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Level 2
Level 3
Level 4
Level 5
Modelling assumptions & SI units
36 questions
Lesson
Not started
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In mechanics, explain what is meant by modelling a body as a "particle", and give one advantage of doing so.
●●
●●●
Level 2
2 marks
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→
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A string is modelled as "light and inextensible". State what each of these assumptions means.
●●
●●●
Level 2
2 marks
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→
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State what is meant by modelling a surface as "smooth", and how this affects the forces in a problem.
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●●●
Level 2
2 marks
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A carpenter is making a
2
2
2
metre length of wooden handrail.
The cross-section of the handrail is symmetrical about the
x
x
x
-axis and is shown in the diagram below, where
x
x
x
and
y
y
y
are measured in centimetres.
O
x
y
The table below gives values of
y
y
y
for the lower half of the cross-section.
x
x
x
0
0
0
1.2
1.2
1.2
2.4
2.4
2.4
3.6
3.6
3.6
4.8
4.8
4.8
6.0
6.0
6.0
y
y
y
0
0
0
−
2.53
-2.53
−
2.53
−
3.01
-3.01
−
3.01
−
2.89
-2.89
−
2.89
−
2.20
-2.20
−
2.20
0
0
0
Use the trapezium rule, with the values shown in the table above, to find the best estimate for the volume of wood in the handrail.
(5 marks)
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●●
Level 3
5 marks
Start
→
Mark as done
Two particles are connected by a light inextensible string passing over a pulley. Explain why the two particles have the same acceleration.
●●●
●●
Level 3
1 mark
Start
→
Mark as done
State the SI units used for
(a)
force,
(b)
mass and (c) acceleration.
●●●
●●
Level 3
3 marks
Start
→
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A ball is kicked and its flight is modelled. State two modelling assumptions commonly made, and the effect of each.
●●●
●●
Level 3
2 marks
Start
→
Mark as done
A small stone is projected from a point
O
O
O
on horizontal ground by a catapult.
In an initial model
the stone is modelled as a particle moving freely under gravity
the stone is projected from
O
O
O
with speed
24.5
m s
−
1
24.5\ \text{m s}^{-1}
24.5
m s
−
1
at an angle
α
\alpha
α
to the horizontal, where
tan
α
=
1
2
\tan\alpha = \frac{1}{2}
tan
α
=
2
1
the stone hits the ground at the point
A
A
A
For the motion of the stone from
O
O
O
to
A
A
A
at time
t
t
t
seconds, the horizontal distance of the stone from
O
O
O
is
x
x
x
metres
at time
t
t
t
seconds, the vertical distance of the stone above the ground is
y
y
y
metres
Using the model,
(a)
show that
y
=
x
2
−
x
2
98
y = \frac{x}{2} - \frac{x^2}{98}
y
=
2
x
−
98
x
2
(5 marks)
(b)
Use the answer to part (a), or otherwise, to find the distance
O
A
OA
O
A
.
(2 marks)
The model does not include air resistance.
(c)
State one other limitation of the model.
(1 mark)
●●●●
●
Level 4
8 marks
Start
→
Mark as done
At a science fair, a small ball is projected from a point
O
O
O
on horizontal ground by a compressed-air launcher.
The ball is projected with speed
U
m s
−
1
U\ \text{m s}^{-1}
U
m s
−
1
at an angle
α
\alpha
α
to the horizontal, where
tan
α
=
3
4
\tan\alpha = \frac{3}{4}
tan
α
=
4
3
The ball is modelled as a particle moving freely under gravity.
Using the model, the greatest height of the ball above the ground is
7.2
7.2
7.2
m.
(a)
Show that
U
2
=
392
U^2 = 392
U
2
=
392
(3 marks)
(b)
Find the time taken for the ball to return to the ground.
(2 marks)
(c)
Find the horizontal distance travelled by the ball in this time.
(2 marks)
(d)
State one limitation of the model that could affect the answer to part (c).
(1 mark)
●●●●
●
Level 4
8 marks
Start
→
Mark as done
[For a cone with base radius
r
r
r
, height
h
h
h
and slant height
l
l
l
, the following formulae are given.
Curved surface area,
S
=
π
r
l
S = \pi rl
S
=
π
r
l
Volume,
V
=
1
3
π
r
2
h
V = \tfrac13\pi r^2h
V
=
3
1
π
r
2
h
]
A sweet shop makes paper cones, open at the top, to hold sweets. Each cone has base radius
r
r
r
cm and holds a volume of
V
V
V
cm
3
^3
3
when full. Each cone uses
25
π
25\pi
25
π
cm
2
^2
2
of paper.
(a)
Show that
V
=
1
3
π
r
625
−
r
4
V = \tfrac13\pi r\sqrt{625 - r^4}
V
=
3
1
π
r
625
−
r
4
.
(4 marks)
(b)
In this question you must show detailed reasoning.
It is given that
V
V
V
has a maximum value for a certain value of
r
r
r
.
Find the maximum value of
V
V
V
, giving your answer correct to 3 significant figures.
(5 marks)
●●●●
●
Level 4
9 marks
Start
→
Mark as done
A water company is investigating how the amount of water used in a town depends on the air temperature. The investigation runs for several weeks during the summer. A model is proposed for the amount of water used per day,
W
W
W
megalitres, and the mean daily temperature,
T
T
T
°C, at time
t
t
t
weeks after the start of the investigation.
In the model
W
=
25
e
0.04
t
W = 25e^{0.04t}
W
=
25
e
0.04
t
and
T
=
15
+
6
sin
(
0.2
t
)
T = 15 + 6\sin(0.2t)
T
=
15
+
6
sin
(
0.2
t
)
, where
0.2
t
0.2t
0.2
t
is in radians.
The model assumes that
W
W
W
and
T
T
T
can be treated as continuous variables.
(a)
State the meaning of
d
W
d
T
\dfrac{dW}{dT}
d
T
d
W
.
(1 mark)
(b)
Determine
d
W
d
T
\dfrac{dW}{dT}
d
T
d
W
when
t
=
5
t = 5
t
=
5
.
(4 marks)
(c)
Suggest a reason why this model may not be valid for values of
t
t
t
greater than
15
15
15
.
(1 mark)
●●●●
●
Level 4
6 marks
Start
→
Mark as done
Three points
A
A
A
,
B
B
B
and
C
C
C
lie in that order on a straight horizontal road.
A bus moves along the road from
A
A
A
to
C
C
C
.
The bus passes
A
A
A
with speed
3
m s
−
1
3\ \text{m s}^{-1}
3
m s
−
1
The bus moves with constant acceleration from
A
A
A
to
B
B
B
, reaching
B
B
B
with speed
V
m s
−
1
V\ \text{m s}^{-1}
V
m s
−
1
The bus then moves with constant deceleration from
B
B
B
until it comes to rest at
C
C
C
The bus is modelled as a particle.
(a)
Sketch a speed-time graph for the motion of the bus from
A
A
A
to
C
C
C
.
(3 marks)
The distance
A
B
AB
A
B
is
165
165
165
m and the distance
B
C
BC
B
C
is
120
120
120
m.
(b)
Find, in terms of
V
V
V
,
(i) the time taken by the bus to travel from
A
A
A
to
B
B
B
(ii) the time taken by the bus to travel from
B
B
B
to
C
C
C
(4 marks)
The bus takes
42
42
42
s to travel from
A
A
A
to
C
C
C
.
(c)
Find the value of
V
V
V
.
(2 marks)
(d)
Find the deceleration of the bus as it moves from
B
B
B
to
C
C
C
.
(2 marks)
●●●●
●
Level 4
11 marks
Start
→
Mark as done
A car moves along a straight horizontal road. At time
t
=
0
t = 0
t
=
0
, the car passes the point
O
O
O
with speed
5
m s
−
1
5\ \text{m s}^{-1}
5
m s
−
1
10
25
35
0.8
−0.2
−1
t (s)
a (m s⁻²)
Figure 1 (not accurately drawn)
Figure 1 shows the acceleration-time graph for the motion of the car in the interval
0
⩽
t
⩽
35
0 \leqslant t \leqslant 35
0
⩽
t
⩽
35
, where
t
t
t
seconds is the time after the car passes
O
O
O
. The acceleration of the car is
0.8
m s
−
2
0.8\ \text{m s}^{-2}
0.8
m s
−
2
for
0
<
t
<
10
0 < t < 10
0
<
t
<
10
−
0.2
m s
−
2
-0.2\ \text{m s}^{-2}
−
0.2
m s
−
2
for
10
<
t
<
25
10 < t < 25
10
<
t
<
25
−
1
m s
−
2
-1\ \text{m s}^{-2}
−
1
m s
−
2
for
25
<
t
<
35
25 < t < 35
25
<
t
<
35
The car is modelled as a particle.
(a)
Show that the speed of the car when
t
=
25
t = 25
t
=
25
is
10
m s
−
1
10\ \text{m s}^{-1}
10
m s
−
1
(3 marks)
(b)
Sketch a speed-time graph for the motion of the car in the interval
0
⩽
t
⩽
35
0 \leqslant t \leqslant 35
0
⩽
t
⩽
35
(3 marks)
(c)
Find the total distance travelled by the car in the interval
0
⩽
t
⩽
35
0 \leqslant t \leqslant 35
0
⩽
t
⩽
35
(4 marks)
●●●●
●
Level 4
10 marks
Start
→
Mark as done
A lorry moves along a straight horizontal road with constant acceleration.
The lorry passes the point
A
A
A
with speed
u
m s
−
1
u\ \text{m s}^{-1}
u
m s
−
1
and,
16
16
16
s later, passes the point
B
B
B
with speed
22
m s
−
1
22\ \text{m s}^{-1}
22
m s
−
1
The distance
A
B
AB
A
B
is
256
256
256
m.
The lorry is modelled as a particle.
(a)
Show that
u
=
10
u = 10
u
=
10
(2 marks)
(b)
Find the time taken for the lorry to travel from
A
A
A
to the midpoint of
A
B
AB
A
B
.
(5 marks)
The mass of the lorry is
2400
2400
2400
kg. As the lorry moves from
A
A
A
to
B
B
B
, the resistance to the motion of the lorry is modelled as a constant force of magnitude
900
900
900
N.
(c)
Find the magnitude of the driving force produced by the engine of the lorry as it moves from
A
A
A
to
B
B
B
.
(3 marks)
●●●●
●
Level 4
10 marks
Start
→
Mark as done
A small stone is projected vertically upwards with speed
16.8
m s
−
1
16.8\ \text{m s}^{-1}
16.8
m s
−
1
from a point
A
A
A
at the top of a vertical cliff. The point
A
A
A
is
38.5
38.5
38.5
m vertically above the point
B
B
B
. The point
B
B
B
lies on horizontal ground at the foot of the cliff.
The stone moves freely under gravity until it hits the ground at
B
B
B
with speed
V
m s
−
1
V\ \text{m s}^{-1}
V
m s
−
1
. After hitting the ground the stone does not rebound.
(a)
Find the value of
V
V
V
.
(3 marks)
(b)
Find the time taken for the stone to reach
B
B
B
.
(3 marks)
The point
C
C
C
is
9.8
9.8
9.8
m vertically above
A
A
A
.
(c)
Find the length of time for which the stone is above
C
C
C
.
(4 marks)
(d)
Sketch a speed-time graph for the motion of the stone from the instant it is projected to the instant it reaches
B
B
B
.
(No further calculations are required.)
(2 marks)
●●●●
●
Level 4
12 marks
Start
→
Mark as done
The value £
V
V
V
of a machine
t
t
t
years after it is bought is modelled by
V
=
a
b
t
V = ab^{t}
V
=
a
b
t
where
a
a
a
and
b
b
b
are constants.
When
t
=
2
t = 2
t
=
2
the value is £
12
800
12\,800
12
800
, and when
t
=
5
t = 5
t
=
5
the value is £
6553.60
6553.60
6553.60
.
(a)
Show that
b
=
0.8
b = 0.8
b
=
0.8
, and find the value of
a
a
a
. (4 marks)
(b)
Explain what the values of
a
a
a
and
b
b
b
represent in this model. (2 marks)
(c)
The machine is replaced as soon as its value falls below £
5000
5000
5000
.
Find the least whole number of years after purchase at which this happens. (4 marks)
●●●●
●
Level 4
10 marks
Start
→
Mark as done
A ball is dropped from a height of
2
2
2
m onto horizontal ground. After each bounce the ball rises to
65
%
65\%
65%
of the height it reached before that bounce.
(a)
Find the height the ball reaches after the
5
5
5
th bounce, giving your answer to
3
3
3
significant figures. (2 marks)
(b)
Find the total distance travelled by the ball from the moment it is dropped until the moment it hits the ground for the
6
6
6
th time, giving your answer to
3
3
3
significant figures. (4 marks)
(c)
Assuming the ball continues to bounce indefinitely, find the total distance it travels, giving your answer to
3
3
3
significant figures. (3 marks)
●●●●
●
Level 4
9 marks
Start
→
Mark as done
A tank is draining. At time
t
t
t
minutes the depth of water in the tank is
h
h
h
metres, and the depth is modelled by the differential equation
d
h
d
t
=
−
k
h
\dfrac{\mathrm{d}h}{\mathrm{d}t} = -k\sqrt{h}
d
t
d
h
=
−
k
h
where
k
k
k
is a positive constant. When
t
=
0
t = 0
t
=
0
the depth is
4
4
4
m.
(a)
Solve the differential equation to show that
h
=
2
−
k
t
2
\sqrt{h} = 2 - \dfrac{kt}{2}
h
=
2
−
2
k
t
. (4 marks)
(b)
Given that the depth is
2.25
2.25
2.25
m when
t
=
10
t = 10
t
=
10
, find the value of
k
k
k
. (2 marks)
(c)
Find the depth of water in the tank when
t
=
20
t = 20
t
=
20
. (2 marks)
(d)
Find the time at which the model predicts the tank will be empty. (2 marks)
●●●●
●
Level 4
10 marks
Start
→
Mark as done
Figure 5 shows an open-topped storage tank in the shape of a cuboid. The base is a square of side
x
x
x
metres and the height of the tank is
h
h
h
metres.
x m
h m
x m
Figure 5
(not accurately drawn)
The tank must hold exactly
32
32
32
m³ of liquid. The base is made from a reinforced material costing £
15
15
15
per m², and the four vertical sides are made from a lighter material costing £
6
6
6
per m².
(a)
Show that the total cost, £
C
C
C
, of the material used to make the tank is given by
C
=
15
x
2
+
768
x
C = 15x^2 + \dfrac{768}{x}
C
=
15
x
2
+
x
768
(4 marks)
(b)
Use calculus to find the value of
x
x
x
for which
C
C
C
is a minimum, giving your answer to
3
3
3
significant figures, and prove that your value gives a minimum. (5 marks)
(c)
Hence find the minimum cost of the material, to the nearest pound. (1 mark)
●●●●
●
Level 4
10 marks
Start
→
Mark as done
Figure 5 shows a container in the shape of a right circular cone with its vertex pointing downwards. The cone has radius
6
6
6
cm and height
18
18
18
cm.
6 cm
r
18 cm
h
Figure 5
(not accurately drawn)
Water is poured into the container at a constant rate of
8
8
8
cm³ s⁻¹. At time
t
t
t
seconds the depth of the water is
h
h
h
cm and the radius of the water surface is
r
r
r
cm.
(a)
Show that the volume
V
V
V
cm³ of water in the container is given by
V
=
π
h
3
27
V = \dfrac{\pi h^3}{27}
V
=
27
π
h
3
. (3 marks)
(b)
Find the rate at which the depth of the water is increasing at the instant when
h
=
6
h = 6
h
=
6
, giving your answer to
3
3
3
significant figures. (4 marks)
(c)
Find the rate at which the area of the water surface is increasing at that same instant, giving your answer to
3
3
3
significant figures. (3 marks)
[The volume of a cone of radius
r
r
r
and height
h
h
h
is
1
3
π
r
2
h
\tfrac13\pi r^2 h
3
1
π
r
2
h
.]
●●●●
●
Level 4
10 marks
Start
→
Mark as done
A population of insects is modelled by the differential equation
d
P
d
t
=
0.4
P
cos
(
0.5
t
)
\dfrac{\mathrm{d}P}{\mathrm{d}t} = 0.4P\cos(0.5t)
d
t
d
P
=
0.4
P
cos
(
0.5
t
)
where
P
P
P
is the population in thousands and
t
t
t
is the time in weeks. When
t
=
0
t = 0
t
=
0
,
P
=
6
P = 6
P
=
6
.
(a)
Solve the differential equation to show that
P
=
6
e
0.8
sin
(
0.5
t
)
P = 6\mathrm{e}^{0.8\sin(0.5t)}
P
=
6
e
0.8
s
i
n
(
0.5
t
)
(6 marks)
(b)
Find the maximum population predicted by this model, giving your answer to
3
3
3
significant figures. (2 marks)
●●●●
●
Level 4
8 marks
Start
→
Mark as done
In this question use
g
=
9.8
m s
−
2
g = 9.8\ \text{m s}^{-2}
g
=
9.8
m s
−
2
A machine for cricket catching practice fires cricket balls upwards. Each ball leaves the machine at ground level with a speed of
11.2
m s
−
1
11.2\ \text{m s}^{-1}
11.2
m s
−
1
The manufacturer claims that the balls can reach a maximum height of
6.4
6.4
6.4
metres above the ground.
(a)
Suppose that the machine fires the balls vertically upwards.
(a)
(i) Verify the manufacturer's claim.
(ii) State two modelling assumptions you have made in verifying this claim.
(4 marks)
(b)
In fact the machine fires the balls in a direction anywhere between
0
0
0
and
30
30
30
degrees from the vertical.
The range of maximum heights,
h
h
h
metres, above the ground which can be reached by the balls may be expressed as
k
<
h
≤
6.4
k < h \le 6.4
k
<
h
≤
6.4
Find the value of
k
k
k
(4 marks)
●●●●
●
Level 4
8 marks
Start
→
Mark as done
A forester models the mean height,
y
y
y
metres, of a species of young tree, up to the age of
25
25
25
years, by the formula
y
=
a
+
b
log
10
x
y = a + b\log_{10} x
y
=
a
+
b
lo
g
10
x
where
x
x
x
is the age of the tree in years, and
a
a
a
and
b
b
b
are constants.
The table shows the mean heights of trees of this species aged
4
4
4
years and
20
20
20
years.
Age,
x
x
x
(years)
4
4
4
20
20
20
Mean height,
y
y
y
(metres)
2.1
2.1
2.1
7.4
7.4
7.4
(a)
The forester uses the data for trees aged
4
4
4
years to write the correct equation
2.1
=
a
+
b
log
10
4
2.1 = a + b\log_{10} 4
2.1
=
a
+
b
lo
g
10
4
(a)
(i) Use the data for trees aged
20
20
20
years to write a second equation.
(ii) Show that
b
=
5.3
log
10
5
b = \frac{5.3}{\log_{10} 5}
b
=
lo
g
10
5
5.3
(iii) Find the value of
a
a
a
.
Give your answer to two decimal places.
(5 marks)
(b)
Use a suitable value for
x
x
x
to determine whether the model can be used to predict the mean height of trees of this species that are six months old.
(2 marks)
●●●●
●
Level 4
7 marks
Start
→
Mark as done
The temperature
θ
\theta
θ
°C of the air inside a sauna
t
t
t
minutes after its heater is switched on can be modelled by the equation
θ
=
5
(
18
−
14
e
−
k
t
)
\theta = 5\left(18 - 14\mathrm{e}^{-kt}\right)
θ
=
5
(
18
−
14
e
−
k
t
)
where
k
k
k
is a positive constant.
Initially the sauna is at room temperature.
The maximum temperature of the sauna is
T
T
T
°C
The temperature predicted by the model is shown in the graph below.
O
t
θ
(a)
Find the room temperature.
(2 marks)
(b)
Find the value of
T
T
T
(2 marks)
(c)
The sauna reaches a temperature of
52
52
52
°C ten minutes after the heater is switched on.
(c)
(i) Find the value of
k
k
k
.
(ii) Find the time it takes for the temperature of the sauna to be within
1
1
1
°C of its maximum.
Give your answer to the nearest minute.
(4 marks)
●●●●
●
Level 4
8 marks
Start
→
Mark as done
Priya, a wildlife officer, believes that the number of seals,
S
S
S
, at a breeding colony has grown exponentially since
2012
2012
2012
.
Priya models the incomplete data, shown in the table, using the formula
S
=
a
×
b
N
S = a \times b^N
S
=
a
×
b
N
where
N
N
N
is the number of years since
2012
2012
2012
and
a
a
a
and
b
b
b
are constants.
Year
2012
2012
2012
2015
2015
2015
2018
2018
2018
2021
2021
2021
2024
2024
2024
N
N
N
0
0
0
3
3
3
6
6
6
9
9
9
12
12
12
S
S
S
262
262
262
410
410
410
529
529
529
(a)
Priya wishes to determine the values of
a
a
a
and
b
b
b
. She plots a graph of
log
10
S
\log_{10} S
lo
g
10
S
against
N
N
N
and then draws a line of best fit as shown in the diagram below.
2
4
6
8
10
12
2.3
2.4
2.5
2.6
2.7
2.8
2.9
N
log₁₀ S
The equation of Priya's line of best fit is
log
10
S
=
0.035
N
+
2.41
\log_{10} S = 0.035N + 2.41
lo
g
10
S
=
0.035
N
+
2.41
(a)
(i) Use the equation of Priya's line of best fit to show that, correct to three significant figures,
a
=
257
a = 257
a
=
257
(ii) Use the equation of Priya's line of best fit to find the value of
b
b
b
.
Give your answer to three significant figures.
(2 marks)
(b)
According to Priya's model, state the yearly percentage increase in the number of seals.
(1 mark)
(c)
(i) Use Priya's model to predict the number of seals at the colony in
2030
2030
2030
(ii) Explain why the prediction made in part (c)(i) may be unreliable.
(3 marks)
●●●●
●
Level 4
6 marks
Start
→
Mark as done
A football is kicked from a point
O
O
O
on horizontal ground and first hits the ground again at the point
A
A
A
, where
O
A
=
30
OA = 30
O
A
=
30
m.
In an initial model
the ball is modelled as a particle moving freely under gravity
the ball is projected from
O
O
O
with speed
U
m s
−
1
U\ \text{m s}^{-1}
U
m s
−
1
at an angle
α
\alpha
α
to the horizontal, where
0
<
α
<
90
∘
0 < \alpha < 90^\circ
0
<
α
<
9
0
∘
Using the model,
(a)
show that
U
2
sin
α
cos
α
=
147
U^2\sin\alpha\cos\alpha = 147
U
2
sin
α
cos
α
=
147
(6 marks)
Using the model, the greatest height of the ball above the ground is
10
10
10
m.
(b)
Show that
U
2
=
306.25
U^2 = 306.25
U
2
=
306.25
(4 marks)
In a refinement to the model, the effect of air resistance is included. The motion of the ball, from
O
O
O
to
A
A
A
, is now modelled as that of a particle projected at the same angle
α
\alpha
α
whose initial speed is
V
m s
−
1
V\ \text{m s}^{-1}
V
m s
−
1
. This refined model is used to calculate a value for
V
V
V
.
(c)
State which is greater,
U
U
U
or
V
V
V
, giving a reason for your answer.
(1 mark)
(d)
State one further refinement to the model that would make the model more realistic.
(1 mark)
●●●●●
Level 5
12 marks
Start
→
Mark as done
A small ball is projected with speed
14
m s
−
1
14\ \text{m s}^{-1}
14
m s
−
1
from a point
O
O
O
on horizontal ground in a park.
After moving for
T
T
T
seconds, the ball passes through the point
A
A
A
, which is the top of a tree.
The point
A
A
A
is
8
8
8
m horizontally and
8
8
8
m vertically from the point
O
O
O
.
The motion of the ball from
O
O
O
to
A
A
A
is modelled as that of a particle moving freely under gravity.
Given that the ball is projected at an angle
α
\alpha
α
to the horizontal, use the model to
(a)
show that
T
=
4
7
cos
α
T = \dfrac{4}{7\cos\alpha}
T
=
7
cos
α
4
(2 marks)
(b)
show that
tan
2
α
−
5
tan
α
+
6
=
0
\tan^2\alpha - 5\tan\alpha + 6 = 0
tan
2
α
−
5
tan
α
+
6
=
0
(5 marks)
(c)
find the greatest possible height of the ball above the ground.
(3 marks)
The model does not include air resistance.
(d)
State one other limitation of the model.
(1 mark)
●●●●●
Level 5
11 marks
Start
→
Mark as done
A small pebble is lying on horizontal ground. The pebble is struck by the blade of a lawnmower and is projected from the point
O
O
O
where it was lying.
In an initial model
the pebble is modelled as a particle
P
P
P
moving freely under gravity
the pebble is projected from
O
O
O
with speed
21
m s
−
1
21\ \text{m s}^{-1}
21
m s
−
1
at an angle
α
\alpha
α
to the horizontal, where
tan
α
=
1
3
\tan\alpha = \frac{1}{3}
tan
α
=
3
1
the pebble hits the ground at the point
A
A
A
For the motion of
P
P
P
from
O
O
O
to
A
A
A
at time
t
t
t
seconds, the horizontal distance of
P
P
P
from
O
O
O
is
x
x
x
metres
at time
t
t
t
seconds, the vertical distance of
P
P
P
above the ground is
y
y
y
metres
(a)
Using the model, show that
y
=
x
3
−
x
2
81
y = \frac{x}{3} - \frac{x^2}{81}
y
=
3
x
−
81
x
2
(6 marks)
(b)
Use the answer to part (a), or otherwise, to find the length
O
A
OA
O
A
.
(2 marks)
Using the model, the greatest height of the pebble above the ground is found to be
H
H
H
metres.
(c)
Use the answer to part (a), or otherwise, to find the value of
H
H
H
.
(2 marks)
The model is refined to include air resistance. Using this refined model, the greatest height of the pebble above the ground is found to be
K
K
K
metres.
(d)
State which is greater,
H
H
H
or
K
K
K
, justifying your answer.
(1 mark)
(e)
State one limitation of this refined model.
(1 mark)
●●●●●
Level 5
12 marks
Start
→
Mark as done
A manufacturer makes cylindrical food cans of radius
r
r
r
cm. Each can is closed at both ends and has a volume of
250
π
250\pi
250
π
cm
3
^3
3
. The metal for the curved surface costs
0.2
0.2
0.2
pence per cm
2
^2
2
and the metal for the two ends costs
0.6
0.6
0.6
pence per cm
2
^2
2
.
(a)
Show that the total cost,
C
C
C
pence, of the metal for one can is given by
C
=
1.2
π
r
2
+
100
π
r
−
1
C = 1.2\pi r^2 + 100\pi r^{-1}
C
=
1.2
π
r
2
+
100
π
r
−
1
.
(4 marks)
(b)
Use calculus to determine the minimum cost of the metal for one can, giving your answer correct to the nearest penny. You should justify that it is a minimum.
(6 marks)
(c)
Suggest one reason why the manufacturer may not make cans with the radius found in part (b).
(1 mark)
●●●●●
Level 5
11 marks
Start
→
Mark as done
The points
A
A
A
and
B
B
B
lie on a straight horizontal track.
10
30
V
t (s)
speed (m s⁻¹)
Figure 1 (not accurately drawn)
A go-kart
P
P
P
starts from rest at
A
A
A
at time
t
=
0
t = 0
t
=
0
and moves along the track towards
B
B
B
. Figure 1 shows the speed-time graph for the motion of
P
P
P
.
Go-kart
P
P
P
accelerates uniformly for
10
10
10
seconds until its speed is
V
m s
−
1
V\ \text{m s}^{-1}
V
m s
−
1
. It then moves at constant speed
V
m s
−
1
V\ \text{m s}^{-1}
V
m s
−
1
. When
t
=
30
t = 30
t
=
30
,
P
P
P
passes
B
B
B
.
Given that the distance
A
B
AB
A
B
is
300
300
300
m,
(a)
show that
V
=
12
V = 12
V
=
12
(3 marks)
(b)
Find the acceleration of
P
P
P
between
t
=
0
t = 0
t
=
0
and
t
=
10
t = 10
t
=
10
(2 marks)
Go-kart
P
P
P
continues to move along the track in the same direction at the same constant speed,
V
m s
−
1
V\ \text{m s}^{-1}
V
m s
−
1
When
t
=
8
t = 8
t
=
8
, a second go-kart
Q
Q
Q
starts from rest at
A
A
A
and moves along the track in the same direction as
P
P
P
. Go-kart
Q
Q
Q
accelerates uniformly for
T
T
T
seconds until its speed is
18
m s
−
1
18\ \text{m s}^{-1}
18
m s
−
1
. It then moves at constant speed
18
m s
−
1
18\ \text{m s}^{-1}
18
m s
−
1
Go-kart
Q
Q
Q
catches up with
P
P
P
when
t
=
32
t = 32
t
=
32
The go-karts are modelled as particles.
(c)
On a copy of Figure 1, sketch a speed-time graph showing the motion of both go-karts for the interval
0
⩽
t
⩽
32
0 \leqslant t \leqslant 32
0
⩽
t
⩽
32
(3 marks)
(d)
Find the value of
T
T
T
(5 marks)
●●●●●
Level 5
13 marks
Start
→
Mark as done
A closed cylinder of radius
r
r
r
and height
h
h
h
is inscribed in a sphere of radius
6
6
6
cm, so that both circular ends of the cylinder touch the sphere.
r
6
h
(a)
Show that the volume of the cylinder,
V
V
V
cm³, is given by
V
=
36
π
h
−
π
h
3
4
V = 36\pi h - \dfrac{\pi h^3}{4}
V
=
36
π
h
−
4
π
h
3
. (4 marks)
(b)
Use calculus to find the value of
h
h
h
that maximises
V
V
V
, justifying that your value gives a maximum. Give the maximum volume in the form
k
3
π
k\sqrt{3}\,\pi
k
3
π
cm³. (6 marks)
(c)
Show that the ratio of the maximum cylinder volume to the volume of the sphere is
3
:
3
\sqrt{3} : 3
3
:
3
. (4 marks)
●●●●●
Level 5
14 marks
Start
→
Mark as done
The depth of water,
D
D
D
metres, in a harbour at a time
t
t
t
hours after midnight is modelled by
D
=
8
+
3
cos
(
0.5
t
)
−
4
sin
(
0.5
t
)
,
0
⩽
t
⩽
24
D = 8 + 3\cos(0.5t) - 4\sin(0.5t), \qquad 0 \leqslant t \leqslant 24
D
=
8
+
3
cos
(
0.5
t
)
−
4
sin
(
0.5
t
)
,
0
⩽
t
⩽
24
where the angle
0.5
t
0.5t
0.5
t
is measured in radians.
(a)
Express
3
cos
θ
−
4
sin
θ
3\cos\theta - 4\sin\theta
3
cos
θ
−
4
sin
θ
in the form
R
cos
(
θ
+
α
)
R\cos(\theta + \alpha)
R
cos
(
θ
+
α
)
, where
R
>
0
R > 0
R
>
0
and
0
<
α
<
π
2
0 < \alpha < \dfrac{\pi}{2}
0
<
α
<
2
π
.
Give the exact value of
R
R
R
and the value of
α
\alpha
α
to
4
4
4
decimal places. (3 marks)
(b)
Find the maximum depth of water in the harbour, and the value of
t
t
t
at which this maximum first occurs, giving
t
t
t
to
2
2
2
decimal places. (4 marks)
(c)
A ship can enter the harbour only when the depth of water is at least
10
10
10
metres.
Find the two values of
t
t
t
between
t
=
6
t = 6
t
=
6
and
t
=
18
t = 18
t
=
18
for which the depth is exactly
10
10
10
metres, giving each value to
2
2
2
decimal places. (5 marks)
●●●●●
Level 5
12 marks
Start
→
Mark as done
A ball is kicked from a point
O
O
O
on horizontal ground with speed
20
20
20
m s⁻¹ at an angle
α
\alpha
α
above the horizontal, where
tan
α
=
3
4
\tan\alpha = \dfrac34
tan
α
=
4
3
.
The ball is modelled as a particle moving freely under gravity. Take
g
=
9.8
g = 9.8
g
=
9.8
m s⁻².
(a)
Find the time of flight of the ball, giving your answer to
3
3
3
significant figures. (3 marks)
(b)
Find the horizontal distance from
O
O
O
at which the ball lands, giving your answer to
3
3
3
significant figures. (2 marks)
(c)
A wall of height
5
5
5
m stands on the ground at a horizontal distance of
30
30
30
m from
O
O
O
, in the plane of the ball's motion.
Determine whether the ball passes over the wall. (5 marks)
(d)
State one assumption made in the model, and explain how your answer to part (c) might change if this assumption were not made. (3 marks)
●●●●●
Level 5
13 marks
Start
→
Mark as done
A ball is struck from a point
O
O
O
on horizontal ground with speed
21
21
21
m s⁻¹ at an angle
θ
\theta
θ
above the horizontal. The ball is modelled as a particle moving freely under gravity. Taking
O
O
O
as the origin,
x
x
x
metres and
y
y
y
metres are the horizontal and vertical displacements of the ball from
O
O
O
at time
t
t
t
seconds. In this question use
g
=
9.8
g = 9.8
g
=
9.8
m s⁻².
(a)
Show that the equation of the trajectory of the ball is
y
=
x
tan
θ
−
x
2
90
(
1
+
tan
2
θ
)
y = x\tan\theta - \dfrac{x^2}{90}\left(1 + \tan^2\theta\right)
y
=
x
tan
θ
−
90
x
2
(
1
+
tan
2
θ
)
(4 marks)
The ball passes through the point
Q
Q
Q
, which is
30
30
30
m horizontally from
O
O
O
and
2.5
2.5
2.5
m above the ground.
(b)
Find the two possible values of
tan
θ
\tan\theta
tan
θ
. (4 marks)
(c)
Find the speed of the ball as it passes through
Q
Q
Q
, giving your answer in the form
k
2
k\sqrt{2}
k
2
m s⁻¹, and explain why this speed is the same for both possible values of
tan
θ
\tan\theta
tan
θ
. (5 marks)
●●●●●
Level 5
13 marks
Start
→
Mark as done
In this question you must show detailed reasoning.
Figure 1 shows a closed box in the shape of a cuboid with a square base of side
x
x
x
cm and height
h
h
h
cm. The total surface area of the box is
600
600
600
cm².
x cm
h cm
x cm
Figure 1
(not accurately drawn)
(a)
Show that the volume
V
V
V
cm³ of the box is given by
V
=
150
x
−
x
3
2
V = 150x - \dfrac{x^3}{2}
V
=
150
x
−
2
x
3
. (3 marks)
(b)
Find the value of
x
x
x
for which
V
V
V
is a maximum, and find the maximum volume. You must show that your value gives a maximum. (5 marks)
(c)
Find the corresponding value of
h
h
h
, and describe the shape of the box in this case. (2 marks)
●●●●●
Level 5
10 marks
Start
→
Mark as done
The diagram below shows the approximate shape of the vertical cross section of a natural rock arch above a horizontal footpath.
The inside of the arch meets the footpath at the points
O
O
O
and
P
P
P
O
P
Rhiannon models the shape of the inside of the arch using the equation
x
2
+
2
y
2
=
a
x
x^2 + 2y^2 = a\sqrt{x}
x
2
+
2
y
2
=
a
x
where
a
a
a
is a constant, and
x
x
x
and
y
y
y
are the horizontal and vertical distances respectively, in metres, measured from
O
O
O
(a)
The distance
O
P
OP
O
P
is
5
5
5
metres.
Find the exact value of
a
a
a
that Rhiannon should use in the model.
(2 marks)
(b)
Show that the maximum height of the inside of the arch above the footpath is approximately
2.43
2.43
2.43
metres.
(6 marks)
(c)
Suggest one limitation of the model Rhiannon has used.
(1 mark)
●●●●●
Level 5
9 marks
Start
→