3
Answer all questions in the spaces provided.
1 (a) Define velocity and acceleration.
Velocity :
[1]
Acceleration :
(b) Fig. 1.1 shows the variation with time t of the velocity v for an object. v / m s
–1
20
10
0
−10
−20
2.0 4.0 6.0 8.0 10.0
Fig. 1.1
(i) Calculate the displacement of the object at t = 12.0 s.
12.0 t / s
[1]
For
Examiner’s
Use
NYJC 2012 displacement = m [2]
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(ii) Calculate the time the object returns to its starting position.
When the object is back at its starting position, s = 0. Let the time taken be t,
0 = ½ × 4.0 × 20 + ½ × (t − 4.0 + t − 6.0) × ( − 20) t = 7.0 s time = s [1]
(iii) Hence, state the time at which the object is at maximum distance from the starting point. time = 12.0 s [1]
(iv) On Fig. 1.2, sketch a graph to show the variation of the displacement s with time t for the object. (You are expected to label significant values of displacement.) [2] s / m
For
Examiner’s
Use
0 2.0
4.0
6.0
8.0
10.0
12.0
t / s
Fig. 1.2
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2 A military lightweight tank accelerates linearly from rest to a speed of 35 km h
-1
in 12 s on a horizontal railway track. Fig. 2.1 shows the side profile of a tank operator seated in the tank. The arrow labelled ‘R’ represents the support force acting on the operator by the tank. The support force acts at an angle θ (in degrees) above the horizontal. direction of motion
For
Examiner’s
Use
R
Fig 2.1
(a) Given that the operator has a mass of 65 kg, show that the magnitude of the resultant force on him is 53 N.
Acceleration of driver = (v – u) / t
= (35.0 x 1000 /3600 – 0)/ 12.0 = 0.81 m s
-2
Net force = m a = 65.0 x 1.62 = 52.66 = 53 N (2 s.f.)
Note: 1) Conversion of speed into m/s; 2) Evidence of “ F = ma = m (v – u)/t “
[2]
(b) Calculate the value of θ and the magnitude of the support force R acting on the tank operator.
Taking right as positive:
Taking upwards as positive:
R sin θ = 65.0 g
Solving (1) and (2): tan θ =(65.0 g)/(52.7) θ = 85.30
NYJC 2012
θ = magnitude of force =
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N [2]

6
(c) Fig.2.2 shows the tank at rest on a slope inclined at 18 o
to the horizontal. Its gun is tilted upwards such that it makes an angle of 30 o with the horizontal. When engaged in a continuous fire mode, it fires 3 artillery shells, each with a mass of
30.0 kg, consecutively in 2.0 s.
For
Examiner’s
Use direction of fire
30
0
18
0
Fig. 2.2
(i) If the shells are fired with a speed of 60.0 m s
-1
, calculate the average recoil force on the tank during the continuous firing. You may assume that the gravitation force on the shells is very much smaller than the recoil force.
Total force exerted on shells
= (change in momentum)/time = (m ∆ v)/ ∆ t
= (30 x 3 x (60 – 0))/2 = 2700 N
By Newton’s 3rd law, recoil force = 2700 N magnitude of recoil force = N [2]
(ii) In practice, the recoil experienced by the tank varies in magnitude with time.
The tank has a mass of 6000 kg. The maximum instantaneous recoil force is
9000 N and acts in the direction opposite to the shell’s direction of fire.
Calculate the magnitude of the static friction between the tank and the ground such that the tank does not slide downslope due to the maximum recoil force during firing.
Σ F = 0 f– W sin 18
0
– Fshell cos (30 o
-18
0
) = 0 f = 26083 = 26000 N (2 s.f.)
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N [2]

7
3 A Ferris wheel in an amusement park performs uniform circular motion, making one revolution in 10.0 s.
The passengers sit on seats inside each of the cabins shown in fig 3.1. Each passenger rotates in a vertical circle of radius 5.00 m.
For
Examiner’s
Use
C cabins
A
Fig 3.1
(a) The mass of a passenger is 70.0 kg.
(i) Calculate the linear speed of the passenger.
ω = 2 π / T = 2 π / 10.0 = 0.628 rad s
-1 v = r ω = 5.00 (0.628) = 3.14 m s
-1
[1]
B m s
− 1
[2] linear speed =
(ii) Calculate the centripetal acceleration a c
of the passenger. a c
= r ω
2
= (5.00)(0.628)
2
= 1.97 m s
-2
[1] acceleration a c
= m s
(iii) Explain why there is an acceleration even though his speed is constant.
− 1
[1]
The acceleration, which acts perpendicularly to the direction of the velocity[1], will cause the direction of the velocity to
NYJC 2012 change[1], without changing the magnitude of the velocity.
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[2]

8
(b) The cabin at B is halfway between the lowest and highest point in the Ferris wheel.
(i) The passenger in the cabin at B experiences two forces: R is the force exerted on the passenger by the seat and W is the weight of the passenger. On fig 3.2, draw labelled arrows to represent these forces. [1]
For
Examiner’s
Use
Fig 3.2
(ii) Describe the direction of the centripetal force on the passenger in fig 3.2.
The direction of the centripetal force is horizontal and towards the centre of the circular motion.
[1]
(c) The apparent weight of the passenger is the force exerted by the passenger on his seat.
By drawing free body diagrams similar to that in (b)(i) showing the forces acting on the passenger at the highest point C and lowest point A in his circular motion, determine the difference between the apparent weight of a passenger at the highest and lowest point of his circular motion. [2]
Since apparent weight = force exerted by the passenger on his seat.
By Newton’s third law this has the same magnitude as the force exerted ON the passenger BY the seat.
At the highest point let apparent weight be R
1
W – R
1
= 70(1.97)
At the lowest point let apparent weight be R
2
R
2
– W = 70(1.97)
W – R
1
+ R
2
– W = 2(138 )
R
2
– R
1
= 2 (138) = 276 N
N [2]
NYJC 2012 difference in apparent weight =
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4 A constant power supply of 1000 W is used to melt 1.0 kg of ice, to heat the water produced, and finally to turn all the water to steam.
The table below provides some data for ice and water.
Specific heat capacity of water
Specific latent heat of fusion of ice
Specific latent heat of vaporisation of water
Melting point
Boiling point
4 x 10
3
J kg
-1
K
-1
3 x 10
5
J kg
-1
3 x 10
6
J kg
-1
0 o
C
100 o
C
(a) Neglecting heat loss to environment, use the data provided to calculate the time required to
(i) melt 1.0 kg of ice
For
Examiner’s
Use
(ii) heat the water at melting point to boiling point time = s [1]
NYJC 2012
(iii) turn all the water at boiling point to steam time = s [1]
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10
(b) On Fig 4.1, use your answers from (a) to sketch the variation of time with temperature from the time when the heater is switched on until the time when all the water has turned into steam. Label this graph as X and include significant values on the axes. [2]
For
Examiner’s
Use
Temperature / o
C
Time / s
Fig 4.1
(c) The rate of heat loss of a body is proportional to the temperature difference between the body and the surroundings. The temperature of the surroundings is kept constant at 0 o
C.
Considering heat loss to surroundings, on Fig 4.1 sketch the variation of temperature with time from the time when the heater is switched on until the time when all the water has turned into steam. Label this graph as Y. [2]
(d) Using the kinetic theory of matter, state one reason why there is a considerable difference in magnitude between its specific latent heat of fusion and vaporization.
[1]
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5 A 10.0 m wire is wound to form a helical conductor as shown in Fig 5.1.
Fig 5.1
A potential difference of 5.0 V is applied across the conductor and in 1 hour, 500 C of charge passes through the conductor.
(a) Explain the underlined phrase.
5.0 J of energy is converted from electrical to other forms per 1 C of charge passing through the conductor. [2]
[2]
(b) Calculate the power dissipated by the wire.
Total energy dissipated in 1 hour = V Q
= 5.0 × 500 = 2.5 × 10 3 J [1]
Power dissipated = 2.5 × 10
3
/ (1 × 60 × 60)
= 0.69 W [1]
Power dissipated =
(c) Calculate the resistance per unit length of the wire.
Resistance = V2 / P = 5.02 / 0.69
= 36 Ω [1]
Resistance per unit length = 36 / 10.0
= 3.6 Ω m
–1
[1]
W [2]
Ω m
–1
[2]
NYJC 2012
Resistance per unit length =
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For
Examiner’s
Use

12
(d) Deduce and explain what will happen to the resistance of the conductor if it is compressed such that the adjacent turns of the conductor become in electrical contact with each other.
The conductor will effectively become a hollow tube (instead of a helix).
The length of the conductor will become much less than 10 m.[1/2] (wire tube)
The cross-sectional area will become much greater. [1/2] (wire tube)
Since R = ρ L / A, the resistance of the conductor will be much lower. [1]
[2]
6 Fig. 6.1 shows part of an array of wind turbines on farmland.
For
Examiner’s
Use
Fig 6.1
Each turbine converts the kinetic energy of the wind into electrical energy. At a particular time of the day, it is known that 1.16 × 10
7
kg of air travelling at a speed of
20 m s -1 passes through the turbine each minute.
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(a) Calculate the maximum input power of the turbine in MW, stating one assumption made in the calculation.
Assumption: all the kinetic energy of the wind is transferred to electrical enegy.
Power input = average rate of loss of kinetic energy of wind
= ½( 1.16 x 10
7
)20
2
/ 60
= 38.7 MW
For
Examiner’s
Use maximum power = MW [2]
Assumption :
[1]
(b) Blades of length l sweep out an area A with each rotation.
When air of density ρ and wind speed v passes the blades, the maximum power P transferred to the turbine is given by the expression
P = ½ A ρ v
3
By making measurements, values of v and P are collected and a graph of ln P against ln v is plotted as shown in Fig 6.2.
NYJC 2012
Fig 6.2
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(i) Complete Fig. 6.3 by using the graph plotted in Fig 6.3. [2] v / m s -1 P / MW
Fig 6.3
(ii) Determine the vertical intercept of the graph. ln (v / m s − 1
0.50
2.50
) vertical intercept = [1]
(iii) Hence determine the blade length L, given that the density of air is 1.23 kg m
-3
.
For
Examiner’s
Use blade length = m [3]
(iv) Suggest with a reason whether the relation P = ½ A v
3
is still valid when the wind travels at a much higher speed.
[1]
(c) In the wind turbine business there are basically two types of turbines to choose from, vertical axis wind turbines and horizontal axis wind turbines. They both have their advantages and disadvantages.
Unlike the traditional propeller style wind turbines, it consists of blades which rotate around a vertical axis as shown in Fig 6.4. The design below look likes an egg beater.
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Guy wires
(anchored to the ground)
Rotor
Gear box
Fig 6.4
Some specifications of the vertical axis wind turbine from the manufacturer are provided below.
Rated power : 3.0 kW
Rated speed : 230 rev min
-1
Cut-off wind speed : 15 m s
-1
Rotor diameter: 4.0 m
(i) Use the data given above to determine the maximum linear speed of the rotating blades. v = r ω = (2.0)(230 x 2 π /60) = 48 m s
-1
For
Examiner’s
Use
NYJC 2012 linear speed = m s
− 1
[2]
(ii) Suggest one advantage and one disadvantage of the vertical wind turbine compared to traditional wind turbine with reference to Fig 6.4.
Advantage:
[1]
Disadvantage:
[1]
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(iii) Explain briefly how the guy wires provide additional stability to the vertical wind turbine.
For
Examiner’s
Use
[1]
NYJC 2012
End of Paper
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