Solutions to Questions provided by Bruce Walsh, Xavier College
These suggested solutions have been prepared by the AIP (Vic Branch) Education Committee to
assist teachers and students when using this exam paper as a revision exercise. A suggested
marking scheme is provided in italics. Consequentials are indicated as “Conseq on 1”
Every effort has been made to check the solutions for errors and typos.
For the 2008 questions, the average and maximum scores, and the average as a percentage are
included at the end of the each solution in square brackets.
Motion in one and two dimensions
2004
1.
30 m
Initial vertical component of the velocity = 10 sin30° = 5.0 m/s (1). Define
upwards as positive, u = 5.0, a = -10, t = 3.0, s = ?, using s = ut + ½gt2,
s = 5.0 x 3.0 - (10 x 9.0)/2 (1) = -30 m (1), which 30m below initial
position.
3.
B
The vertical component will reduce to zero as the package rises, but the
horizontal component of the velocity will be unchanged, so either A or B.
After maximum height the vertical component of the velocity will increase at
a constant rate, so the next section of the graph should be straight, so the
answer is B. (2)
11. 422 m
Fnet = –800 N and so a = 800/m = 800/(1000+2000) = 800/3000
a = 0.267 m/s² (1), u = 15.0, v = 0, s = ? Using v2 = u2 +2as,
s = 15 x 15 / (2 x 0.267) (1) = 422 m (1).
13. 2.0 x 1030 kg From GM/4π² = R3/T² (1), where R = 1.5 x 1011 and T = 365 x 24 x 3600,
M = [(1.5 x 1011)3 x 4π² ]/[( 365 x 24 x 3600)2 x 6.67 x 10-11] (1)
14. 4.2 m
Loss in GPE = Gain in KE + Work Done (1). mgh = ½mv² + Friction x d (1),
so h = (½ x 30 x 8² + 50 x 6)/300 (1) = 4.2 m (1)
15. 0.253 m
½kx² = ½mv² (1), x² = 30 x 8²/30,000 = 8²/1000 = 0.064 (1),
x = 0.253 m (1)
2005
4.
7.
8.
9.
14.
1.0 m/s2
R = 1.8 N (1), Net force = 0.2 x 10 – 1.8 – 0.2 = ma (1). a = 0.2/0.2 = 1.0 (1)
B
Vector addition of inward force and upward tangential force = B (2).
9.3 m/s
By cons of momentum: 3 x v (1) = (3 + 1) x 7.0 (1), v = 28/3 = 9.3 m/s (1).
By Newton’s 3rd Law: Ftr on car = - Fcar on tr, (1) Time of impact is the same,
so pcar = - ptruck (1), therefore pbefore = pafter Momentum is always conserved.
B (2)
A, C and D are all correct ways of describing the term g.
2006
3.
9.0 m
6.
7.
17 N
38 m
Method 1: Using equations of accelerated motion
Net force is now 260 N (1), producing an acceleration of - 260/(90+40)m/s2
or – 2.0 m/s2. (1) u = 6.0 m/s, v = 0, a = -2.0 m/s2m, s = ?, so using v2 = u2
+2as, s = 9.0m. (1)
Method 2: Using Loss of KE = Work done by opposing forces (1)
Loss in KE = ½ x (90+40) x 6.02 = 260 x s, (1) find s. (1) Conseq on 1
Resultant or Net force = 22 – mg (1) = 22 – 0.5 x 10 = 22 – 5 = 17 N (1)
Net force = 17 N, so the acceleration = F/m = 17/0.5 = 34 m/s2. (1) u = 0, a
= 34, t = 1.5, s = ? using s = ut + ½at2, s = 0 + ½ x 34 x 1.52 = 38.25 m (1)
8.
10.
11.
12.
13.
14.
68 m/s, 130
Horizontal impulse = Ft = 22 x 1.5 =33 Ns which produces a change in
momentum. 22 x 1.5 = 0.5 x v, v = 66 m/s horizontally (1). Meantime
the rocket has fallen under gravity for 1.5 s and using “v=u+at”, reaches a
vertical velocity of 15 m/s (1). So using Pythagoras, the observed speed is
sqrt (662 + 152) = 67.8 m/s (1), the angle to the ground is given by Tan-1
(15/66), which is 12.80. (1)
Using Conservation of Momentum, the momentum of the shuttle before equals the
momentum after of the combined mass (1). So 6000 x 0.50 = (6000 + M) x 0.098 (1), solve
for M. (1)
1500 N
Average Force = Rate of change of momentum of either the shuttle or the
space station. Choose the space station as it is simpler and quicker.
Fav = p/t (1) = 3.00 x 105 x (0 – 0.098) / 20 (1) = 1470 N (1)
Total work done = Gain in PE + Energy loss due to opposing force (1), so
22720 = 13720+300 x L (1), solve for L. L = (22720 – 13720) / 300 = 9000/300 = 30 m. (1)
20 m
The gain in PE = mgh, so 13720 = 70 x 10 x h (1), solve for h.
h = 13720 / (70 x 10) = 19.6 m. (1)
390 N/m
The Grav PE at the top is converted to Elastic PE at the bottom. (1)
Mgh = ½ kx2, where h = 18 m and x = 8.0 m, (1)
so k = 70 x 10 x 18 x 2 /82 = 393.75 N/kg (1)
2007
3.
By conservation of momentum, momentum before = momentum after.
2 x 3 + 1 x 0 = 2 x v + 1 x 4 (1), solve for v (1).
5.
400 N
Average force = Change of momentum / Time taken (1)
Average force = (1 x 4 – 1 x 0) / 0.01 (1) = 400N (1).
9.
0.40 kg
Extension = 0.6 – 0.4 = 0.2 (1) Mg = kx, find M. M x 10 = 20 x 0.20 (1),
M = 0.40 kg (1).
10. 0.50 J
Increase in PE = ½ kx22 - ½ kx12 (1) = ½ x 20 x ((0.30)2 – (0.20)2) (1)
Increase in PE = 0.50 J (1).
11. D
Total energy is conserved. (2)
12. A
Gravitational force = GMm/r2. G, M and m are same for both,
so ratio equals (RP/RE)2 = (6/10.5)2 = 0.33. (2)
13. 574 years
Using Kepler’s 3rd Law: R3/T2 is the same for both objects,
then (TE)2 = (RE/RP)3 x (TP)2 = (10.5/6)3 x 2482 (1) = 329623= 574 years. (1)
17. A, C
A: same max height so time to go up and down is unchanged (1). B: twice
the horizontal distance in the same time, so horizontal component is doubled,
but max height is the same so vertical component in unchanged. So the
initial speed is greater. C: acceleration equals g and unchanged (1). D:
Horizontal component is doubled, so angle is less.
Note: It could be argued that the question includes what happens in the gun
where the acceleration of the paintball is greater, so only giving A.
2008
1.
0.070 m/s2
2.
4.0 m/s
From the graph, the water resistance on ship at 2.0 m/s is 2.0 x 104 N. So,
the net force on ship = 9.0 x 104 - 2.0 x 104 = 7.0 x 104 N. (1) Using Net
Force = ma, the acceleration = 7.0 x 104 / Mass of ship = 7.0 x 104 / 100 x
104 = 0.070 m/s2. (1). [1.0/2, 50%]
At constant speed, the net force equals zero, so the water resistance must
equal the towing force of 9.0 x 104 N. (1)
From the graph this occurs at a speed of 4.0 m/s. (1) [1.2/2, 60%]
7.
8.
9.
12.
13.
14.
15.
9.0 m
First find the time to travel the horizontal distance of 72.0 m using the
horizontal component of the initial speed which remains constant, vH = 30.0
x cos 36.90. (1) Time = 72.0 / (30.0 x cos 36.90) = 3.00 s. Now find the
height after 3.00 s using
s = ut + ½ at2, s = 30.0 x sin 36.90 x 3.00 - ½ x 10 x (3.00)2 (1)
s = 9.03 m = 9.0 m. (1) [1.7/3, 57%]
2.0 m/s
By the conservation of momentum, the momentum of the locomotive before
equals the combined momentum of the locomotive and the four trucks after
collision. So, 20 x 103 x 8.0 = (20 x 103 + 3 x 10 x 103) x v (1), find v.
So v = 2.0 m/s. (1) [1.7/2, 85%]
1.2 x 105 kgm/s to the left.
Impulse to the locomotive equals the change of momentum
of the locomotive, which also equal the magnitude of the change of
momentum of the three trucks. Change is final minus initial,
so p = 20 x 103 x (2.0 - 8.0) (1), p = - 120 x 103 kgm/s. (1) The initial
speed to the right was set as positive, so the impulse is to the left. (1)
Conseq on (8). [1.9/3, 63%]
0.20 J
Energy stored in the spring = ½ kx2, first find ‘k’. At 60 cm, an extension of
20 cm (0.20 m), the weight force, mg, of the toy is balanced by the spring
force, kx. Equating them gives mg = kx, k = 0.20 x 10 / 0.20 = 10 N/kg. (1)
Using this value the energy stored = ½ x 10 x (0.20)2 = 0.20 J. (1) [1.1/3,
37%]
D (2)
At the bottom and the top the toy is stationary. [0.8/2, 40%]
A (2)
At the bottom the gravitational potential energy is zero, then increases
linearly to a maximum at the top. [1.0/2, 50%]
Speed decreases from X to Y. (1) Total energy is constant. (1) [1.0/2, 50%]
The total energy is the sum of the comet’s gravitational potential energy and kinetic energy.
By the conservation of energy this sum remains constant. As the comet approaches the sun
it loses gravitational potential energy and gains an equal amount of kinetic energy and so
travels faster. [1.2/2, 60%]
Electronics and photonics
2004
7.
10.
11.
850 Ω
For a 1.5 V drop across the diode, there will be an 8.5 V drop across R (1) and so
using V=IR, R = 8.5/0.010 = 850 Ω (1)
C
Light on the base increases the collector current. This results in a larger drop across
R and so Vout decreases. This question is about phototransistors which are not on
the course for 2009 - 2012.
The graph will be straight (at 4 mA) between 0 and10. It will drop (somehow?) to 2 mA between 10
and 18. It then rises to 4 mA between 30 and 38. This question refers to phototransistors which are
not on the course for 2009 - 2012.
2006
8.
Modulation: The varying voltage (1) from the microphone is converted into a varying
optical signal (1).
Demodulation: The varying optical signal (1) from the cable is converted into a varying
voltage (1).
2007
8.
750 ohms
9.
decrease (1)
10.
D
11.
C
2008
1.
18 mA
10.
11.
From the figure 6, at 20 lux, Resistance of LDR = 1500 ohms (1). Using
voltage divider equation: 4 = 6 x (1500/(1500+R)), solve for R. R = 750
ohms. (1) Alternatively voltage across LDR = 4V, so voltage across R = 2V,
so R is half the resistance of the LDR.
Need: Vout = 4V for a higher lux value. Greater lux means smaller LDR
resistance, so R will also need to decrease to keep each resistor’s fraction of
the voltage the same. (1)
Only B and D refer to brightness. Brightness cannot go negative, so D.
Note: This answer assumes that the value at the origin of the graphs is zero
and also assumes that circuit for the laser diode is sufficiently biased so that
the laser diode still emits light even when the voltage at W goes negative. (2)
Only A and C refer to voltage. When the photodiode does not conduct the
voltage at Y would be constant, so A is not possible. Note: This answer
assumes that the value at the origin of the graphs is zero. (2)
Voltage across the LED = 2.5 V, so voltage across R = 8.0 - 2.5 = 5.5 V. (1)
Using Ohm’s law, current through R = 5.5 /300 = 0.0183 A = 18 mA. (1)
[1.2/2, 60%]
2000 
At 50C, the thermistor resistance = 4000 ohms. (1) Use the voltage divider
relatioship to determine the value of R. 4 V = R / (4000 + R) and solve for
R.(1) Alternatively, realise that the voltage across R = 4 V, so the voltage
across the thermistor = 8 V, so R must be half of 4000 ohms, i.e, 2000 ohms.
(1)
[1.9/3, 61%]
Increase (1). Lower temp means that the thermistor resistance goes up (1). To turn of the
cooling the voltage need to = 4.0 V, so if the thermistor resistance goes up, then R also
needs to rise to keep the ratio of the resistors the same. (1) [1.2/3, 40%]
Investigating materials and their use in structures
For the multiple choice questions for the 2008 Detailed Studies, the percentage choosing each
answer, the percentage who did not answer and any comment on significant distracters is in a
table at the end.
2004
1.
2.
3.
4.
6.
8.
11.
Young’s modulus is the gradient of the graph in the 3rd quadrant,
which is 20 x 107 / 10 x 10-3 = 2 x 1010 N/m2 (2)
4.0 x 108 N
Max stress = is 20 x 107, so max force = max stress x area (1)
max force = 20 x 107 x 2.0 (1) = 4.0 x 108 N (1)
6
3
1.0 x10 J/m Strain energy per unit volume = Area under graph in 3rd quadrant.
Area = ½ x 20 x 107 x 10 x 10-3 (1) = 1.0 x106 (1)
C
Strain energy = Strain energy per unit volume x Volume.
Volume = 8.0 x 2.0 = 16. (2)
A
The outer parts of the slab will be pushed down by the weight of the house and so
the top surface of the slab will be under tension. Concrete is weak under tension, so
the reinforcements should be placed at the top.
B
Tensile strength is the stress at fracture, so the endpoints of the graphs should differ
by a factor of 2, so D is out. Both have a plastic region, so both graphs should
curve over, so A is out. Toughness is area under the graph, so the Area under X’s
graph should be three times that of Y’s graph. B
Arches rely on the compressive strength of materials (1). Ice is strong under compression (1). The
low wall resists the outward force caused by the weight of the material in the arch pushing outward.
(1)
D
2005
7.
Both arrows towards centre of cable, i.e. an arrow at A from A to B (1) and an arrow at B
from B to A (1)
10. 1000 Nm
Torque = 200 x 10 x 0.5 (1) = 1000 Nm (1)
11. 1.0 m
1000 = 100 x 10 x X (2), X = 1.0 m. (1) Conseq on 10
2006
3.
0.1 m
4.
5.
3150 MPa
1260 kg
6.
510 J
From the graph the Strain is 2% (1), so the extension = 0.02 x 5.0 (1) = 0.1
m. (1)
Young’s modulus = gradient = 63 / 0.02 (1) = 3150 MPa (1)
Stress = 63 MPa, Weight = mg = Force = Stress x Area
mg = 63 x 106 x 2.0 x 10-4. (1) Mass = 126 x 102 / 10 = 1260 kg. (1)
Area under the graph ½ Stress x Strain = Energy density = Energy / Volume.
(1)
Energy = ½ x 63 x 106 x 0.02 x 2.0 x 10-4 x 2.0 (1) = 512 J (1)
2007
2.
Taking torques about T2: T1 x 3L/4 (1) = Mg x L/4 (1). Cancelling and rearranging (1)
T1 = Mg/3.
3.
2
Substitute Mg = 3T1 (1) into the supplied equation T2 = Mg – T1
gives T2 = 3T1 – T1 =2T1 (1)
4.
D
Between the cables the bottom is under tension, to the right of T2, the top is under
tension. (2)
8.
3.8 J
Strain energy = Area under graph x Volume (1)
Strain energy = ½ x 240 x 106 x 0.4 x 10-3 x  x (10/2 x 10-3)2 x 1.0 (1) = 3.8 J (1)
10. 120 N upward component of the force in the rod = Weight. CB x sin600 = Mg (1)
Outward component of the force in the rod = Force in AB. CB x cos600 = AB (1)
Eliminating CB, AB = Mg / tan600 = 20 x 10 / tan 600 = 115.5 N (1)
2008
4.
A
5.
C
12.
B
Question
4
5
12
Strain energy per unit volume = Area under the graph in the 3rd quadrant.
Area = ½ x 8.0 x 107 x 15 x 10-4 = 6.0 x 104 J/m3.
Strain energy = (Strain energy per unit volume) x Volume, so the factor is the
volume, which is 20 x 1.5 = 30.
Taking torques about Y gives 4.000 x 10 x 2.0 = m x 10 x 4.0, m = 2,000 kg.
%
A
67
16
14
%
B
14
21
47
%
C
13
56
27
%
D
6
5
9
%
No Answer
1
1
2
Most common error
Area of rectangle, not triangle
A: by Area, B: by Length
Equal to mass of beam as in a see-saw