chemology unit 3 2011

Chemology Physics Unit 3 Trial Exam-2011

CHEMOLOGY EDUCATION SERVICES

Name:______________________

2011 PHYSICS UNIT 3TRIAL EXAM

Time allowed: 1 hour 30 minutes

QUESTION AND ANSWER BOOKLET

Structure of booklet

Area of study Number of questions

Marks available Suggested time(minutes)

1. Motion in one and two dimensions 19 45 40

2. Electronics and Photonics 15 35 30

3.1 Einstein’s Special Relativity

3.2 Materials and their use in structures

3.3 Further Electronics

10

9

9

20

20

20

20

20

20

Directions to studentsThis booklet is 24 pages long.You should answer all questions in Areas of Study 1 and 2.You should answer all questions in your selected Detailed Study in Area of Study 3.(Note: You should choose only ONE Detailed Study and answer only questions from that Study)You may use an A4 page of notes annotated on both sides.There is a total of 100 marks available. It is suggested that you spend about 1 minute per mark.

© CHEMOLOGY EDUCATION SERVICESP O BOX 477 MENTONE 3194Mobile: 0412 405 403 or 0425 749 520E: [email protected]


SECTION A – CORE

AREA OF STUDY 1 – Motion in One and Two DimensionsSpecific instructions for Area 1

There are 19 questions in Area 1.A total of 45 marks is available for Area 1.Take g = 10 m/s2 wherever necessary in this context.You may ignore air resistance and friction unless otherwise specified.

Armindo and Emilio are both driving their funky cars at a vintage car rally and are waving at the crowds who have come to admire their vehicles. Distracted by his adoring fans, Emilio drives straight into Armindo at a speed of 6m/s, from a direction that is at right angles to his motion. Following the collision, the two cars stick together and travel on as shown in Figure 1 below. Each car has a mass of 1200kg and Armindo has a mass of 74kg.

Figure 1

Question 1

If Emilio’s mass is 82kg, calculate the total momentum of the system.

2 marks

Question 2

At what speed was Armindo travelling prior to the collision?

2 marks

Question 3

At what speed do the cars travel on together after their collision?

2 marksArea 1 - continued

25o

Path after collisionArmindo

Emilio

kgm/s

m/s

m/s


Question 4

Was the collision elastic or inelastic? Show calculations to justify your answer.

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

4 marks

Emilio disengages his car from Armindo’s vehicle and drives it away around a curve that is approximately circular with a radius of 153m. The curve is banked at 200. Figure 2 shows the REAR of Emilio’s car as it takes this curve. He is driving fast enough that there is friction between his tyres and the road, acting sideways down the slope.

Figure 2

Question 5

On the copy of Figure 2 below, draw arrows to represent the following forces: weight (mg), normal reaction force (N) and Friction (Fr).

3 marks

Emilio now slows down until he is travelling at the precise speed where sideways friction is no longer needed to help the car turn the corner.

Area 1 - continued


Question 6

Which of the following best represents the relationship between N and mg?

A. N = mgB. N = mgsin20C. N = mgcos20D. mg = Nsin20E. mg = Ncos20

2 marks

Question 7

What is the magnitude of the net force on Emilio’s car towards the centre of the curve now?

3 marks

Question 8

At what speed would Emilio be travelling to reduce the sideways friction to zero?

2 marks

Catherine has cooked up some of her famous salted caramel fudge and throws a piece to her friend Emma, who is sitting on the floor at a horizontal distance of 4m from the point at which Catherine releases the fudge, as shown in Figure 3 below.The fudge is a vertical distance of 1.6m from the ground when Catherine throws it and has a mass of 50g. It lands at Emma’s feet.

Figure 3Area 1 - continued

N

m/s


Question 9

What is the total time for which the piece of fudge was in the air?

2 marksQuestion 10

Calculate the initial horizontal velocity with which Catherine threw the fudge.

2 marks

Question 11

Calculate the speed at which the fudge lands at Emma’s feet.

2 marks

Question 12

Explain what happens to the kinetic energy and momentum of the fudge as it lands and comes to rest at Emma’s feet.

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

2 marks

Scotty is a champion pole vaulter who is about to attempt to vault a bar 7.3m high. He achieves the height for his vault by converting stored elastic potential energy in the pole into gravitational potential energy. Scotty’s mass is 85kg, and 90% of his potential energy needs to be supplied from the elastic potential energy in the pole.

Question 13

How much elastic potential energy needs to be stored in the pole?

2 marks

Area 1 - continued

s

m/s

m/s

J


After his vault, Scotty falls onto the mats behind the pole. As he hits the mats, he is travelling vertically at 13m/s. The compression-force graph for the mats is shown in Figure 4 below.

Figure 4

Question 14

What is the maximum compression Scotty causes in the mats?

3 marks

Sean and Danielle are intrepid astronauts who are travelling the Universe in search of a black hole. They are currently flying their spacecraft Longun past the mysterious planet of Anika, as shown in Figure 5 below.

Figure 5

Area 1 - continued

m


The following data applies to the questions below:Universal gravitational constant = 6.7 x 10-11 Nm2kg-2

mass of Anika = 5.5 x 1022kg radius of Anika = 4.9 × 10 4 mtotal mass of Longun and her crew = 7.6 x 107 kg.

At a point which is 4 x 109 m from the planet’s centre, Danielle turns off Longun’s engines and travels solely under the influence of Anika’s gravity. Figure 6 below shows how the force on Longun varies with its distance from the centre of the planet.

Figure 6

Question 15

What is the gain in potential energy of Longun in the first 2 x 109m of its motion away from Anika without engine power?

2 marks

Sean finds that she when they arrives at this point 2 x 109m further on, their kinetic energy is zero.

Question 16

What was Longun’s speed at the moment of turning off its engines?

3 marksArea 1 - continued

J

0

1

2

3

4

5

6

7

0 1 2 3 4 5 6 7 8 9 10

Fo

rce

on

Long

un(1

04 N)

Distance of Longun from centre of planet (109m)

m/s


Danielle places Longun into orbit at a height of 3600km above the surface of Anika.

Question 17

What is the gravitational field strength at this orbit?

3 marks

Question 18

At what speed does Longun orbit Anika?

2 marks

Question 19

Calculate the period of Longun’s orbit.

2 marks

Area 1 - continued

N/kg

m/s

s


AREA 2 – Electronics and Photonics

Specific information for Area 2:Planck’s constant = 6.63 x 10-34 JsSpeed of light = 3x108m/sCharge on an electron = 1.6 x 10-19 C

Sophie is experimenting with a toaster, using a switch, S, to vary the power dissipated in the toaster by selecting one of three resistors, as shown in Figure 1. Power is supplied via a battery of voltage 15V RMS. The toaster has a resistance, RT, of 15Ω, and the values of R1, R2 and R3 are 100 Ω, 50 Ω and 15 Ω respectively.

Figure 1

Question 1

What is the RMS voltage across the toaster if the switch is connected to R1?

2 marks

Area 2 - continued

Specific Instructions for Area 2Area 2 consists of 15 questions. Answer all the questions in the answer booklet.A total of 35 marks is available for Area 2.

V


Question 2

If the switch is connected to R2, what is the value of the ratio of the RMS current through R2 to the RMS current through RT?

2 marks

Question 3

If the switch is connected to R3, how much electrical power is dissipated in the toaster?

2 marks

Emma, Ben and Gary are very concerned about the environment and global warming and are investigating alternative energy sources. They obtain a solar cell, a diode-like device that converts light energy into electrical energy, with the amount of electrical energy produced depending on the intensity of the sunlight incident on the cell. They plot an I-V graph for the solar cell, with the results shown in Figure 2 below.The cell produces a maximum voltage on “open circuit” – that is, when the load is effectively an infinite resistance

Figure 2

Area 2 - continued

A

W


Question 4

How much electrical power does the cell deliver on “open circuit”?

2 marks

Question 5

What is the maximum current this solar cell can deliver?

2 marks

When a certain resistor, R, is used as a load, the current flowing through it is measured as 1.0mA

Question 6

With resistor R in use, what is the voltage across the terminals of the solar cell?

2 marks

Question 7

What power is dissipated in resistor R under the above conditions?

2 marks

Question 8

Calculate the value of resistor R.

2 marks

Question 9

Why will the power output of the cell never be as large as the product of the maximum voltage and maximum current?

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

3 marks

Area 2 – continued

W

μA

V

W

Ω


Ben is fascinated by amplifiers and decides to investigate one more closely, sketching a graph of input and output characteristics for the amplifier. The results are shown in Figure 3 below.

Figure 3


Input to amplifier

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0 1 2 3 4 5 6 7 8

time (arbitrary units)

volt

age

(V)

Output from amplifier

-10

-8

-6

-4

-2

0

2

4

6

8

10

0 1 2 3 4 5 6 7 8

time (arbitrary units)

volt

age

(V)


Question 10

What is the gain of the amplifier?

2 marks

Question 11

On the axes provided, sketch the output voltage of the amplifier as a function of the input voltage.

3 marks

Question 12

Define the term clipping as it applies to an amplifier.

__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

2 marks


-20

-15

-10

-5

0

5

10

15

20

-1.5 -1 -0.5 0 0.5 1 1.5

Ou

tpu

t V

olt

age

(V)

Input Voltage (V)

Amplifier characteristics


In an npn transistor circuit, the collector current is 18mA and 93% of the electrons emitted reach the collector.

Question 13

What is the base current?

3 marks

Question 14

Define the term “de-coupling capacitor”________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

3 marks

Question 15

Briefly describe the operation of a phototransistor.

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

3 marks

Area 2 - continued

A


SECTION B – DETAILED STUDIES

AREA 3.1 – Einstein’s Special Relativity

Specific Instructions for Area 3.1

Area 3.1 consists of 10 questions.There are 20 marks available for Area 3.1.Specific information for Area 3.1:c = 3 x 108m/s

Philip and Dennis are accomplished astronauts on a 5-year mission to boldly go where no one has gone before.Even astronauts need relaxation time, and Philip and Dennis are currently laying back in their reclining chairs and watching Masterchef on their in-cabin colour television. An electron in the television tube has a rest mass of 9.1 x 10-31kg. Philip’s mass is 76kg.

Question 1

What is the energy equivalent to the mass of an electron at rest?

A. 8.2 x 10-14 JB. 6.7 x 10-14 JC. 3.2 x 10-14 JD. 4.9 x 10-14 J 2 marks

The electron moves so quickly in the television tube that its mass energy is 19% greater than its mass energy at rest.

Question 2

What is the kinetic energy of the electron?

A. 3.2 x 10-14 JB. 4.1 x 10-14 JC. 1.6 x 10-14 JD. 2.2 x 10-14 J 2 marks

Question 3

What is the speed of the electron in the television tube?

A. 3.46 x 108 m/sB. 1.64 x 108 m/sC. 2.63 x 108 m/sD. 3.33 x 108 m/s 2 marks

Area 3.1 - continued


Question 4

How much work must be done to increase the electron’s speed from rest to 0.6c?

A. 1.4 x 10-14JB. 2.2 x 10-14JC. 4.5 x 10-14JD. 1.8 x 10-14J 2 marks

Philip and Dennis have left their puppy Bosley behind on Earth as they zoom through the universe completing their mission. Occasionally, their spaceship passes close to the Earth and faithful little Bosley always sits outside to watch them fly past. On a particular fly-over, they zoom past Earth at a speed of 0.4c, while Bosley watches rapturously down on Earth.

Question 5

What is Philip’s mass on the spaceship?

A. 77kgB. 83kgC. 96kgD. 42kg 2 marks

Question 6

If Philip and Dennis spent the whole of their 5-year mission travelling at 0.4c, how much time would have elapsed on Earth when they returned?

A. 5.5 yearsB. 2.6 yearsC. 7.4 yearsD. 5 years 2 marks

Question 7

Which of the following best describes a major difference between the Newtonian model and Einstein’s postulates?

A. The Newtonian model assumes that the speed of light is the absolute maximum speed possible, wheras Einstein suggests that it is not.

B. The Newtonian model assumes that the time interval between events is independent of the motion of the observer, wheras Einstein does not.

C. The Einsteinian model assumes that the length of a body is independent of the motion of the observer, wheras the Newtonian model does not.

D. The Einsteinian model assumes that the mass of a body is independent of the motion of the observer, wheras the Newtonian model does not.

2 marks



Caz and Ali are housemates who live on a planet far, far away, on which the speed on light is 100km/hr in all reference frames.

Question 8

Which of the following best describes how flat sheet of paper , which appears square when at rest, will appear to Caz if it moves directly towards her (face on) at 20km/hr?

A. exactly the same as when it is at restB. square but smaller than when it was at restC. rectangular and the same size as it was when at restD. rectangular and smaller than it was when at rest 2 marks

Question 9

Which of the following best describes how Ali appears to Caz as she whizzes past her on a bike?

A. exactly the same as when at restB. thinner than when at restC. fatter than when at restD. taller than when at rest 2 marks

Caz heads off to work in the morning, driving at 80km/hr, and returns home in the evening at the same speed.

Question 10

Upon her return home, Caz and Ali notice that:

A. neither Caz nor Ali have aged.B. Caz is younger than she was when she departed.C. Caz has aged more than Ali.D. Ali has aged more than Caz. 2 marks



AREA 3.2 – Materials and their use in Structures

Specific instructions for Area 3.2There are 9 questions in Area 3.2There are 20 marks available for Area 3.2Specific data for Area 3.2:

Young’s modulus for steel = 2.0 x 1011 N/m2

Stress at the elastic limit for steel = 8.0 x 108 N/m2

Tensile strength of steel = 1.2 x 109 N/m2

Take g=10N/kg where required

Bella is poised on the end of a diving board, ready to perform her world-famous signature dive, The Comet. The diving board on which she stands is supported by a bracket at points P and Q, as shown in Figure 1. Bella’s mass acts through point S at the end of the diving board.Bella’s mass is 62kg and the mass of the diving board is 55kg.

Figure 1

Question 1

Which of the following is true of the forces on the diving board?

A. The board is under compression.B. The board is under tensionC. The top of the board is under tension and the bottom is under compressionD. The bottom of the board is under tension and the top is under compression

2 marksQuestion 2

What is the magnitude of the force acting on the board at point P?

A. 1.0 x 104NB. 3.6 x 103NC. 1.0 x 10-4ND. 3.6 x 10-3N

2 marks



Bella sees that someone is already in the water below her, so decides to delay her dive and walks back along the board from S to R.

Question 3

Which of the following best describes the forces acting on the board at points P and Q as Bella walks back along it?

A. The force at P will increase while the force at Q decreases.B. The force at Q will increase while the force at P decreases.C. The force at both P and Q will increase.D. The force at both P and Q will decrease.

2 marks

Dynae is a brilliant makeup artist who, needing close access to her makeup supplies as she works on her clients, has attached a shelf to her wall, within easy reach. The shelf is supported by two identical steel cables, of cross-sectional area 2.0 x 10-5m, each of which supports a half of the total weight of the shelf, as shown in Figure 2.

Figure 2

Question 4

What is the smallest force that must be applied to one of the cables to exceed the elastic limit of steel?

A. 4.6 x 105 NB. 1.6 x 104 NC. 8.0 x 108 ND. 3.2 x 104 N

2 marksArea 3.2 – continued


The tension in each cable when it is supporting the shelf is 110N.

Question 5

If the unstretched length of each cable is 50cm, by how much does each cable stretch in supporting one half of the weight of the shelf?

A. 3.2 x 10-5 mB. 4.8 x 10-5 mC. 3.5 x 10-5 mD. 1.4 x 10-5 m 3 marks

Question 6

What is the mass of the shelf?

A. 24.3kgB. 15.6kgC. 43.7kgD. 42.2kg 2 marks

The tensile stress-strain characteristics of a particular material are shown in Figure 3 below. The material has a Young’s modulus of 3.2 x 1011.

Figure 3

Question 7

If the strain at Point A is 0.4%, the stress in the material at Point A is:

A. 1.7 x 109 PaB. 1.3 x 109 PaC. 3.1 x 109 PaD. 3.7 x 109 Pa 2 marks



Question 8

What is the energy stored per unit volume in the material at Point A?

A. 2.6 x 106 J/m3

B. 3.2 x 108 J/m3

C. 3.3 x 1011 J/m3

D. 1.7 x 107 J/m3 3 marks

Question 9

Which of the following materials is most likely represented by the characteristics in Figure 3?

A. steelB. glassC. concreteD. marble 2 marks



AREA 3.3 – Further Electronics

Stuart loves musicals and has a miniature model of the set from The Phantom of the Opera, which runs on mains electricity and has moving parts including a huge chandelier that lights up at the appropriate time. Unfortunately, the model runs off a voltage of 220V RMS and has a power rating of 80W.Stuart, determined to get the model to work, connects the circuit in Figure 1 below.

Figure 1Question 1

What is the RMS current of the model at its stated rating?

A. 0.54AB. 0.63AC. 0.38AD. 0.36A 2 marks

Question 2

Why might the circuit shown enable the model to function?

A. The resistor will reduce the power rating of the model.B. The resistor will reduce the potential difference across the model.C. The resistor will increase the power rating of the model.D. The resistor will increase the potential difference across the model. 2 marks

Area 3.3 – continued

Specific Instructions for Area 3.3Area 3.3 consists of 9 questions. A total of 20 marks is available for area 3.3.


Question 3

What value of resistor, R, must be used to ensure that the voltage across the doll is 220V?

A. 56B. 64C. 124D. 256 2 marks

Question 4

Why would Stuart have been better to use a transformer to reduce the voltage across the model?

A. The resistor would be more expensive.B. The transformer would be more readily available.C. The transformer would be more energy efficient.D. The resistor would lead to a short circuit. 2 marks

Question 5

In order to step the voltage across the doll down from 240V to 220V, how many turns must be in the secondary winding of the transformer if it has 120 turns in the primary winding?(You may consider it an ideal transformer)

A. 120B. 110C. 220D. 210 2 marks

Question 6

What is the peak current in the secondary coil of the transformer?

A. 0.74AB. 0.55AC. 0.83AD. 0.26A 3 marks

Sam is a journalist who films interviews with celebrity guests. Her current guest is so famous, and the interview so top-secret, that he insists she be the only one in the room for their live interview. Sam is stumped for a minute and then comes up with a clever plan to use the timing circuit in Figure 2, which will allow a short time delay after she presses Record on the camera, to give her time to get to her seat with dignity and begin interviewing her guest. When Sam presses Record, switch S closes and the capacitor begins to charge. Vout must be 20V for the recording to begin.



Figure 2

Question 7

If the time delay is 30 seconds, what is the value of capacitor C?

A. 4000FB. 3000FC. 4500FD. 3500F 3 marks

Question 8

After the switch S is closed, which of the following best describes the variation in the voltage across the capacitor with time?

A The voltage decreases to a minimum value close to zeroB The voltage remains at a constant valueC The voltage variation follows a sinusoidal patternD The voltage increases until it reaches a certain constant value 2 marks

Sam finds that 30 seconds is not long enough for her to return to her seat and settle down in order to give the cleanest possible start to the live interview.

Question 9

Which of the following changes could be made in order to increase the length of the time delay?

A. Increase the value of the resistorB. Decrease the value of the capacitorC. Decrease the value of the resistorD. Change the positions of the capacitor and resistor in relation to the battery

2 marks


CHEMOLOGY EDUCATION SERVICESName:______________________

PHYSICS Unit 4MULTIPLE CHOICE ANSWER SHEET

Colour the box after the letter corresponding to your answer.3.1 Einstein’s Special Relativity 3.2 Materials and their use in structures

1. A B C D 1. A B C D

2. A B C D 2. A B C D

3. A B C D 3. A B C D

4. A B C D 4. A B C D

5. A B C D 5. A B C D

6. A B C D 6. A B C D

7. A B C D 7. A B C D

8. A B C D 8. A B C D

9. A B C D 9. A B C D

10. A B C D


CHEMOLOGY EDUCATION SERVICESName:______________________

PHYSICS Unit 4MULTIPLE CHOICE ANSWER SHEET

Colour the box after the letter corresponding to your answer.3.3 Further Electronics

1. A B C D

2. A B C D

3. A B C D

4. A B C D

5. A B C D

6. A B C D

7. A B C D

8. A B C D

9. A B C D


CHEMOLOGY EDUCATION SERVICES

P O BOX 477 MENTONE 3194Mobile: 0412 405 403 or 0425 749 520

[email protected]

SUGGESTED SOLUTIONS TO 2011 PHYSICS UNIT 3 TRIAL EXAM

AREA OF STUDY 1 – Motion in one and two dimensions.Question 1

We know that the momentum of Emilio and his car was:p(Emilio)= mv = 1282 x 6 = 7692 kgm/s Thus, we have a vector diagram as below (repositioning original vectors):

Using trig, the resultant momentum is given by:p(total)=7692/sin25=18200kgm/s≈ 1.8x104 kgm/s ANS

Question 2

p(total) = 18200 kgm/s (from above) = p(Emilio)+p(Armindo)Thus, 18200 = 7692 + p(Armindo)So, p(Armindo) = 10508 kgm/s = mv = 1274vThus, v(Armindo) = 10508/1274= 8.3m/s ANS

Question 3

p(total) = 18200 kgm/s = mv (after collision)Thus, 18200 = (1282+1274)vSo, v = 7.1m/s ANS

Question 4

An elastic collision is one in which kinetic energy is conserved.Total KE(before) = ½mv2 = ½ (1274)(8.3) 2 + ½ (1282)(6) 2 = 66960Total KE(after) = ½mv2 = ½ (1274+1282)(7.1) 2 = 64424

KE has been lost, and thus the collision is inelastic.

25o

Path after collision

Armindo

Emilio

v=6m/s, p=7692kgm/s


Question 5

Question 6

Resolving N vertically and horizontally, we find that the vertical component of N (Ncos20) will be equal to mg.

Thus, mg = Ncos20 E ANS

Question 7

The vertical part of N (Ncos20) equals mg and thus the only unbalanced (net) force on the car is the horizontal component of N, Nsin20.

mg = Ncos20Thus, 1282 = Ncos 20So, N = 1364Thus, the net force = Nsin20 = 1364(sin20) = 467N ANS

Question 8

To reduce sideways friction to zero, Nsin15 = Fnet = = mv2/rThus, 467 = (1282) v2/153So, v = 7.5m/s ANS

Question 9

Considering vertical motion:d=1.6 a=10 u=0 t=? d = ut + ½ at2

Thus, 1.6 = 0 + ½ (10)t2

t = 0.57 secs ANSQuestion 10

Considering horizontal motion (constant speed, no accel):d=4 t=0.57

v = d/t = u/t = 4/0.57u = 7m/s ANS (Catherine can really throw that fudge!)


Question 11

Using energy equations:

Initial total energy = PE + KE = mgh + 1/2 mv2

= 0.05(10)(1.6) + 1/2 (0.05) x (7)2

= 2.025JThus, final energy (as fudge lands) = 2.025JAs the fudge hits the ground, PE = 0 and thus total energy is all KE

Thus, KE = 2.025 = 1/2 mv2 = 1/2 (0.05)v2

So, v = 9m/s ANS

Question 12

Momentum is always conserved, so the momentum of the fudge is transferred via the floor to Earth. The kinetic energy of the fudge is transformed into other forms of energy, for example elastic potential energy as the fudge deforms (squashes down) and then back to kinetic energy as it bounces back up. Small amounts of energy are also lost as sound and heat as the collision of the fudge with the floor takes place until the fudge comes to rest.

Question 13

90% of his potential energy = 90% of mgh = 90/100 (85)(10)(7.3) = 5585J ANS

Question 14

Slope of F-d graph gives k (F=kx),Thus k = 4000/3

Total energy on landing is kinetic = 1/2 mv2 = 1/2 (85)(13)2 = 7182.5Maximum compression is when all KE is converted to PE (compression of mats),

Thus 1/2 mv2 = 1/2 kx2 = 7182.5 = ½ (4000/3)x2

So, max compression is 3.3m ANS

Question 15Longun has travelled from a distance of 4x109m from Anika to 6x109m.Change in E = work done = area under F-d graph

= (2x10 9) x (3x104) + ½(2x109) x (1x104) = 7 x 1013 J ANS

Question 16

Gain in PE = loss in KEThus, want KEinital = 7 x 1013 J (want KEfinal = 0) KE = ½ mv2

7 x 1013 = ½ (7.6x107) v2

v = 1.4 x 103 m/s ANS


Question 17

Gravitational field strength, g = F/m = GM/R2

= (6.7x10-11) (5.5x1022)/(4.9x104+3.6x106) 2

= 1.9 N/kg ANS

Question 18

Circular motion, so:Use a = v2 / RThus, 1.9 = v2 /(4.9x104+3.6x106) So, v = 2630 m/s ANS

Question 19

GMm/R2 = mv2/R = m42R/T2.Thus, T2 = 42R3/GMSo, T2 = 42 (4.9x 104+3.6x106)3/(6.7x10-11) (5.5x1022) T = 2.3 x 104 seconds ANS

AREA OF STUDY 2 – Electronics and Photonics

Question 1

Vout = (R1/(R1+R2))VsupplyThus, VT = (15/15+100) x 15 = 2V ANS

Question 2

The resistors are in series, and thus the current through them is the same. Thus, ratio of current through R2 to current through toaster = 1 ANS

Question 3

Since the resistors have the same value, the power is shared equally.Total power is given by P = V2/R = (15) 2/30 = 7.5WThus, power dissipated in toaster is 3.75W ANS

Question 4

On “open circuit”, load is effectively infinite, and so current will be 0. Thus, power: P = VI = V x 0 = 0W ANS

Question 5

The maximum current is produced when the load resistance is zero. Thus, we can read the value of maximum current from the graph when V=0.Imax = 1.7mA ANS


Question 6

The value can be read from the graph at the point where I=1.0mA.Thus, V = 4.8V ANSQuestion 7

P= VI = (4.8)(1.0 x 10-3) = 4.8 x 10-3 W = 4.8mW ANS

Question 8

V = IRThus, 4.8 = (1.0 x 10-3) RSo, R = 4.8kΩ ANS

Question 9

When the voltage is at a maximum, the current is zero (and thus, P = VI = 0).When the current is at a maximum, the voltage is zero (and again, P = 0).Non-zero values for P only occur when I and V are both less than their maximum values.

Question 10Using values read from the graphs:

Gain = Vout / Vin 7.5/0.5 = 15 ANS

Question 11

The amplifier is a non-inverting type, with a gain of 15. Thus, the graph will be linear, with a gradient of +15.


Question 12

Clipping is a form of amplifier distortion. When the input voltage goes beyond a limiting value and puts the amplifier into either cut-off or saturation mode (ie: beyond the linear region of the graph), the output voltage will be clipped. This results in the peak voltages of the output waveform being clipped off. In an audio amplifier, this would result in distortion of the sound output.

Question 13

First, find the emitter current.Ic = 0.93I = 18mAThus, I = 19.35mASo, Ib = I – Ic = 19.35-18 = 1.35mA ANS

Question 14

A decoupling capacitor (also known as a coupling capacitor) is one which is used to remove any constant (DC) component from a time-varying (ac) signal. This capacitor couples (passes) the ac voltage component of a signal from one point to another in a circuit but blocks any constant DC component in the same signal. The blocked component appears across the charged capacitor plates. It effectively decouples, or separates, the DC from the ac portion of the signal. This is useful when it is required to amplify a small ac signal that is superimposed on a large DC voltage.

Question 15

A phototransistor is a transistor that has a photosensitive collector-base pn junction. Unlike a normal transistor, it only needs two terminals – the collector and the emitter. When light is incident on the device, the current produced by the photoelectric effect generates the base current of the phototransistor. Thus, an increase in incident light intensity will produce an increase in collector current.

AREA OF STUDY 3.1 – Einstein’s Special Relativity

Question 1

E = mc2 Thus, E = (9.1 x 10-31)(3 x 108) 2 = 8.2 x 10-14 J Thus, A ANS

Question 2

The 19% “extra” energy is the KE.Thus, 19% of 8.2 x 10-14 = 1.6 x 10-14 J Thus, C ANS

Question 3

Using the equation m = m’/√(1-v2/c2)119/100 x (9.1 x 10-31) = 9.1 x 10-31/√(1-v2/(3 x 108) 2

Thus (1/1.19) = √(1-v2/(3 x 108) 2 , So v = 1.64 x 108 m/s Thus, B ANS


Question 4

We know energy of an electron at rest = 8.2 x 10-14J (from above)New energy at 0.6c is gained from E=mc2

Enew = m(at this speed) c2 = (9.1 x 10-31/√(1-(0.6c)2/c 2)) c 2 = 1.0 x 10-13

Thus, work = change in energy = (1.0 x 10-13) – (8.2 x 10-14) = 1.8 x 10-14J Thus, D ANS

Question 5

m = m’/√(1-v2/c2) = 76/√(1- (0.4c)2/c2) = 76/√0.84 = 83kg Thus, B ANS

Question 6

t = t’/√(1-v2/c2) = 5yrs/√(1- (04c)2/c2) = 5yrs/√0.84 = 5.5 years Thus, A ANS(lucky for poor Bosley that they weren’t travelling faster!)

Question 7

Classical (Newtonian) mechanics assumes that the time interval between events and length of a body are independent of the motion of the observer and that the speed of light is relative. Einstein states that space and time are relative and do depend on the motion of the observer, and that the speed of light is absolute.Thus, B ANS

Question 8

If it moves directly towards Caz, no length contraction occurs and the sheet of paper looks exactly the same as when at rest. Thus, A ANS

Question 9

There will be length contraction and so Ali will look thinner than when at rest. Thus, B ANS

Question 10

Time dilation will have occurred, and time will have passed more slowly in Caz’s frame of reference. Thus, Ali will have aged more than Caz. D ANS

AREA OF STUDY 3.2 – Materials and their use in Structures

Question 1

The board will bend down under Bella’s weight. Thus, the top is under tension and the bottom under compression. So, C ANS

Question 2

Taking moments about Q:Fp x 0.2 = (550x1.05) + (620x2.3)Thus, Fp = 10018N = 1.0 x 104N Thus, A ANS


Question 3

As Bella walks back along the board, her distance from P, Q and R decreases and thus, using similar calculations to the ones above, Fp will decrease. Hence, since Fp+620+550 = Fq, Fq will also decrease. Thus, D ANS

Question 4

Stress at elastic limit for steel = 8.0 x 108

and stress = F/AThus, 8.0 x 108 = F / (2.0 x 10-5 )So F = 1.6 x 104 N = 16kN Thus, B ANS

Question 5

Young’s modulus, E = stress / strain E = (F/A) / (ΔL/L) E = (F x L) / (A x ΔL) ΔL = (F x L ) / ( A x E )

Substituting values:ΔL = (110)(0.5) / (2.0 x 10-5) (2.0 x 1011)

Thus, ΔL = 1.4 x 10-5 m ANS Thus, D ANS

Question 6

From the force diagram below:

Thus, T cos 45 = mgSo 110 x 2 x cos45 = 10mAnd m = 15.6 kg Thus, B ANS

Question 7

E = stress/strainThus, 3.2 x 1011 = stress/0.004So, stress = 1.3 x 109 Pa Thus, B ANS


Question 8

Energy per unit volume = area under the graph = ½ x stress x strain (area of a triangle) = ½ x E x (strain)2 (since E = stress/strain)= ½ (3.2 x 1011) (0.004) 2 = 2.6 x 106 J/m3 Thus, A ANS

Question 9

All of the options except for steel are brittle (ie: have very little plastic region in their graphs).Thus, A ANS

AREA OF STUDY 3.3 – Further Electronics

Question 1

At the stated rating, P = 80W and V = 220VThus, PRMS = VRMS IRMS

80 = 220 x IRMS

IRMS = 0.36A Thus, D ANS

Question 2

The voltage must be reduced from the mains 240V in Australia to 220V for the model. Thus, Stuart is hoping that the resistor will reduce the potential difference across the model. So, B ANS

Question 3

Want 220V across the model, so need 20V across the resistor (in series).The model needs 0.36A to function (calculated in Q1), and since this is a series circuit the current is constant throughout the circuit. Thus, current through the resistor is also 0.27A.So, using V = IR across the resistor,

Vres = IRres 20 = 0.36 Rres Rres = 56 Thus, A ANSQuestion 4

The resistor will dissipate the remaining energy as heat, and thus energy will be lost. An ideal transformer would have exactly the same power at its primary and secondary, and all the available energy will be used by the model. Thus, there would be no energy wastage.So, C ANS


Question 5

N1 = 120 turns N2 = ?V1 = 240V V2 = 220VN1/N2 = V1/V2Thus, 120/N2 = 240/220 N2 = 110 turns So, B ANS

Question 6

At any time, Vpeak = √2 Vrms = √2 x 220and P = VIThus, P = VpeakIpeakPower = 80W for either side of the ideal transformer (no power loss) 80 = (√2 x 220) Ipeak Ipeak = 0.26 A Thus, D ANS

Question 7

The voltage out required is 20V. This is equal to the input voltage, and thus capacitor must be fully charged. This occurs after 5 time constants have elapsed.A time delay of 30 seconds is required. Thus, 30 seconds = 5 time constants.

5τ = 30 Thus, τ = 6secs = RC 6 = 1500C C = 0.004 = 4000F Thus, A ANS

Question 8

The charge across a capacitor increases until it reaches a certain final value, so the correct answer is D ANS

Question 9

The time delay depends on the time constant for the timing circuit, which in turn depends on the values of R and C. Increasing either of these values; that is, using a larger value of resistor or capacitor, will increase the delay time. Thus, the only option that will increase delay time is A ANS

chemology unit 3 2011

Documents