Along the flat and up the hill
Question 240W: Warm-up Exercise
1. Net force must be zero, so 5 N
2.
Work  force  distance
 5m 5N
 25 J
3.
Net force must be zero, so 8 N
4.
Work  force  distance
 5m 8N
 40 J
5.
Energy to lift  energy supplied  energy dissipated
 185 J  25 J
 160 J
6.
Energy transferred  m  g  h
160 J
h 
80 N  10 N kg 1
 0 .2 m
7.
Energy supplied  energy to lift  energy dissipated
 160 J  40 J
 200 J
8.
Energy transferred  force  distance
force 
200 J
5m
 40 N
9.
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drag
retarding force
due to gravity
Calculated steps
Question 20S: Short Answer
1.
a
check point
WILTSHIRE
10 km
2.
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b
check point
WILTSHIRE
10 km
3.
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c
b
a
check point
WILTSHIRE
10 km
4. Use A =  r 2 to get: 1 hour, 50.3 km2; 2 hours, 201.1 km2; 3 hours, 452.4 km2. Yes, the area to be
searched is proportional to the time2.
5. Use A =  r 2 to get: 3 km hr–1, 28.3 km2; 5 km hr–1, 50.3 km2; 5 km hr–1, 78.6 km2. Yes, the area
to be searched is proportional to the speed2.
6. The faster you go, the more often you need to check where you are, in order to retain the same
predictive power about where you will be next.
Coping with graphs
Question 30S: Short Answer
1. The gradient of the speed-time graph is constant
2. gradient of graph = acceleration
gradient 
15 m s –1
 1.5 m s – 2
10 s
or
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a
v u
t
a
15 m s –1  0 m s –1
 1. 5 m s – 2
10 s
3. at 4s, speed = slope of tangent

62
 6.2 m s –1
10
4. at 10s, speed

106
 15.1 m s –1
7
5. average speed= 10.7 m s –1
6. average acceleration

15.1 m s –1  6.2 m s –1
 1.48 m s – 2
6s
7. Answers correspond well, but difficult to draw tangents accurately.
Calculating accelerated steps
Question 50S: Short Answer
Use the relationships v = a  t, x = v  t, and the approximation (average speed = (initial
speed + final speed) / 2) to get these values:
Time
interval / s
Speed at
start of
interval /
m s–1
Acceleration
during
interval /
m s–2
Speed at
end of
interval /
m s–1
Average
speed
during
interval /
m s–1
Distance
covered
during this
interval / m
0–0.5
5.00
–4.00
3.00
4.00
2.00
0.5–1.0
3.00
–3.00
1.50
2.25
1.125
1.0–1.5
1.50
–2.00
0.50
1.00
0.50
1.5–2.0
0.50
–1.00
0.00
0.25
0.125
The platform must go 3.75 m beyond the trackside bollard.
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Uniform acceleration
Question 60S: Short Answer
1. v = u + a t, 60 m s–1 from zero at 4 m s–2 so t = 15 s.
2. s = ½ a t2. Substitute in this equation to get 450 m.
3. The speed of air over the wings is the key. If air is already moving over the wings (head winds)
you do not need as high a forward velocity as in still air.
4.
25
20
15
10
5
0
0
2
4
6
8
10
time / s
5. Acceleration is uniform up to t = 4 s and decreases from t = 4 to t = 8 s, after which car moves
with constant speed
6. As car speed increases, air resistance increases with speed, resulting accelerating force
decreases, so acceleration decreases. When resistive force = driving force, resultant force = 0
and car moves with uniform speed.
7. In first 5s, distance travelled = area under curve
1
  15.4 m s –1  5 s  38.5 m
2
From 5s to 9s, area of trapezium
 15.4 m s –1  23 m s –1 
  4 s  76.8 m



2


Total distance in first 9s = 38.5 + 76.8 = 115.3m
8. Estimate acceleration = average gradient
15.0 m s –1  0 m s –1

 3.0 m s – 2
5s
Braking distance and the Highway Code
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Question 70S: Short Answer
1. For example, 50 km h–1 = 50 1000 m / 3600 s = 13.9 m s–1 etc.
2. For example 13.9 m s–1  t = 9.7m thus t = 0.7 s.
3. Use new reaction times and add appropriate braking distance. 30.6 m + 71.2 m = 101.8 m and
39.8 m + 71.2 m = 111 m.
4. 13.92 =193 m, use the graph to decide which surface gives a braking distance of 13 m.
5. Use v2 = u2 + 2 as. v = 0 and a is negative here so, u2 = 2 as, so plot initial speed squared
against braking distance, assuming a is constant, which should give a linear graph through the
origin.
Throwing a ball
Question 80S: Short Answer
1. time in air

range
46 m

 3.1s
horizontal component 15 m s –1
2. 15 m s–1, same as horizontal component because angle = 45°
3. time to maximum height = half total time = 1.5 s
4.
υ 2  u 2  2gs υ  0, u  15 m s –1 so g 
(15 m s –1 ) 2
u2

 9 .8 m s – 2
2s
2  11.5 m
A slide
Question 90S: Short Answer
1. Right, as both cars have the same amount of kinetic energy on leaving the track. A good answer
would relate this to the same drop in height, explicitly linking the change in potential energy to the
change in kinetic energy.
2. Right, as the bottom arrangement gains kinetic energy more quickly, making the average speed
greater. A good answer would relate this more rapid change in kinetic energy to the steepness of
the slope–top arrangement gets faster sooner!
3. Same, assuming the frictional forces are really negligible as the change in kinetic energy equated
to the change in potential energy soon gives:
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v  2gh
which is independent of the mass.
Accurate archery
Question 110S: Short Answer
1. Time of flight
40 m

 0.67 s
60 m s –1
Vertical drop
2
h
1 2 1
2 
gt   9.81m s – 2   s   2.2 m
2
2
3 
2. Aim above the target.
3. Similar calculations
2
h50
h60
1
5 
  9.81m s – 2   s   3.4 m
2
6 
1
2
  9.81m s – 2  1 s  4.9 m
2
4. Twice the trip time so four times fall, at 60 m now drops 20 m.
Lifting a car
Question 150S: Short Answer
1. Applying Newton's second law, where T is the force due to tension in the rope, on the car.
T – mg = ma
T – 1500 kg × 9.8 m s–2 = 1500 kg × 3.0 m s–2
T = 1.9 × 104 N
3. T cos 15° = mg
T = 1500  9.8 / (cos 15°)
T = 1.5 104 N.
Newton's second law
Question 160S: Short Answer
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1.
a
F
200 N

 2 .5 m s – 2
m
80 kg
2. Slope is less than because at this angle, force down slope = weight / 2, which is greater than
200 N.
3.
a
2000 N
 20 m s – 2
100 kg
4. Twice
5. 3000 N
6. Four times
7. Astronauts will be brought to rest at the end of the dive which may require the application of a
large force resulting in injury.
8. F = ma = 0.050 kg × 1000 ms–2 = 50 N
9. Not very large – as you can probably lift 250 N with one arm.
10.
m
200 N
0.002 m s – 2
 1.0  10 5 kg
A skateboarder
Question 220S: Short Answer
1. 320 J, from the intercept, when x = 0 then all of the energy is kinetic.
2. Yes, from intercept with y-axis, kinetic energy would be zero at about x = 2.1.
3. Look for the slope, same rate of exchange of energy, and intercept kinetic energy = 0 when x =
1.6 m.
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300
200
100
0
0.0
0.5
1.0
1.5
x/m
4. Graph shows that you need 250 J:
1
2
mv 2  250 J
v2 
250 J
2  60 kg
v  2 .9 m s 1.
5. Same starting and end points as the last graph, but now kinetic energy changes more rapidly with
x when x is larger.
300
200
100
0
0.0
0.5
1.0
1.5
x/m
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The bow as a spring
Question 250S: Short Answer
1. Assuming energy stored
1
1
 Fe  ke 2
2
2
W 
1
 82 N  0.7 m  28.7 J
2
2.
28.7 J 
1
 0.026 kg  υ 2
2
υ
2  28.7 J
 47 m s –1
0.026 kg
k 
F
82 N

 117 N m –1
x
0 .7 m
3.
Rolling up and down slopes
Question 260S: Short Answer
1. The lower line.
2. PE = mgh
135 J = 5 kg × 10 m s–2 × h
h
135 J
50 kg  50 m s  2
 2 .7 m
PE = mgh
100 J = 5 kg × 10 m s–2 × h
h
100 J
50 kg  50 m s 2
2m
Working out with a cycle
Question 280S: Short Answer
1.
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D
W
2.
F  1000 N  cos( slope )
1
F  1000 N 
26
 196 N
3.
Energy transferred  force  distance
 200 m  196 N
Energy transferred  40 kJ
4.
Energy transferred  force  distance
 200 m  15 N
Energy transferred  3 kJ
5.
Energy transferred  3 kJ  39.2 kJ
42.2 kJ
Power 
75 s
Power  563 W
6. The graph gives the retarding force as 45 N.
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70
60
50
45 N
40
30
20
10
0
2
4
6
8
10
12
speed / m s–1
14
16
18
7. Since speed is constant, the driving force, provided by the component of gravity down the slope,
must also be 45 N. Resolve to get:
1000 N  sin   45 N
 45N 

  sin 1
 1000 N 
 2.6
Displacement–time graphs
Question 10M: Multiple Choice
1. The concept of displacement as a vector should guide you to B. This graph implies that the
velocity is positive then abruptly changes to negative and constant, most of the time. The
transition from positive to negative is almost instantaneous.
2
Graph D has the parabolic shapes we might expect for the way the height of a bouncing ball
might vary with time, however in reality on each bounce the ball would lose some energy and so
not reach the same height as on the previous bounce.
Acceleration–time graphs
Question 40M: Multiple Choice
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1. The vehicle has constant velocity (approximately) between collisions so its acceleration between
collisions is zero. During successive collisions its acceleration must be positive and negative to
produce the change from positive to negative velocity as the vehicle changes direction. The
acceleration will build up to a maximum and then decrease to zero as the buffer deforms. Hence
the best graph is C.
Adding forces graphically
Question 120M: Multiple Choice
1. The 3 N and horizontal 4 N forces together give a resultant of (32 + 42) = 5 N in the opposite
direction to the remaining 4 N force. So the net force is 1 N in the D direction.
A loading problem
Question 130M: Multiple Choice
1.
(2502 + 6002) = 650 N
Landing an aircraft
Question 140M: Multiple Choice
1.
F = 2 × T × (3/5) = 6T/5
F= ma:
Some tricky problems!
Question 180M: Multiple Choice
1. C
2. D
Slowing down a bicycle
Question 100E: Estimate
1. Mass of cyclist plus bike perhaps 100 kg, speed about 4 m s–1
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kinetic energy 
1
2
mv 2

1
2
 100 kg  ( 4 m s 1 ) 2
 800 J.
A bouncing ball
Question 200E: Estimate
1. Drop height 1 m, rebound height 0.25 m.
mghrebound
fraction of energy 
mghdrop
0.25
1
 25%.

Landing heavily
Question 210E: Estimate
1. Height of drop 3 m.
kinetic energy  potential energy
1
2
mv 2  mgh
v 2  2  10 N kg 1  3 m
v  8 m s 1.
2. Impact hole 0.1 m deep, knees bend by 0.5m, mass of man 80 kg
work done  potential energy
F  d  mgh
80 kg  10 N kg 1  3 m
0 .6 m
F  4 kN .
Note that this is five times the man's weight and a dangerously large force.
F 
3. Landing speed no change
4. Average force will be somewhat large, as the concrete does not yield. So, as work done must be
the same, force is larger still.
Analysing motion sensor data
Question 50D: Data Handling
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1. t = 0.05 s; frequency = 1 / t =1 / 0.05 s = 20 Hz; distance resolution is to the nearest 1 mm
2.
Accelerated and then braking motion
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5
time / s
3. The formula works out the distance two clicks ahead and subtracts the distance two clicks ago
and then divides by the time interval t = 0.20 s.
The velocities are negative because the trolley is getting closer to the motion sensor, i.e. its
distance from the sensor is decreasing.
4. For cells C4 and C3 there are not enough cells before to read a formula value; for cells C66 and
C67 there are not enough cells after to read a formula value.
Since the trolley is stationary both at the start and at the end of the motion the value 0 m s–1 can
be entered in all these cells because the distance is not changing around these times.
5. Around 3.2 s; the trolley travels about (1.455 – 1.055) = 0.40 m during the acceleration;
acceleration = change in velocity / change in time = (–0.825 – 0) m s–1 / (3.2 – 2.3) s = –0.92 m
s–2 (negative sign means towards the sensor); displacement = (0.177 – 1.455) m = –1.278 m
(negative sign means towards the sensor).
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
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Open the Excel Worksheet
6. The deceleration of the trolley is low at first and increases (greater gradient on the velocity-time
graph as it approaches rest). It is not constant.
7. With smaller time intervals the trolley travels less far in the time interval. Values of the velocity
now have larger random fluctuations. This shows up in the velocity time graph which is ‘noisier’
and less smooth than with the larger time interval.
Uncertainties in measuring g
Question 100D: Data Handling
1. Interval = 1 / 50 s = 0.020 s
2.
spread in total distance dropped after 0.3 s
4
3
2
1
0
36
37
38 39 30
distance / cm
41
42
At t = 0.30 s the mean distance dropped = 39.7  2.0 cm (or 5%) ignoring one low outlier value
at 36.3 cm.
3.
d mean
0.0
0.2
0.4
1.2
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d mean
2.3
3.8
5.6
7.8
10.8
14.1
17.4
21.3
25.8
29.9
34.5
39.7
g by tickertape
45
40
35
30
25
20
15
10
5
0
0.0
0.1
0.2
time / s
0.3
0.4
y = 518 × 2.1 or distance = constant  time2.1
4. For constant acceleration we know that s = ut + ½at 2 but here u = 0 so we expect a relationship
of the form s = ½at 2 for a mass accelerating from rest.
5.
time2 / s2
0.0000
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g linear graph
45
40
35
30
25
20
15
10
5
0
0.00
0.02
0.04 0.06
time2 / s2
0.08
0.10
y = 447x – 0.4
Using s = ½gt 2 gives an estimate for g of:
(12 m s -1 )2  (8 m s -1 )2  208 m 2 s -2  14.0 m s -1
6. This value is systematically low by at least 4%, because the fall is not entirely free because of
drag from the tickertape and printer.
7. t 2= 0.4 / 447 = 0.000895 s2 ; this corresponds to a time
t  0.000895  0.030 s
This confirms that there may have been some difficulty determining when the mass had ‘started’
to drop, by about one and a half ticks (1.5  0.02 s = 0.03 s) on average.
Open the Excel Worksheet
Variation in braking distance with road conditions
Question 110D: Data Handling
1.
Open the Excel Worksheet
2. Dry, unladen:
mean braking distance = 45.1 m
range = 1.4 m
spread = ± 0.7 m = ± 2% rounded up.
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Dry, laden:
mean braking distance = 46.4 m
range = 1.3 m
spread = ± 0.6 m = ± 2% rounded up.
Wet, unladen:
mean braking distance = 58.0 m
range = 7.2 m
spread = ± 3.6 m = ± 6% rounded.
Wet, laden:
mean braking distance = 52.1 m
range = 8.5 m
spread = ± 4.3 m = ± 8% rounded.
3. In wet conditions the variability of braking distance is much larger than in dry conditions. In dry
conditions, braking distance is slightly greater for laden than unladen. In wet conditions, braking
distance is larger for unladen than laden.
4. Dry, unladen:
mean deceleration = 8.6 m s–2
Dry, laden:
mean deceleration = 8.3 m s–2
Wet, unladen:
mean deceleration = 6.7 m s–2
Wet, laden:
mean deceleration = 7.4 m s–2
The deceleration for dry conditions is close to the acceleration due to gravity. The deceleration for
wet conditions is around 20% less.
5. The % spread in deceleration is the same as the % spread in braking distance s, because
deceleration only depends upon s-1 (not on any higher or lower power).
Dry, unladen:
spread = ± 2% rounded up
Dry, laden:
spread = ± 2% rounded up
Wet, unladen:
spread = ± 6% rounded up
Wet, laden:
spread = ± 8% rounded up
Variation in braking distance between cars
Question 120D: Data Handling
1.
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Open the Excel Worksheet
2. From 90 km h–1
mean braking distance = 41 m
range = 22 m
spread = ± 11 m = ± 27
From 120 km h–1:
mean braking distance = 73 m
range = 36 m
spread = ± 18 m = ± 25%
3.
Open the Excel Worksheet
The decelerations average about 7.7 m s–2, and so are nearly 80% of the acceleration due to
gravity.
4. The ratio of the braking distances for the two speeds is very nearly constant (about 1.8). This
indicates that the deceleration is similar in the two cases, for a given car. This suggests a
constant braking force, with braking distance proportional to the initial kinetic energy.
Cycling through air
Question 190D: Data Handling
Force against average speed
20
++
15
+
10
+
+
5
++
+
++
0
0
21
1
2
3
4
average speed / m s–1
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5
6
power against average speed
120
++
100
80
60
+
40
+
+
20
++
+
0
0
1
++
2
3
4
average speed / m s–1
5
6
A good answer would include reference to the likely sources of retarding forces, and the likely
predominance of terms due to sliding friction, rather than drag, at the low velocities. Deploying the
equation relating P, F and v is likely to help to relate these to the power outputs required to go at
different speeds.
Retarding a cyclist
Question 270D: Data Handling
1. Use power = force  velocity, to calculate required power outputs, then plot power/speed and use
the (given) power output of the cyclist to find his speed.
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1200
1000
800
600
400
200
12.5 m s–1
0
2
4
6
8
10
12
speed / m s–1
14
16
18
2. First calculate the power required to lift him 2 m s–1 (vertically remember), then subtract this from
his power output, to give the power available to push him through the air.
power required to lift  force  velocity
 800 N  2 m s 1
 160 W
power to push through air  375 W – 120 W
Then use the graph again:
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1200
1000
800
600
400
200
10.25 m s–1
0
2
4
6
8
10
12
speed / m s–1
14
16
18
3.
drag
weight
Use the graph to get a retarding force of 28 N, for a speed of 12 m s–1.
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70
60
50
40
30
28 N
20
10
0
2
4.
4
6
8
10
12
speed / m s–1
14
16
18
Since speed is constant, the driving force, provided by the component of gravity down the slope,
must also be 28 N. Resolve to get:
800 N  sin   28 N
 28N 

  sin 1 
 800 N 
 2
Inertia reel seat belts
Question 170X: Explanation–Exposition
1. A good answer would relate the changes in the car's motion to the changes in the occupants'
motion, arguing that a seat belt allows the occupants to share an appropriate part of the car's
motion, and not the motion of the bumper.
2. A good answer would focus on the role of the gripper, that which allows the belt to run free or not.
The explanation would probably connect a certain freedom of movement relative to the reel
containing the spare belt to the inertial properties of the gripper, thus showing how it grips when
accelerating, but not when at constant velocity.
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