-track vehicle moves freely to and fro along an air track, colliding elastically with the buffers at
each end of the track.
1. Which one of the graphs A to E best represents how the acceleration of the air-track vehicle
varies with time?
Adding forces graphically
Question 120M: Multiple Choice
Teaching Notes | Key Terms | Hints | Answers
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Instructions
You have to be able to add forces as vectors to find a resultant force. This question encourages you
to do the addition graphically.
Vector addition of forces
The vector diagram below shows three forces in a plane, acting on an object at P.
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3N
P
4N
4N
A
B
C
D
E
zero
2N
1N
1N
1N
1. Which one of the force vectors A to E above represents the resultant of all three forces?
A loading problem
Question 130M: Multiple Choice
Teaching Notes | Key Terms | Hints | Answers
Instructions
This problem will test your understanding of vector addition of forces.
Adding forces
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Quick Help
crane
T
250 N
quay
quay
600 N
barge
The diagram (not to scale) shows a crane being used to load a crate into a barge. A man standing on
the deck of the barge uses a rope to apply a horizontal force of 250 N to the rope supporting a 600 N
crate.
1. What is the best estimate of the tension T in the rope from the crane: A: 250 N; B: 310 N; C: 600
N; D: 650 N; or E: 850 N?
Landing an aircraft
Question 140M: Multiple Choice
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Instructions
This multiple choice item will test your ability to resolve forces.
Rapid deceleration
When a jet lands on the deck of an aircraft carrier it is brought to rest by an elastic cord. The cord is
held between points X and Y. When the aircraft has stopped, the cord has been stretched to the
position XZY. The tension in it is T.
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X
T
Aircraft
Z
30 m
T
Y
1. What is the final effective stopping force exerted by the cord on the plane:
A: 2 T
B: 8 / 5 T
C: 6 / 5 T
D: 4 / 5 T
E: 3 / 5 T?
F = ma:
Some tricky problems!
Question 180M: Multiple Choice
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Skills
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Instructions
These two questions will test your understanding of F = ma. You need to be very clear about the term
resultant force and confident about adding vectors.
Constant acceleration with and without air drag
1. The graphs A to E show five ways in which a force F can vary with speed v. Which of the graphs
A to E best represents how the force F needed to accelerate a car at a uniform rate varies with its
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speed, v? Assume the effect of resistive forces is negligible.
A
B
F
F
0
0
v
0
C
F
0
v
0
E
D
F
F
0
0
v
0
2
v
0
0
v
The graph F shows how the resistive force on a car travelling at speed v varies with v. Which of
the graphs A to E best represents how the total force F needed both to accelerate the car at a
uniform rate and also to overcome the resistive force, varies with its speed, v ?
F
F
0
0
v
Slowing down a bicycle
Question 100E: Estimate
Teaching Notes | Key Terms | Answers
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In this question you are expected to make sensible estimates for quantities and then combine them to
calculate the required answer.
Give your answer to a justifiable number of significant figures and show the units of your estimated
quantities and of your final answer.
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Write the answers in the spaces provided.
Find
How much energy must be dissipated to stop a bicycle, using the brakes?
Estimates
Calculations
A bouncing ball
Question 200E: Estimate
Teaching Notes | Key Terms | Answers
Quick Help
In this question, you are expected to make sensible estimates for quantities and then combine them
to calculate the required answer.
Write the answers in the spaces provided.
To find
The percentage of its kinetic energy that is left when a tennis ball bounces on hard ground.
Estimates
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Calculations
Landing heavily
Question 210E: Estimate
Quick Help
Teaching Notes | Key Terms | Answers
In these questions you are expected to make sensible estimates for quantities and then combine
them to calculate the required answer.
Give your answer to a justifiable number of significant figures and show the units of your estimated
quantities and of your final answer.
Write the answers in the spaces provided.
Situation
A man jumps from the first-floor window of a house and lands on his feet upon soft earth. He bends
his knees as he lands.
Estimate:
1. His approximate landing speed.
Expression and calculation
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Estimates used
2. The average force exerted by the ground on the man as he is brought to rest.
Expression and calculation
Estimates used
A changed situation
Would your answers be significantly different if the man landed on concrete? In each case explain
why / why not.
3. Landing speed.
4. Average force.
Analysing motion sensor data
Question 50D: Data Handling
Teaching Notes | Key Terms | Answers
Quick Help
These questions are about the use of Excel to represent distance versus time data from a motion
sensor and how to analyse the data to produce a velocity versus time graph.
Accelerated motion down an inclined plane onto a rough braking surface
A dynamics trolley with an attached reflecting plate (for ultrasound motion sensing) was released
down an inclined plane onto a rough horizontal surface where it is brought to rest in front of the
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motion sensor. The logged data is given in the Excel file below
Open the Excel Worksheet
Distance–time graph
1. Examine the data and determine the time interval t between ultrasound clicks of the motion
sensor. Calculate the frequency of the clicks. State the distance resolution of the motion sensor
data.
2. Plot a graph of distance from the motion sensor on the y-axis against time on the x-axis.
Calculating velocities
3. Next you will calculate the velocity of the trolley and plot a graph of velocity against time. To
calculate the velocity you need to complete the third column in the Excel table, as described
below.
Remember two things:
velocity = distance / time
changes are always reckoned as (after – before) so that positive (+) changes are increases
and negative (–) changes are decreases.
You will use an Excel formula for velocity, so type in cell C5 =(B7-B3) / (A7-A3) and ENTER, then
fill down (click and hold on cell C5 so that it is highlighted; while still keeping the mouse button
depressed, track the mouse down to cell C65 and then release the mouse).
Explain how this formula calculates the velocity centred on time = 2.15 s.
State the time interval t that you have used.
Explain why the velocities have a negative (–) value.
4. The cells C4, C3 and C66, C67 cannot be filled by this formula.
Explain why this is the case, and state what values you can type directly into these cells by
studying the data.
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Distance and velocity versus time graphs
5. Now try to plot on the same graph axes distance and speed versus time graphs. (This happens to
be possible with these data because the values have similar numerical magnitudes even though
they have different meanings. It is useful because you can easily see that the velocity graph is the
gradient of the distance versus time graph.)
From your graphs (or data table) find the following values:
the time when the trolley stops accelerating down the ramp
its displacement while accelerating
the average acceleration of the trolley during this time.
the total displacement of the trolley.
Accelerated and then braking motion
2.0
distance
velocity
1.5
1.0
0.5
0.0
–0.5
–1.0
2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
time / s
Open the Excel Worksheet
6. Describe the deceleration of the trolley. Is it constant?
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Extension: effect of using a smaller time interval
7. Try taking a smaller time interval (t = 0.10 s) by typing in cell C4 =(B5-B3) / (A5-A3). Fill down and
plot the graphs again.
Comment on the graph compared with the first one.
Uncertainties in measuring g
Question 100D: Data Handling
Teaching Notes | Key Terms | Answers
Quick Help
These questions are about the use of uncertainty (‘error’) bars in Excel to represent experimental
uncertainties, and give practice in interpreting gradients and intercepts of lines of best fit. They also
emphasise the importance of getting data plots into linear format to look for relationships and
systematic error.
Measuring the acceleration of gravity g by (nearly) free fall with tickertape
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paper tape
to a.c. low
voltage supply
0.1kg mass
A class of 14 students performed the same experiment, dropping a 0.1 kg mass and observing its
acceleration under gravity. Each was given a 45 cm length of tickertape to record the motion.
Tickertape machines print 50 dots s–1 onto paper tape, run from the mains frequency and assumed
accurate.
1. State the time interval between the printed dots on tickertape.
How the measurements were made
Students were instructed to start the measurement of distance dropped at the first dot after which
there is a clear gap of 1 or 2 mm to the next dot on the tickertape record.
This may have given rise to some systematic zero error in the time of drop of the order of one or two
tickertimer ‘ticks’ each of 1 / 50 s = 0.02 s.
All students attempted to measure the distances of drop for each dot on the tape to a precision of 0.1
cm and these are collected together in the spreadsheet below.
Open the Excel Worksheet
Estimating the largest source of uncertainty
The major source of percentage uncertainty in the data is the spread in distance values at equivalent
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times, which can be assessed from the table.
2. Estimate the uncertainty of the experiment by plotting a histogram of the total distance dropped
after the final time shown of 0.30 s.
The use of 0.5 cm bin intervals is recommended.
3. Use Excel to find the mean distance dropped at each time interval. Plot a suitable graph to
display the variation of mean distance dropped on the y-axis with time on the x-axis. Add
appropriate y-uncertainty (‘error’) bars to your graph points using your uncertainty estimate. Add a
best fit trendline to the mean data and record the equation of the line using the power law function
(left click on points, right click / Add trendline / Power law / Options / Display equation on chart).
4. Give an explanation of why the relationship has the approximate form
distance = constant  time2
5. To test the relationship in linear form create a column of t 2 values in Excel. Plot a graph of
distance dropped on the y-axis against t 2 on the x-axis. Add the best fit linear trendline, record it
and use it to estimate a value for g the acceleration due to gravity. Estimate the uncertainty of the
result.
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6. Comment on your value for g compared with the accepted value of 9.8 m s–2.
Trying to account for the small residual systematic error
With these data we are still left with the small problem that the best-fit straight line through the data
does not pass exactly through the origin. There is a small offset on this linear graph.
7. Use the linear trendline equation to calculate the x intercept (when y = 0), and suggest a reason
why the intercept may not be zero.
Variation in braking distance with road conditions
Question 110D: Data Handling
Teaching Notes | Key Terms | Answers
Quick Help
Plot and look: level and variation of values
Transport engineers need to check the braking distance of vehicles. You are provided here with data
for the braking distance of a car. The braking distance at a single fixed speed was measured under
four different sets of conditions:
dry road, car unladen
dry road, car laden
wet road, car unladen
wet road, car laden.
The braking distance measurement was repeated 10 times under each set of conditions. The data
are in the spreadsheet here and in the table below:
Open the Excel Worksheet
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Braking distances from 100 km h–1 (27.8 m s–1) for a Pontiac Grand AM
Conditions
Braking distances / m
1
2
3
4
5
6
7
8
9
10
Dry unladen
44.8
45.0
45.0
45.3
45.5
45.5
45.3
44.4
44.4
45
Dry laden
46.8
45.1
46.5
46.7
46.4
46.4
46.0
46.7
46.6
46
Wet unladen
60.1
58.2
56.1
58.9
54.3
55.9
59.3
57.5
57.8
61
Wet laden
50.2
52.3
49.8
51.9
50.9
51.1
55.0
54.9
48.3
56
You will find the mean and spread of braking distances for each set of conditions, and decide whether
the braking distance varies with the conditions. You can also calculate the mean and spread of the
deceleration of the car, and see how that varies under different conditions.
Remember that the data here are for braking distance, not the overall stopping distance, which needs
to include the ‘thinking’ distance.
Braking distance
1. Use the table above, or the spreadsheet, to make dot-plots of the braking distance data, for each
set of conditions.
2. For each set of conditions, calculate the mean braking distance, and the range of values.
Estimate the spread as ± half range. Estimate the percentage spread.
3. Consider whether the braking distance is affected by the conditions.
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Deceleration
4. Use the equation v 2 = u 2 + 2as, with final velocity v = 0, initial velocity u = 27.8 m s–1 (100 km
h–1) and s = braking distance, to calculate the mean deceleration a of the car in each set of
conditions. Compare the values with the acceleration due to gravity, g = 9.8 m s–2.
5. Use the percentage spread in braking distances to estimate the percentage spread in the values
of mean deceleration.
You have practised
1. Asking questions about the level and variability in a batch of data.
2. Calculating the mean of a batch of data as a measure of the level.
3. Estimating the spread (half range) as a measure of the variability and expressing this as a ± % of
the mean.
4. Using dot-plots as a way of graphically showing the variation in a batch of data.
Variation in braking distance between cars
Question 120D: Data Handling
Teaching Notes | Key Terms | Answers
Quick Help
Plot and look: level and variation of values
Transport engineers need to check the braking distance of vehicles. You are provided here with data
for the braking distance of a number of different cars. The braking distance was measured for each
car from two different initial speeds. The tests were done on dry roads, with unladen cars.
The data are in the spreadsheet here, and in the table below:
Open the Excel Worksheet
Braking distances of different cars:
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Braking distances from 90 km h–1 (25 m s–1) and 120 km h–1 (33.3 m s–1)
Braking distance /
m (at 90 km h–1)
Braking distance / m
(at 120 km h–1)
Honda Integra GS-R
42.0
74.4
Audi A4
43.5
80.7
BMW Z3 (2.8)
36.9
64.5
Ferrari 550 Maranello
33.6
59.7
Lexus ES300
42.0
73.8
Lexus LS400
45.3
78.0
Mazda MX-5
45.6
76.8
Mazda Protege
47.4
86.1
Mercedes C36
36.0
63.0
Mercedes SLK230 K
36.0
62.7
Nissan Maxima
42.0
72.9
Nissan 200SX
38.7
68.4
Saab 9000 Aero
36.6
66.3
Subaru Liberty RX
40.8
70.8
Toyota Camry V6
43.5
82.2
Toyota Corolla
55.8
95.7
Porsche 911 Carrera 4
37.8
66.9
Model
Remember that the data here are for braking distance, not the overall stopping distance, which needs
to include the ‘thinking’ distance.
1. Use the table above, or the spreadsheet, to make dot-plots of the braking distances, for the two
initial speeds.
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2. For each initial speed, calculate the mean braking distance, and the range of values. Estimate the
spread of braking distances at each speed as ± half range. Estimate the percentage spread at
each speed.
3. For each car, and each initial speed, calculate the deceleration of the car. Use the equation v 2 =
u 2 = 2as, with s = braking distance, final velocity v = 0, initial velocity u = 25.0 m s–1 (90 km h–1)
or initial velocity u = 33.3 m s–1 (120 km h–1). Compare the values with the acceleration due to
gravity g = 9.8 m s–2.
4. Calculate the ratio of braking distances for each car, at the two initial speeds. Explain why this
quantity might be more or less constant.
You have practised
1. Asking questions about the level and variability in a batch of data.
2. Calculating the mean of a batch of data as a measure of the level.
3. Estimating the spread (± half range) as a measure of the variability and expressing this as a ± 
of the mean.
4. Using dot-plots as a way of graphically showing the variation in a batch of data.
Cycling through air
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Question 190D: Data Handling
Quick Help
Teaching Notes | Key Terms | Answers
This question asks you to process some data and to draw sensible conclusions from it.
Data
A team of five people work together to find out how the effective power of a cyclist varies with the
speed at which he is travelling. The diagram shows how they do it.
measuring spring balance
freewheeling
dragging
timing from X to Y
20 m
Run
number
Time
taken / s
Average
speed /
m s–1
Force / N
1
13.4
2.1
2
8.0
4.0
3
20
1.4
4
5.0
8.5
5
5.7
6.5
6
3.65
18
7
4.45
11
8
12.5
2.0
9
8.3
4.2
10
3.6
18.3
Analysis
Complete the table.
Draw up graphs of force / speed and power / speed.
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Power / W
Interpret each graph in terms of the physics that you know.
Retarding a cyclist
Question 270D: Data Handling
Teaching Notes | Key Terms | Hints | Answers
Quick Help
Think carefully, and calculate a little.
Two speeds, one principle
Here is some data for the retarding forces acting on a cyclist, travelling through still air on a smooth
level surface.
Speed / m s–1
Retarding force / N
2.24
4.9
4.47
7.6
8.94
18
17.9
61
A fit professional 69 kg racing cyclist can provide a steady power output of 375 W, mounted on his 11
kg bicycle.
1. Find the speed at which he can travel on the level.
2. On a steady climb he is reported as gaining height at the rate of 120 m / minute. Find his speed
now.
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3. On another day he is travelling at steady 12 m s–1 down a hill, without pedalling. Draw a precise
diagram to show the retarding and gravitational forces acting on the bicyclist and his bicycle,
making the connections between them plain.
4. Find the steepness of the hill.
Inertia reel seat belts
Question 170X: Explanation–Exposition
Teaching Notes | Key Terms | Answers
Quick Help
You will need to do some careful thinking and research to answer these questions.
A seat belt
By law, all British cars must be fitted with seat belts. This is so as to reduce the number of deaths and
serious injuries caused by traffic accidents, and the subsequent medical costs to the taxpayer.
1. If a car is in a head-on collision with a wall, how does a seat belt help to reduce injuries to an
occupant of the car?
An inertia reel
Inertia-reel seat belts are those where, when the car is at rest (or travelling with a constant velocity), it
is easy to pull more belt off the reel which houses the belt not in use. However, when the car is
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accelerating or decelerating, the belt is firmly gripped by the reel.
2. Explain how the reel system must work, including a sketch or sketches in your explanation.
The mysteries of mass
Reading 10T: Text to Read
Teaching Notes | Key Terms
Quick Help
This is to stimulate your thinking and curiosity about something it is easy to take for granted: mass.
Thinking about mass
If you kick it, it hurts your foot. The more there is of it the harder it is to get it moving. It was Sir Isaac
Newton who first put into a logical and clear form these vague ideas about what everything seems to
be made of. At least, everything that is anything: stones, people, water, the Moon and the Earth, and
so on. When he did this, sometime in the late seventeenth century, he couldn't do it by giving a simple
independent definition of what he was talking about. He named the 'stuff' of the Universe mass. He
could only define it in terms of what it did. It did two things, one of which is where this paragraph
starts. Mass has the property that it will not change how it is moving (or not) unless pushed. To be as
precise as Newton, the mass we are talking about is what you get when you divide a force by the
acceleration it produces. So we can't really understand what mass 'is' unless we also understand
what a force is and what acceleration is. We can then tell how much mass an object has by working it
out from the equation m = F / a.
We'll come to the second property that mass has later.
Don't push me around
This first property of mass, a kind of unwillingness to be pushed around, is called inertia. So if you
read about inertial mass it isn't a special new kind of mass, it's just the ordinary kind, but the writer
wants you think about the unwillingness-to-change-movement aspect of it.
This may all seem very reasonable, but think about it. How do you know how much force there is,
say, acting on an object? Well, you measure the acceleration and multiply it by the mass, obviously.
But we don't know what the mass is unless we already know what the force is (see the end of the first
paragraph). We seem to be going round in circles here. The only way we can break out of the circle is
to get a lump of something and say 'This is what I mean by mass. Here is 1 kilogram of it. Hang on to
it for dear life'. So there exists a lump of platinum–indium alloy, carefully isolated from the common
herd in an air-conditioned cellar in Sèvres, France. This is the actual and only basic international
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standard of mass.
Newton was aware of the problem, and called mass simply 'quantity of matter'. But what is matter?
We now know, as Newton didn't, that matter is made up of atoms, which in turn are made up of a
small number of elementary particles: electrons and quarks are enough to make everyday materials.
The simplest substance, hydrogen, has a nucleus (with three quarks) and a single electron. One way
of defining mass would be to say that this atom represents one unit, two represents two units and so
on. The mass of a bag of sugar would be represented by an astronomical number, of course, so it
isn't a very practical unit. But we could call a number of hydrogen atoms, say for the sake of argument
6.0  1026, a kilogram, and the everyday world wouldn't be any the wiser, or sorrier.
An appeal to gravity
But we still don't know what mass actually is. Why should a hydrogen atom – or anything else – have
this unwillingness to change how it moves? Perhaps the other property of mass will give us a clue.
This property was also identified by Newton. He realised that the Moon went around the Earth
because the two bodies attracted each other with a force he named as gravity. He used this idea,
and the inertia property, to calculate with good accuracy how long the Moon should take in travelling
around its orbit. He used the idea that the force between the two masses depended on how far apart
they were – and how much mass each had. This seemed to be the same kind of 'mass' as the inertial
kind. He could cancel the mass out in his calculations and still get the right answers. But gravitational
mass is linked to force by a different relationship than inertial mass is. The gravitational relationship is
F G
Mm
r2
.
Here M and m are the two masses (each with its own inertia), r is the distance between them and G
is a constant needed to get all the numbers to work out right. This relationship defines gravitational
mass: the property of mass that allows it to attract another mass across empty space (and non-empty
space, come to think of it – you can't escape from gravity).
Maths will keep creeping in to physics, often making things simpler – but often also hiding the
genuine underlying mystery of it all. Now for Newton the really big mystery was how a force from the
Earth could actually get to the Moon. There was nothing to carry it. 'I just give up' he said (in Latin
'Hypothese non fingo'). But he decided to accept it – that is just how things actually were.
Are the two 'kinds' of mass really the same?
But are inertial mass and gravitational mass 'the same'? Inertia and gravity are such different things
that on the face of it, it would seem to very unlikely if they were. But physicists have checked and
rechecked the idea that inertial mass and gravitational mass can be described by the same numbers.
The question is 'Is a kilogram of inertial mass the same as a kilogram of gravitational mass?'. The
answer is 'yes', to as high an accuracy as we can measure anything. Could this be a coincidence, or
is it a sign that there is something deeper underlying all this?
Well, thank you, Sir Isaac. Now let's give a warm welcome to Albert!
Albert Einstein came to thinking about gravity via a deep study of time and space. He got to thinking
about time and space because he thought that there was something odd about physicists' ideas
about light. So what has light got to do with space and time? That's another story, but we can remind
ourselves that nowadays distance (as measured by surveyors planning roads and railways) is defined
in terms of how long light takes to travel the distance being measured. There used to be a standard
distance (another expensive bit of platinum in the Sèvres cellar, next to the standard kilogram), but
that has become an obsolete museum exhibit. To cut a rather a long story short, Einstein has
managed to convince physicists that mass has a deeper and more fundamental property: it curves
space–time. The Moon doesn't move in a circle through space. It moves in a straight line through
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space–time. But the space–time is curved. Its orbit is a kind of optical illusion because we humans
haven't evolved a sense of seeing space–time as it really is. Which may be just as well for ordinary
existence. In Einstein's theory, called General Relativity, he showed that there was no way to tell the
difference between an acceleration and a uniform gravitational field. This had to mean that inertial
mass and gravitational mass were indistinguishable – and so identical.
Einstein's work also revealed something else. He showed that there is a deep connection between
energy and mass. In General Relativity, it is energy that gravitates, and has inertia. What we call
mass is just a part of the total energy - a part we usually forget about, despite its being enormous.
The mass of an object can be understood as just the so-called 'rest energy' - the energy the object
has when not travelling past us.
Now we get somewhere on what mass 'really is'. The mass of an object is just the energy Erest = mc 2
it has when it is not moving past us: m = Erest / c2 . Does it help? Now the problem is what energy
'really is'. But at least two difficult questions have become one difficult question.
And we get something new; travelling at high speed past us the energy of an object starts to have
noticeable inertia of its own. That is the reason why things can't go past us faster than light.
Gravitons
If you have read this far you should have realised that the property of matter we call mass is not at all
a simple thing to understand. And this is where it gets even worse. Again.
The small particles that everyday matter is made of, electrons and quarks, not to mention the fairly
rare particles like mesons and muons, behave and interact in very strange ways that Newton – or
even Einstein – never dreamt of. Particles of matter like this are held together or pushed apart by
forces and can be accelerated and slowed down like cars and aeroplanes. But on their tiny scale the
modern theory is that the forces are carried by special force-carrying particles. The particles are
called bosons, and the most well-known of these is called a photon. This particle is not just 'a little bit
of light', but also carries the electromagnetic force between the particles. For example, electrons repel
each other because they swap photons, which transfer momentum between the particles. Or in
simple language, they give each other a rapid succession of little pushes. Protons also repel each
other like this. These force-carrying swapped particles exist for a short time only. They only exist at all
because, for a very short time, they are allowed to break the law of conservation of mass–energy.
They aren't real particles in the ordinary sense and so are called virtual particles.
But is this how gravity works? If this model applies to gravitational forces, then we have to have a
boson that carries the force. This is called a graviton. But the graviton has yet to be discovered. If it
exists it would overcome Newton's objection to his own theory – how a force can be carried by empty
space.
Higgs enters the field – and brings his own
Forces like electricity and magnetism have fields – regions where the forces exist and work. Think
about electricity. Its field is caused by the fact that some particles have electric charge. This charge,
amazingly, is the same for every free charged particle in the known universe. Quarks (which are
always found trapped inside bigger particles) carry either one-third or two-third size lumps of this
charge. So charge is a fundamental property of matter. It comes in two flavours: positive and
negative. The only particle of matter that doesn't have a charge of some kind locked in it is the
neutrino. This is a genuinely neutral particle. Other neutral particles like neutrons are neutral because
the opposite charges carried by the quarks inside them just cancel each other out. They add up to
zero.
The theories of fields and particles have become extremely mathematical in the last 40 years, but the
mathematics has had the habit of producing equations which contain odd, awkward things which
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might be interpreted as particles. The earliest particle to be predicted in this way was the positive
electron (positron). This happened in the early 1930s. The advanced theory – and the double-particle
system – was needed to explain the very existence of these kind of particles and fields.
In the 1960s an even more advanced theory about fields and particles emerged. This was the Higgs
field – and its particle the Higgs boson. Peter Higgs is a British physicist, educated in Birmingham
and Bristol, formerly Professor of Theoretical Physics at Edinburgh University. But the Higgs field and
boson were not about matter and its ability to have electric charge, but about matter and its ability to
have mass. At last we are getting to a theory that might explain why matter has this mysterious but
rather useful property.
There's not as much matter in mass as you might think
Now one of the basic objects that make up the universe, the quark, seems to have very little inertial
mass. The three quarks that make up a proton, for example, are measured to have masses that add
up to just a few per cent of the total mass of the proton. The rest of the mass is due to the internal
energy of the proton – remember the Einstein equation: m = E / c2. The quarks are held by very
strong fields (potential energy) and move around a lot (kinetic energy) and keep exchanging virtual
particles (which have mass as long as they last). And it gets worse. The quantum theory of particles
is one of the most successful physical theories of all time. It makes predictions to an extremely high
degree of accuracy, for example. The most modern theory (well – so far) only works if electrons and
quarks have zero mass.
As long ago as 1966 Peter Higgs thought of a way to get around this rather serious contradiction that
mass does actually exist (try kicking a 10 kilogram lump of iron) with this splendid theory that it can't.
The Higgs idea is like this: a truly massless particle such as a photon has to travel at the speed of
light. A particle with mass can't ever travel at this speed. But what if all of space is filled with a
'something' that can grab hold of some particles and slow them down? This space-filling something
would be a field (like the gravitational or electrical fields). It would grab hold of the particles with its
force carrier and have the effect of slowing them down to sub-light speeds. The field is now called the
Higgs field and the grabber particle is the Higgs boson. This means that the 'really' massless particles
like electrons and quarks have an 'effective mass'. They have different effective masses insofar as
they interact differently with the Higgs field.
As we said earlier, these force-carrier particles only exist for extremely short times. The only reason
that they can exist at all is due to the indeterminancy principle which says that the universe doesn't
mind if something impossible happens or exists just as long as the product of energy and time
involved is less than a certain value (6.6  10–34 joule seconds). But as long as they obey this rule we
can have as many of these particles as we like. If, however, we want to 'see' them they have to last
long enough for our instruments to detect them. This means providing a lot of energy in the right place
at the right time. This is what the big particle accelerators at places like CERN are for. The new
collider at CERN – the LHC or Large Hadron Collider – should be able to provide enough energy for
the Higgs boson to be observed. If it exists.
More to read
Bowdery C 1996 The origin of mass Physics Education 31 237–41. This is a good easy-to-read
introduction to modern theories about mass.
Davies P and Gribbin J 1991 The Matter Myth: Towards 21st Century Science (Viking Penguin). A
readable summary of modern physics: goes from the death of materialism via quantum physics, the
origin of the universe and a little string theory to what is life?
Greene B 1999 The Elegant Universe (Jonathan Cape). All you want to know about practically
everything; deals with string theory, which has been ignored in this reading. Take a week off to read
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the book and find out why.
Revision Checklist
I can show my understanding of effects, ideas and relationships by
describing and explaining:
the meaning of relative velocity
A–Z References: relative velocity
Summary Diagrams: Relative velocity
motion under constant acceleration and force
A–Z References: acceleration, mass, Newton's laws of motion
Summary Diagrams: Stepping through uniform acceleration, Logic of motion 1, Logic of motion
2, Graphs for constant acceleration, Graphs for realistic motion, Computing uniform acceleration
the parabolic trajectory of a projectile
that the horizontal and vertical components of the velocity of a projectile are independent
that a force changes only the component of velocity in the direction of the force
A–Z References: projectile
Summary Diagrams: A parabola from steps
that work done = force × displacement in the direction of the force
power as the rate of transfer of energy (energy transferred per second)
A–Z References: work, kinetic energy, potential energy, power
Summary Diagrams: Calculating kinetic energy, Kinetic and potential energy, Flow of energy to
a train, Power, force and velocity
I can use the following words and phrases accurately when
describing the motion of objects:
relative velocity
A–Z References: relative velocity
Summary Diagrams: Relative velocity
acceleration, force, mass
A–Z References: acceleration, mass, Newton's laws of motion
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Summary Diagrams: Stepping through uniform acceleration
gravitational force, weight, acceleration of free fall
A–Z References: weight, mass, free fall
kinetic energy, potential energy, work done, power
A–Z References: kinetic energy, potential energy, work, power
Summary Diagrams: Calculating kinetic energy, Kinetic and potential energy, Flow of energy to
a train, Power, force and velocity
I can interpret:
vector diagrams showing relative velocities
A–Z References: relative velocity
Summary Diagrams: Relative velocity
graphs of speed–time and distance–time for accelerated motion, including the area under a
speed-time graph and the slope of a distance–time and speed–time graph
A–Z References: acceleration
Summary Diagrams: Graphs for constant acceleration, Graphs for realistic motion
I can calculate:
the resultant vector produced by subtracting one vector from another
A–Z References: relative velocity
Summary Diagrams: Relative velocity
speed from the gradient (slope) of a distance–time graph
distance from the area under a speed–time graph
Summary Diagrams: Graphs for constant acceleration, Graphs for realistic motion
the unknown variable, when given other relevant data, using the kinematic equations:
v  u  at ;
v 2  u 2  2as ;
s  ut  21 at 2 ;
s
(u  v )
t
2
Summary Diagrams: Logic of motion 1, Logic of motion 2
the unknown quantity, given other relevant data, using the equation F = m a
A–Z References: Newton's laws of motion
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the path of a moving body acted upon by the force of gravity when the body is (a) moving
vertically and (b) moving both vertically and horizontally, including use of the kinematic
equations and of step by step changes of velocity and displacement in short time intervals
A–Z References: free fall, projectile
Summary Diagrams: Stepping through uniform acceleration, Computing uniform acceleration, A
parabola from steps
the kinetic energy of a moving body using KE = ½ m v2
A–Z References: kinetic energy
Summary Diagrams: Calculating kinetic energy
the change in potential energy when a massive body changes height in a gravitational field
using  PE = m g h
A–Z References: potential energy
Summary Diagrams: Kinetic and potential energy
the work done (energy transferred) E = F s
force, energy and power: power = E / t = F v
A–Z References: work, power
Summary Diagrams: Power, force and velocity, Flow of energy to a train, Kinetic and potential
energy,
I can show my ability to make better measurements by:
measuring force, acceleration, velocity, kinetic and potential energy, work, power with known
uncertainty
A–Z References: accuracy, systematic error, uncertainty
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