Solutions to Questions provided by Bruce Walsh, XavierCollege

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 mInitial vertical component of the velocity = 10sin30° = 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.BThe 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 mFnet = –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 kgFrom GM/4π² = R3/T² (1), where R = 1.5 x 1011 and T = 365 x 24 x 3600,

M = [(1.5 x 1011)3x 4π² ]/[( 365 x 24 x 3600)2x 6.67 x 10-11] (1)

14.4.2 mLoss 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 m½kx² = ½mv² (1), x² = 30 x 8²/30,000 = 8²/1000 = 0.064 (1),

x = 0.253 m(1)

2005

4.1.0 m/s2R = 1.8 N (1), Net force = 0.2 x 10 – 1.8 – 0.2 = ma (1). a = 0.2/0.2 = 1.0 (1)

7.BVector addition of inward force and upward tangential force = B (2).

8.9.3 m/sBy cons of momentum: 3 x v (1) = (3 + 1) x 7.0 (1), v = 28/3 = 9.3 m/s (1).

9.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.

14.B (2)A, C and D are all correct ways of describing the term g.

2006

3.9.0 mMethod 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

6.17 NResultant or Net force = 22 – mg (1) = 22 – 0.5 x 10 = 22 – 5 = 17 N (1)

7.38 mNet 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)

8.68 m/s, 130Horizontal 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)

10.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)

11.1500 NAverage 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)

12.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)

13.20 mThe gain in PE = mgh, so 13720 = 70 x 10 x h (1), solve for h.

h = 13720 / (70 x 10) = 19.6 m. (1)

14.390 N/mThe 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 NAverage force = Change of momentum / Time taken (1)

Average force = (1 x 4 – 1 x 0) / 0.01 (1) = 400N (1).

9.0.40 kgExtension = 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 JIncrease 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.DTotal energy is conserved. (2)

12.AGravitational 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 yearsUsing 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, CA: 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/s2From 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%]

2.4.0 m/sAt 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%]

7.9.0 mFirst 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%]

8.2.0 m/sBy 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%]

9.1.2 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%]

12.0.20 JEnergy 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%]

13.D (2)At the bottom and the top the toy is stationary. [0.8/2, 40%]

14.A (2)At the bottom the gravitational potential energy is zero, then increases linearly to a maximum at the top. [1.0/2, 50%]

15.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

7.850 ΩFor 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)

10.CLight 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.

11.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 ohmsFrom 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.

9.decrease (1)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)

10.DOnly 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)

11.COnly 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)

2008

1.18 mAVoltage 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%]

10.2000 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%]

11.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.DYoung’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)

2.4.0 x 108 NMax 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)

3.1.0 x106 J/m3Strain 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)

4.CStrain energy = Strain energy per unit volume x Volume.

Volume = 8.0 x 2.0 = 16. (2)

6.AThe 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.

8.BTensile 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

11.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)

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 NmTorque = 200 x 10 x 0.5 (1) = 1000 Nm (1)

11.1.0 m1000 = 100 x 10 x X (2), X = 1.0 m. (1) Conseq on 10

2006

3.0.1 mFrom the graph the Strain is 2% (1), so the extension = 0.02 x 5.0 (1) = 0.1 m. (1)

4.3150 MPaYoung’s modulus = gradient = 63 / 0.02 (1) = 3150 MPa (1)

5.1260 kgStress = 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)

6.510 JArea 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.

3.2Substitute Mg = 3T1(1) into the supplied equation T2 = Mg – T1

gives T2 = 3T1 – T1 =2T1(1)

4.DBetween the cables the bottom is under tension, to the right of T2, the top is under tension. (2)

8.3.8 JStrain 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 Nupward 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.AStrain 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.

5.CStrain energy = (Strain energy per unit volume) x Volume, so the factor is the volume, which is 20 x 1.5 = 30.

12.BTaking torques about Y gives 4.000 x 10 x 2.0 = m x 10 x 4.0, m = 2,000 kg.

Question / % A / % B / % C / % D / %
No Answer / Most common error
4 / 67 / 14 / 13 / 6 / 1 / Area of rectangle, not triangle
5 / 16 / 21 / 56 / 5 / 1 / A: by Area, B: by Length
12 / 14 / 47 / 27 / 9 / 2 / Equal to mass of beam as in a see-saw