300828 physics 1 examples of quiz questions and past exam papers with solutions 2016

8/15/2019 300828 Physics 1 Examples of Quiz Questions and Past Exam Papers With Solutions 2016

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300828 Physics 1

Examples of Quiz Questions and Past Exam Papers with Solut ions

This document contains:

1. A brief discussion about how to study physics

2. An example of all questions in the quiz banks for the four online assessablequizzes with solutions or suggestions as to how to solve them.

3. A sample Data Analysis Test which you will complete in your final laboratory class.

4. Exam papers from the last five years with solutions for the written questions andanswers for the multiple choice questions


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Contents

How to Study Physics .................................................................................................................................................... 4

Examples of Multiple Choice Questions for Part A ......................................................................................................... 6

Pass level questions on SI prefixes and Units Conversion ................................................................. 6

Advanced level questions on SI prefixes and Units Conversion ......................................................... 7

Advanced level questions on rounding off of calculated values and units. ......................................... 8

Pass level questions for introductory kinematics (Part A). ................................................................... 9

Advanced level questions for introductory kinematics (Part A). ........................................................ 14

Advanced questions on vector representation in kinematics (Part A) .............................................. 18

Examples of Multiple Choice Questions for Part B ....................................................................................................... 22

Pass level questions on calculating forces involving weight, acceleration and friction. ................... 22

Advanced level Questions in Dynamics for Part B ............................................................................. 28

Pass level questions on Work and Kinetic Energy. ............................................................................ 33

Advanced level Questions on Work, Energy and Power for Part B .................................................. 36

Pass level questions on Momentum and Impulse for Part B ............................................................. 40

Examples of Multiple Choice Questions for Part C ...................................................................................................... 44

Pass level questions on DC electricity, resistance and Ohm’s Law. ................................................. 44

Advanced level questions on electric fields, electrical potential and forces involving charges ....... 50

Pass level questions on capacitance and storage of charge an energy. .......................................... 52

Advanced level questions on charging and discharging in resistor-capacitor circuits. .................... 53

Pass level questions on magnetic fields and electric currents. .........................................................

55

Advanced level questions on magnetic fields and electric currents. ................................................. 57

Examples of Multiple Choice Questions for Part D ...................................................................................................... 61

Advanced level questions on induced EMF and Faraday’s Law....................................................... 61

Advanced level questions on inductance and resistor-inductor circuits ............................................ 63

Pass level questions on reactance, impedance, resonance and AC circuits ................................... 65

Pass level questions on wave motion ................................................................................................. 68


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Pass level questions on refraction and refractive index. .................................................................... 69

Advanced level questions on refraction and refractive index ............................................................ 71

Pass level questions on lenses and image formation. ....................................................................... 75

Advanced level questions on lenses and image formation................................................................ 77

Pass level questions diffraction of light by narrow slits and diffraction gratings. .............................. 79

Sample Data Analysis Test ......................................................................................................................................... 86

Sample Data Analysis Test ANSWERS ............................................................................................. 91

2011 Exam Paper ....................................................................................................................................................... 97

2011 Exam Answers to Multiple Choice Questions ......................................................................... 115

2011 Exam Solutions for Written Questions ..................................................................................... 116

2012 Exam Paper ..................................................................................................................................................... 125

2012 Exam Answers to Multiple Choice Questions ......................................................................... 143

2012 Exam Solutions for Written Questions ........................................................................................ 144

2013 Exam Paper ...................................................................................................................................................... 153

2013 Exam Answers to Multiple Choice Questions ............................................................................. 171


2014 Exam Paper ...................................................................................................................................................... 180



2015 Exam Paper ...................................................................................................................................................... 207




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How to Study Physics

You cannot learn physics by just listening to lectures or reading a text book in the sameway that you cannot learn to play a piano by just reading a book on the subject. You need

to practice solving physics problems. The more you practice the more you learn and thequicker you get at solving problems. By solving many problems you will begin tounderstand and appreciate the various ways that physical situations can memathematically modelled and solved.

In this document a number of problems with solutions are given.

On the vUWS web site there are a number of practice tests containing multiple choicequestions. After attempting these tests you can then look at your submissions and alsosee descriptions of how to solve them. This provides good practice at solving a range ofquestions in a limited time.

Additional problems from the text book for self study are suggested at the end of each setof lecture slides. Lectures are used to present the significant points of each topic but thechallenge of a university education is to learn how to teach yourself. This is why a goodtext book is prescribed for this subject.

The best way to use the text book to do this is to start by looking at the example problemsfor each lecture to find out what sort of problems you will be expected to be able to solve.Depending on your previous experience, you may:

Know how to do the problems. Students who have studied 2 unit physics for the

HSC can probably do this for many simpler problems, particularly in the first part ofthe subject. Be aware that some formulae will appear different from what you havepreviously learned (eg in HSC physics you may have written s = ut + ½at 2 whereaswe use x = x0 + v0t + ½at

2). Work through the problems for practice. You may thinkyou know how to do them but you need to be sure of this. Make sure that you alsoread the relevant sections and look over the examples in the text.

Have some idea how to do the problems. If this is the case then work through therelevant examples given in the text book for practice and then reattempt theproblems. You should also read the relevant sections of the text.

Have no idea how to solve the problems. In this case you should try to solve oneproblem at a time. Read the relevant lecture notes and section of the text book andwork through any relevant examples. The way to "read" a physics text book is tohave a pen and piece of paper with you to sketch diagrams, write out equationsand work through the theory as it is discussed. This helps you to identify themeanings of the symbols used. You should also try to fill in any missing lines ofworking between equations. For example as part of a derivation the text may say"... we can obtain an expression that does not contain time by substituting thevalue of t from equation 3.7 into equation 3.9. This gives .... equation 3.11]". Youshould actually work through this derivation on your sheet of paper. Once you

understand enough to solve one problem then do so. You can then move on to thenext problem.


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Note the order of events

First read the problem,

then look at the examples

then read the text.

It is an advantage to work through problems with one or two other students. Not only mayother students help you but the process of explaining how to do a problem to someoneelse reinforces your own understanding. One of the best ways to learn is to teach!


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Examples of Multiple Choice Questions for Part A

Pass level questions on SI prefixes and Units Conversion

The unit of power is the Watt, symbol W. 153.2 μW is equivalent to:a. 1.532×10−7 Wb. 1.532×10−1 Wc. 1.532×10−4 Wd. 1.532×10−3 We. 1.532×10−5 W Answer: 153.2 μW = 153.2×10−6 W = 1.532×10−4 W. SI prefixes can be foundinside the front cover and also in table 1.4 of the text.

The unit of energy is the Joule, symbol J. 5.792×107 J is equivalent to:

a. 57.92 MJb. 579.2 MJc. 5.792 MJd. 5.792 GJe. 57.92 GJ Answer: 5.792×107 J = 57.92×106 J = 57.92 MJ.

An imperial unit of length is the mile, abbreviation mi. 1 mi=1609 m and 1 hr=3600s. Anold car has a maximum speed in 1st gear of 6.0 mi.hr −1, converted to m.s−1 this is nearestto:

a. 34.7 m.s−1

b. 13.4 m.s−1 c. 0.37 m.s−1 d. 2.68 m.s−1 e. 0.074 m.s−1

Answer: 6.0mi

hr

1609

1

m

mi

1 hr

2.68

3600

m

s s


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An imperial unit of mass is the ounce, abbreviation oz. During 1975 the price of gold wasquoted as US$172 per oz. At that time the exchange rate was A$1.00=US$1.32. Given1.0000 oz = 0.02835 kg, the price of gold in A$.kg−1 at that time was nearest to:

a. A$4596 per kgb. A$6067 per kg

c. A$8008 per kgd. A$6437 per kge. A$3694 per kg

Answer:$

172US

oz

1.00 $

1.32 $

A

US

1.00000 oz

$4596

0.02835

A

kg kg

An old unit of length is the foot, abbreviation ft. The flow rate of a glacier in some oldrecords was recorded as 3.25 ft.yr −1. 1 ft = 0.3048 m and 1 yr = 3.156×107 s. Convertedto m.s−1 the flow rate of this glacier would be nearest to:

a. 5.20x10−8 m.s−1

b. 3.38x10−7

m.s−1

c. 1.03x10−7 m.s−1 d. 2.97x10−9 m.s−1 e. 3.14x10−8 m.s−1

Answer: 3.25 ft

yr

0.3048

1

m

ft

1 yr

8

73.14 10

3.156 10

m

s s

Advanced level questions on SI prefixes and Units Conversion

An imperial unit of mass is the ounce, abbreviation oz, and an imperial unit of length is theinch, abbreviation in. 1 oz = 0.02835 kg and 1 in = 0.0254 m. In an old book the density ofan alloy is given as 4.75 oz.in−3. Converted to SI units this is nearest to:

a. 208.7 kg.m−3 b. 5.302 kg.m−3 c. 5295 kg.m−3 d. 5302 kg.m−3 e. 8218 kg.m−3

Answer 4.75oz

3in

3 31 in

3 3

0.02835

0.0254 1

kg

m oz

3

8218kg

m

An imperial unit of length is the yard, abbreviation yd. 1 yd=0.9144 m and 1 min=60s. Insome old records the flow rate of a river is described as 3800 yd3.min−1. Expressed in SIunits this is nearest to:

a. 58 m3.s−1 b. 53 m3.s−1 c. 21 m3.s−1 d. 48 m3.s−1 e. 69 m3.s−1

Answer3

3800 yd

min

1 min

3 3

3 3

0.9144

60 1

m

s yd

3

48m

s


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An imperial unit of length is the foot, abbreviation ft. 1 ft =0.3048 m and1 day = 86400 s. Gas production from an oil well is described as 2 300 000 ft3.day−1.Expressed in SI units this is nearest to:

a. 0.75 m3.s−1 b. 2.47 m3.s−1

c. 8.11 m3

.s−1

d. 45.2 m3.s−1 e. 60.6 m3.s−1

Answer:3

2300000 ft

day

1 day

3 3

3 3

0.3048

86400 1

m

s ft

3

0.75m

s

Advanced level questions on rounding off of calculated values and units.

If v = (21.4 ± 1.8) m.s−1 and t = (6.2 ± 0.7) s then the distance travelled by a particle

starting from rest x = vt is correctly expressed with its precision as:a. (132.7 ± 1.3) mb. (132.7 ± 2.5) mc. (133 ± 26) md. (132.68 ± 0.20) me. (133 ± 11) m

Answer:1.8 0.7

132.68 26.1421.4 6.2

x v t v t x x m

x v t v t

The uncertainty should then be rounded to 2 significant figures, because theleading digit is 3 or less, and the value rounded to match giving

x = (133 ± 26) m.

See pages 6 to 9 of the lab manual for further examples of calculation ofuncertainties.

If x = (21.4 ± 1.8) m.s−1 and t = (6.2 ± 0.7) s then the velocity of a particle starting fromrest v = x/t is correctly expressed with its precision as:

a. (3.5 ± 1.3) m.s−1 b. (3.5 ± 2.5) m.s−1 c. (3.5 ± 0.7) m.s−1

d. (3.45 ± 0.20) m.s−1

e. (3.45 ± 0.29) m.s−1

Answer: 11.8 0.7

3.452 0.680 .21.4 6.2

v x t x t v v m s

v x t x t

The uncertainty should then be rounded to 1 significant figure, because the leadingdigit is 4 or greater, and the value rounded to match givingv = (3.5 ± 0.7) m.s−1.

The length of the side of a cube is x = (8.3 ± 0.3) cm. The volume of the cube V = x3 iscorrectly expressed with its precision as:

a. (570 ± 60) cm3 b. (571 ± 21) cm3


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c. (571.8 ± 0.9) cm3 d. (570 ± 40) cm3 e. (570 ± 70) cm3

Answer: 30.3

3 3 3 571.79 62.008.3

V x xV V cm

V x x

The uncertainty should then be rounded to 1 significant figure, because the leadingdigit is 4 or greater, and the value rounded to match givingV = (570 ± 60) cm3.

Pass level questions for introductory kinematics (Part A).

0.10 s after the start of a race a sprinter's speed is 1.25 m.s−1. 0.90 s after the start thespeed is now 6.50 m.s−1. The average acceleration during this interval is nearest to:

a. 5.3 m.s−2 b. 12.5 m.s−2

c. 7.2 m.s−2

d. 9.7 m.s−2 e. 6.6 m.s−2 Answer:

22 1

2 1

change in velocity 6.50 1.25average acceleration= . 6.6 .

change in time 0.90 0.10

v vva m s

t t t

On a car trip the odometer shows a distance of 25 km 0.50 hours after the start of the journey. The odometer then shows 140 km 2.00 hours after the start. The average speedbetween these two measurements is nearest to:

a. 46 km.hr

−1

b. 50 km.hr −1 c. 70 km.hr −1 d. 77 km.hr −1 e. 110 km.hr −1 Answer:

12 1

2 1

change in position 140 25average speed= . 77 .

change in time 2.00 0.50

x x xv km hr

t t t

A car is travelling at 22 m.s−1 (approx 79 kph). The car is travelling towards a set of trafficlights and 85 m from the lights the driver glances at a mobile phone for 2.50 s. When the

driver looks back to the road ahead the distance to the traffic lights will now be nearest to:a. 76 mb. 55 mc. 140 md. 44 me. 30 m Answer: First calculate how far the driver travels while distracted

22 2.50 55 x vt m

then calculate the distance left to the traffic lightsd = 85 – 55 = 30 m.


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A sports car accelerates from rest with an average acceleration of 3.2 m.s−2 for 4.6 s. Theaverage velocity of the car during this time is nearest to:

a. 0.7 m.s−1 b. 14.7 m.s−1 c. 7.4 m.s−1

d. 29.4 m.s−1

e. 33.9 m.s−1 Answer: This First calculate the distance covered by the car

2 210 0 2

0 0 0.5 3.2 4.6 33.86 x x v t at m

then calculate average velocity

1 133.86 7.36 . 7.4 .4.6

xv m s m s

t

An aircraft touches down on a runway at 75 m.s−1 and immediately begins to brake with adeceleration of 3.2 m.s−2. 15 s after touchdown its velocity will be nearest to:

a. 123 m.s−1

b. 27 m.s−1 c. 48 m.s−1 d. 75 m.s−1 e. 16 m.s−1 Answer:There are three basic equations of uniformly accelerated straight line motion.

210 0 2

0

2 2

0 02

x x v t at

v v at

v v a x x

In this question you are given v0 = 75 m.s−1, a = −3.2 m.s−2 (note the negative sign,deceleration) and t = 15 s. Choose the appropriate equation and calculate v . (27m.s−1)

An aircraft touches down on a runway at 77 m.s−1 and immediately begins to brake with adeceleration of 2.9 m.s−2. 7.5 s after touchdown the distance it has travelled down therunway will be nearest to:

a. 659 mb. 496 mc. 82 m

d. 159 me. 578 m Answer:See the three basic equations of uniformly accelerated motion above.In this question you are given v0 = 77 m.s

−1, a = −2.9 m.s−2 (note the negative sign,deceleration) and t = 7.5 s. You assume x0 = 0 m at the touchdown point. Choosethe appropriate equation and calculate x . (496 m)


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An aircraft touches down on a runway at 73 m.s−1 and immediately begins to brake with adeceleration of 3.9 m.s−2. After it has travelled a distance of 440 m from the point oftouchdown its velocity will be nearest to:

a. 44 m.s−1 b. 94 m.s−1

c. 59 m.s−1

d. 14 m.s−1 e. 69 m.s−1 Answer:There are three basic equations of uniformly accelerated straight line motion.

210 0 2

0

2 2

0 02

x x v t at

v v at

v v a x x

In this question you are given v0 = 77 m.s−1, a = −3.9 m.s−2 (note the negative sign,

deceleration) and x = 440 m. You assume x0 = 0 m at the touchdown point. Choosethe appropriate equation and calculate v2, then remember to take the square rootto get v.( 44 m.s−1)

A car can accelerate at 2.1 m.s−2. It is initially travelling at 16 m.s−1 when it passesthrough an intersection and it then accelerates for 3.5 s. The distance of the car from theintersection at the end of this time is nearest to:

a. 69 mb. 56 mc. 13 md. 43 m

e. 82 m Answer:See the three basic equations of uniformly accelerated motion above.In this question you are given a = 2.1 m.s−2, v0 = 16 m.s

−1, and t = 3.5 s. Youassume x0 = 0 m at the intersection. Choose the appropriate equation andcalculate x. (69 m)

A car can accelerate at 2.1 m.s−2. It is initially travelling at 16 m.s−1 when it passesthrough an intersection and it then accelerates for 3.5 s. The velocity of the car at the endof this time is nearest to:

a. 21.6 m.s−1

b. 7.4 m.s−1 c. 8.7 m.s−1 d. 28.9 m.s−1 e. 23.4 m.s−1 Answer: See the three basic equations of uniformly accelerated motion above.In this question you are given a = 2.1 m.s−2, v0 = 16 m.s

−1, and t = 3.5 s. Choosethe appropriate equation and calculate v . (23.4 m.s−1)


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A freight train accelerates at a constant rate from rest and covers a distance of 150 m in44 s. The acceleration of the train is nearest to:

a. 1.705 m.s−2 b. 3.409 m.s−2 c. 0.039 m.s−2

d. 0.155 m.s−2

e. 6.453 m.s−2 Answer:There are three basic equations of uniformly accelerated straight line motion.

210 0 2

0

2 2

0 02

x x v t at

v v at

v v a x x

Here you are given v0 = 0 m.s−1 as the train starts from rest, x = 150 m and

t = 44 s. You can also assume x0 = 0 m. You are asked to find a so choose the

equation containing x, x0, v0, a and t , substitute in the values you know andrearrange it to make a the subject then calculate a. (0.155 m.s−2)

A freight train accelerates at a constant rate from rest and reaches a velocity of 16 m.s−1 after covering a distance of 120 m. The acceleration of the train is nearest to:

a. 2.133 m.s−2 b. 7.500 m.s−2 c. 0.008 m.s−2 d. 1.067 m.s−2 e. 0.133 m.s−2 Answer: See the three basic equations of uniformly accelerated motion above.

Here you are given v0 = 0 m.s−1

, the train starts from rest, v = 16 m.s−1

and x = 120 m. You can also assume x0 = 0 m. You are asked to find a so choose theequation containing v, v0, x, x0 and a, substitute in the values you know andrearrange it to make a the subject and calculate it. (1.067 m.s−2)


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The acceleration due to gravity is 9.8 m.s−2. A rock is thrown vertically downwards fromthe top of a 25 m high building and hits the ground with a velocity of 29 m.s−1. Thevelocity with which the rock was thrown is nearest to:

a. 36.5 m.s−1 b. 22.1 m.s−1

c. 18.7 m.s−1

d. 24.4 m.s−1 e. 26.8 m.s−1 Answer: See the three basic equations of uniformly accelerated motion above.Here you are given a = g = 9.8 m.s−2, x = 25 m and v = 29 m.s−1 and. You can alsoassume x0 = 0 m at the top of the building. You are asked to find v0 so choose theequation containing v, v0, x, x0 and a, substitute in the values you know andrearrange it to make v0 the subject. (18.7 m.s

−1)

The acceleration due to gravity is 9.8 m.s−2. A rock is thrown vertically downwards fromthe top of a 25 m high building and takes 1.9 s to reach the ground. The velocity with

which the rock was thrown is nearest to:a. 2.6 m.s−1 b. 2.3 m.s−1 c. 22.5 m.s−1 d. 3.8 m.s−1 e. 13.2 m.s−1 Answer: Start with one of the three basic equations of uniformly acceleratedstraight line motion.

210 0 2

0

2 20 02

x x v t at

v v at

v v a x x

Here you are given a = g = 9.8 m.s−2, x = 25 m and t = 1.9 s. You can also assume x0 = 0 m at the top of the building. You are asked to find v 0 so choose the equationcontaining x, x0, v0, t and a, substitute in the values you know and rearrange it tomake v0 the subject. (3.8 m.s

−1)

The acceleration due to gravity is 9.8 m.s−2. A rock is thrown vertically downwards fromthe top of a building with a velocity of 5.2 m.s−1 and strikes the ground with a velocity of20.6 m.s−1. The height of the building is nearest to:

a. 34.0 m

b. 23.0 mc. 40.5 md. 12.1 me. 20.3 m Answer: See the three basic equations of uniformly accelerated motion above.Here you are given a = g = 9.8 m.s−2, v0 = 5.2 m.s

−1 and v = 20.6 m.s−1. You canalso assume x0 = 0 m at the top of the building. You are asked to find x so choosethe equation containing v, v0, x, x0 and a, substitute in the values you know andrearrange it to make x the subject. (20.3 m)


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A car is cruising at 16 m.s−1 when it begins to accelerate to overtake a truck. 5 s afterstarting to accelerate it has covered a distance of 105 m. The acceleration of the car isnearest to:

a. 0.76 m.s−2

b. 1.22 m.s−2

c. 3.20 m.s−2 d. 8.40 m.s−2 e. 2.00 m.s−2 Answer: See the three basic equations of uniformly accelerated motion above.

Here you are given v0 = 16 m.s−1, t = 5 s and x = 105 m. You can also assume x0 =

0 m. You are asked to find a so choose the equation containing x, x0, v0, a and t ,substitute in the values you know and rearrange it to make a the subject. (2.00m.s−2)

Advanced level questions for introductory kinematics (Part A).

A car has an initial speed of 16 m.s−1 when a truck rolls out of a driveway 15 m ahead ofit. The driver takes 0.35 s to react before applying the brakes and the car then skids adeceleration of 8.0 m.s−2 before hitting the truck. The velocity of the car as it hits the truckis nearest to:

a. 13.2 m.s−1 b. 10.3 m.s−1 c. 22.3 m.s−1 d. 12.9 m.s−1

e. 4.0 m.s−1

Answer: This is a multi step question.First calculate the distance the car goes at constant velocity while the driver reacts.This is just v0t = 16x0.35 = 5.6 m. At the point the brakes are applied you now have

x0 = 5.6 m, x = 15 m, v0 = 16 m.s−1 and a = −8.0 m.s−2, note the negative sign

because it is decelerating. Using 2 20 02v v a x x determine v. (10.3 m.s−1)

A car has an initial speed of 15 m.s−1 when a truck rolls out of a driveway 18 m ahead ofit. The driver takes 0.35 s to react before applying the brakes. If the car is to just be ableto stop before hitting the truck the deceleration of the car will be nearest to:

a. 6.3 m.s

−2

b. 8.8 m.s−2 c. 3.1 m.s−2 d. 4.8 m.s−2 e. 0.6 m.s−2 Answer: This is a multi step question.First calculate the distance the car goes at constant velocity while the driver reacts.This is just v0t = 15x0.35 = 5.25 m. At the point the brakes are applied you nowhave x0 = 5.25 m, x = 18 m, v0 = 15 m.s

−1 and v = 0 m.s−2, the final velocity is zero

because the car just stops. Using 2 20 02v v a x x determine a. (8.8 m.s−2)


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A rock is dropped down a very deep mine shaft and the sound of it hitting the bottom isheard 4.2 s later. If the acceleration due to gravity is 9.8 m.s−1 and the speed of sound is341 m.s−1 the depth of the mine shaft is nearest to:

a. 146 mb. 86 m

c. 17 md. 81 me. 77 m Answer: This is a multi−step problem. Assume the depth of the mine is x m. The time taken to fall down the mine is t 1 andis related to x by 21

12 gt .

The time taken for the sound to come back is t 2 and is related to x by 2 s x v t

where v s is the speed of sound.The total time of flight is t 1 + t 2 = 4.2 s.You now have to solve three equations with three unknowns, x, t 1 and t 2. The

simplest method is to equate the first 2 equations to get 2 212 1 2 122

s

s

g v t gt t t v

.

At this point plug in the numbers and substitute into the 3rd equation whichbecomes a quadratic equation in t 1. Solve for t 1 and use this result back into thefirst equation to calculate x. (77 m)

A rock is thrown vertically upwards with a velocity of 30 m.s−1 from the top of a 25 m highbuilding. When it falls back down it just misses the building and falls to the ground. Theacceleration due to gravity is 9.8 m.s−2. The total time of flight of the rock is nearest to:

a. 5.13 sb. 0.74 sc. 6.87 sd. 7.49 se. 6.12 s Answer: There are several ways to do this problem.You could find the time taken to reach maximum height, then calculate this heightand find the time taken to fall back to the ground, adding these two times gives thetotal time of flight. Alternatively, taking positive as upwards and setting the origin at the base of thebuilding you are given v0 = 30 m.s

−1, x0 = 25 m, a = −9.8 m.s−2, note the sign, and x

= 0, final position is the base of the building. Substitute these values into21

0 0 2 x x v t at . You can then solve the quadratic equation to find t . The positiveroot is the correct result. (6.87 s)


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A road train consists of a truck and 4 trailers and has a total length of 52 m. It is travellingalong a straight stretch of road at 30 m.s−1 (approx 108 kph) when a car travelling at 33m.s−1 (119 kph) approaches from behind. The car pulls out to pass when it is 10 m behindthe back of the road train and then pulls back in after passing it when it is 10 m in front ofthe road train. The total distance covered by the car while it is overtaking is nearest to:

a. 682 mb. 72 mc. 720 md. 792 me. 572 m Answer: There are a couple of ways to solve this problem. One is to write thedistance covered by the truck as x = vT t = 30t . During this time the car has to travelthis distance plus the length of the road train plus an additional 20 m. I.e. x + 52 +20= vC t = 33t . This gives 2 equations and 2 unknowns. Solve for x, the distancetravelled by the truck, but remember to add on the extra distance covered by thecar. (792 m)

A ball is thrown with a velocity of 12 m.s−1 horizontally from the roof of a 17 m tall building.If the acceleration due to gravity is 9.8 m.s−2, the distance from the base of the buildingwhere the ball hits the ground is nearest to:

a. 13.8 mb. 41.6 mc. 20.8 md. 12.9 me. 22.4 m Answer: The initial vertical component of the velocity of the ball is zero so thesimplest way to do this question is to firstly calculate the time taken for the ball tofall a distance of 17 m (t = 1.863 s) and then calculate the distance travelledhorizontally at 12 m.s−1 during this time (22.4 m)

A ball is thrown with a velocity of 12 m.s−1 horizontally from the roof of a 17 m tall building.If the acceleration due to gravity is 9.8 m.s−2, the velocity of the ball when it hits theground is nearest to:

a. 21.8 m.s−1 b. 18.3 m.s−1 c. 30.3 m.s−1 d. 17.6 m.s−1

e. 24.0 m.s

−1

Answer: The initial vertical component of the velocity is zero so firstly calculate thevertical component of the velocity after the ball has fallen 17 m(18.3 m.s−1)vertically downwards, and then calculate the magnitude of the total velocity as

2 2 2 2 112 18.3 21.8 . x yv v v m s


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A soccer ball is kicked at an angle of 26° above the horizontal and it travels a distance of38 m before hitting the ground. If air friction is ignored and the acceleration due to gravityis taken as 9.8 m.s−2, the initial velocity of the ball is nearest to:

a. 20.6 m.s−1 b. 47.3 m.s−1

c. 24.5 m.s−1

d. 17.1 m.s−1 e. 21.7 m.s−1 Answer: There are several ways to do this problem. The simplest is to follow theworking in example 3−8 of the text which derives a horizontal range formula

20 0sin 2v R

g

and rearrange it to make v0 the subject,

1

0

0

38 9.821.7 .

sin 2 sin 2 26

Rg v m s

.

A soccer ball is kicked at an angle of 26° above the horizontal with an initial velocity of 17m.s−1. If air friction is ignored and the acceleration due to gravity is taken as 9.8 m.s−2, thetime of flight of the ball until it hits the ground is nearest to:*a. 1.52 sb. 0.76 sc. 1.73 sd. 3.47 se. 3.12 s

Answer: There are several ways to do this problem. The simplest way is to first findthe initial vertical component of the velocity of the ball

10 0 sin 17sin 26 7.452 . yv v m s and then calculate how long it takes to reachmaximum height, i.e. how long for the vertical component of the velocity to slow to

zero 000 7.452

0.7609.8

v vv v at t s

a

, note the sign of the acceleration.

The total time of flight will be double this because the object then has to return tothe ground. t total = 2t = 1.52 s.


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Advanced questions on vector representation in kinematics (Part A)

A ship sails 54 km 35° West of North and then alters course and sails 71 km 55° South ofWest. The magnitude of its displacement from the starting point is nearest to:

a. 125 kmb. 73 kmc. 17 kmd. 103 kme. 63 km Answer: This is a vector addition problem. You can do it by sketching the twovectors as shown below, calculating the angle θ between them and using thecosine rule to find the magnitude of the resultant R. Alternatively you can calculatethe x and y components of each vector, add them to get the components of theresultant and then calculate the magnitude of the resultant. (73 km), See furtherexamples in Lecture 3.

A ship sails 54 km 42° West of North and then alters course and sails 71 km 31° South ofWest. Its direction from the starting point is nearest to:

a. 17° East of Southb. 73° East of Southc. 88° West of Northd. 72° North of Easte. 82° West of North Answer: This is also a vector addition problem. You can do it by sketching the twovectors as shown below, calculating the angle θ between them and using thecosine rule to find the magnitude of the resultant. You then need to use either thesine or cosine rule to calculate the angle shown and use this together with the

direction of that vector to determine the direction of the resultant . (88° West ofNorth). Alternatively you can calculate the x and y components of eachvector, add them to get the components of the resultant and then calculate the

angle of the resultant with the x axis using tan y

x

R

R .

54 km

71 km 35

55

R

54 km

71 km 35

55

R

N

S

EW

N

S

EW

R

54 km

71 km42

31

=88

R

54 km

71 km42

31

=88

N

S

EW

N

S

EW


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A ship sails 54 km 35° West of North and then drops its anchor. A second ship sails 71km 55° South of West from the same starting position. The magnitude of thedisplacement between the ships is now nearest to:

a. 125 kmb. 103 km

c. 17 kmd. 73 kme. 63 km Answer: This is a vector subtraction problem. You can do it by sketching the twovectors, calculating the angle θ between them and using the cosine rule to find themagnitude of the distance D between their end points (103 km). Alternatively youcan calculate the x and y components of each vector, subtract them to get thecomponents of the resultant and then calculate the magnitude of the resultant.

The x component of a velocity is 15.2 m.s−1 and the y component is −13.3 m.s−1. Thevelocity can also be described as:

a. 20.2 −41.2° m.s−1 b. 20.2 41.2° m.s−1 c. 20.2 −48.8° m.s−1 d. 28.5 −48.8° m.s−1 e. 28.5 −61.0° m.s−1 Answer: The magnitude of the vector is given by

2 2 2 2 115.2 13.3 20.2 . x yv v v m s

The direction is given by 113.3

tan tan 41.215.2

y

x

v

v

.

(20.2 −41.2° m.s−1). Pay particular attention to the sign of the x and y components to make sure that you get the quadrant of the answer correct.

71 km

35

55

D

71 km

35

55

D

N

S

EW

N

S

EW


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A velocity is described in polar coordinates as 21.0 163° m.s−1. The x and ycomponents of this velocity are:

a. v x = −6.1 m.s−1, v y = −20.1m.s

−1 b. v x = 6.1 m.s

−1, v y = −20.1 m.s−1

c. v x = 20.1 m.s−1, v y = −6.1 m.s

−1

d. v x = −6.1 m.s−1

, v y = 20.1 m.s−1

e. v x = −20.1 m.s

−1, v y = 6.1 m.s−1

Answer: The components are given by

1

1

cos 21.0cos 163 20.1 .

sin 21.0sin 163 6.1 .

x

v

v v m s

v v m s

An aircraft is heading east at 133 m.s−1 relative to the air. The air is moving west at 42m.s−1 relative to the ground. A car on a road below the aircraft is heading east at 28 m.s−1 relative to the ground. The magnitude of the velocity of the aircraft relative to the car isnearest to:

a. 63 m.s−1 east

b. 147 m.s−1 eastc. 119 m.s−1 eastd. 203 m.s−1 easte. 105 m.s−1 east Answer: This is a one dimensional vector problem involving relative velocities. Thevelocity of the aircraft relative to the ground is given by

aircraft rel to g round aircraft rel to air air v v v .

The velocity of the aircraft relative to the car is then given by

aircraft rel to car aircraft rel to ground car aircraft rel to air air car v v v v v v

If we take east as the positive direction we can write

1133 42 28 63 .aircraft rel to car v m s . I.e. 63 m.s−1 east


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An aircraft is has a velocity of 78 m.s−1 35° West of North relative to the air, however itsvelocity relative to the ground is 74 m.s−1 25° West of North The magnitude of the velocityof the air relative to the ground is nearest to:

a. 152.0 m.s−1 b. 51.4 m.s−1

c. 13.8 m.s−1

d. 4.0 m.s−1 e. 107.5 m.s−1 Answer This is a two dimensional vector problem. We can start by writing

aircraft rel to ground aircraft rel to air air air aircraft rel to ground aircraft rel to airv = v + v v = v v remember

this is a vector problem so

aircraft relto airv is a vector 78 m.s−1 35° East of South.

The answer is given by sketching the vector sum as shown below, finding theangle θ as shown (10°) and using the cosine rule to find the magnitude of theresultant yielding vair = 13.8 m.s

−1

2535

78 m.s-1

74 m.s-1

vair

2535

78 m.s-1

74 m.s-1

vair

N

S

EW

N

S

EW


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Examples of Multiple Choice Questions for Part B

Pass level questions on calculating forces involving weight, acceleration and friction.

An elevator has a mass of 950 kg, the tension in the cable is 10.5 kN and the accelerationdue to gravity is 9.8 m.s−2. The acceleration of the elevator is nearest to:

a. upwards at 1.25 m.s−2 b. downwards at 1.25 m.s−2 c. upwards at 11.05 m.s−2 d. upwards at 0.11 m.s−2 e. downwards at 0.11 m.s−2 Answer: Taking positive as upwards the nett force acting onthe elevator is 15000 950 9.8 1190 F T W T mg N

the acceleration is then given by21190, 1.25 .

950

F F ma a m s

m

. I.e. 1.25 m.s−2 upwards.

The total mass of a freight train is 1500 tonnes and the acceleration due to gravity is 9.8

m.s−2. The tractive force supplied by the locomotive when hauling the train up a 1 in 33grade is nearest to:

a. 153 kNb. 45 kNc. 455 kNd. 15 kNe. 445 kN Answer: First find the slope

1tan , 1.7357

33

h

l .

The component of the weight forceparallel to the slope that the train needs toovercome is

3 3sin 1500 10 9.8 sin1.7351 445.3 10 445 F mg N kN

T

W = mg

m

T

W = mg

m

mg

h

l

mg sin

mg

h

l

mg sin


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The jet engines of an aircraft supply a thrust of 85 kN on takeoff along a level runway. Ifthe mass of the aircraft is 45 tonnes, the time taken from rest for it to reach takeoff speedof 69 m.s−1 is nearest to:

a. 130 sb. 37 s

c. 19 sd. 73 se. 69 s

Answer: First find the acceleration 285000

, 1.889 .45000

F F ma a m s

m

then find

the time to accelerate to takeoff speed from rest using

0

69, 69 0 1.889 , 37

1.889v v at a a s

A car of mass 1400 kg can accelerate from 16.7 m.s−1 (60 kph) to 22.2 m.s−1 (80 kph) in

2.7 s. The average horizontal force exerted by the road on the tyres is nearest to:a. 687 Nb. 10370 Nc. 8659 Nd. 2852 Ne. 11494 N

Answer: First find the average acceleration, 222.2 16.7

2.037 .2.7

va m s

t

, then

calculate the force 1400 2.037 2852 F ma N

A car travelling at 16.7 m.s−1 (60 kph) crashes into a tree and the front of the car crumples

decelerating it over a distance of 1.20 m. The force exerted by the seatbelt on a driver ofmass 75 kg is nearest to:

a. 8.72 kNb. 112.5 kNc. 0.12 kNd. 12.6 kNe. 0.52 kN Answer: First find the acceleration using

2

2 2 2 2 2

0 0

16.72 0 16.7 2 1.20 0.00 116.2 .

2 1.20v v a x x a a m s

then calculate the force 75 116.2 8715 F ma N . The magnitude of theforce is thus 8.72 kN


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A cricket ball has a mass of 160 g and is hit by a batsman at 36 m.s−1 and is caught by afielder who "gathers" the ball and decelerates it over a distance of 15 cm. The averageforce exerted by the fielder on the ball is nearest to:

a. 15.6 Nb. 6.9 N

c. 691 Nd. 1555 Ne. 6912 N Answer: First find the acceleration of the ball. You are given v0 = 36 m.s

−1,v = 0.0 m.s−1 (the ball is stopped), take x0 = 0.0 m then x = 0.15 m.

2 2

0 0

2 2

0

0

2

2

2

4320 .

v v a x x

v va

x x

m s

Then calculate the magnitude of the force F = ma = 0.160×4320 = 691 N

A car of mass 1500 starting from rest covers a distance of 400 m in 16.5 s. The averagehorizontal force exerted by the road on the tyres is nearest to:Then calculate the magnitude of the force F = ma = 1500×2.94 ≈ 4410 N

a. 1100 Nb. 18200 Nc. 4410 Nd. 14700 Ne. 510 N

Answer: First find the acceleration of the car. You are given v0 = 0 m.s−1 (the carstarts from rest), take x0 = 0.0 m then x = 400 m and t = 16.5 s.

21 0 02

2

2

since 0 and 0

2

2.94 .

x at x v

xa

t

m s


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A locomotive can supply a steady tractive force of 70 kN. It starts pulling a 800 tonnefreight train from rest. The velocity of the train after it has covered 200 m will be nearestto:

a. 3.5 m.s−1

b. 5.9 m.s−1

c. 17.5 m.s−1

d. 8.8 m.s−1 e. 1.1 m.s−1 Answer: First find the acceleration.You are given F = 70×103 N and m = 800×103 kg.

20.0875 .

F ma

F a

m

m s

Then calculate the velocity given v0 = 0 m.s−1, x0 = 0 m, x = 200 m

v2 = v02 + 2a( x – x0) = 35.0 v = 5.9 m.s−1

The coefficient of friction between a box of mass 6.2 kg and a vertical wall is 0.71. If theacceleration due to gravity is 9.8 m.s−2, the minimum necessary horizontal force whichneeds to be applied to the box to prevent it slipping down the wall is nearest to:

a. 61 Nb. 43 Nc. 8.7 Nd. 4.4 Ne. 86 N Answer: The weight force on the box is F = mg = 6.2×9.8 = 60.8 NThe normal force exerted on the box to hold it against the wall must be largeenough to provide friction to oppose this weight force.

60.886

0.71

F N

F N N


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A box has a mass of 23.0 kg and the coefficient of friction between the box and the flooris 0.74. If the acceleration due to gravity is 9.8 m.s−2 and a horizontal force of 220 N isapplied to the box its acceleration across the floor will be nearest to:

a. 0.74 m.s−2 b. 7.25 m.s−2

c. 0.43 m.s−2

d. 2.31 m.s−2 e. 1.74 m.s−2 Answer: The weight force on the box is mg = 23.0×9.8 = 225.4 N.This provides the normal force pushing the box against the floor thus the frictionforce is given by F =μ N = 0.74×225.4 = 166.8 N.This force opposes the motion so the nett force acting on the box is given by F tot = 220.0 – 166.8 = 53.2 N.The acceleration is then found by writing

253.2 2.31 .23.0

F ma

F a m s

m

The coefficient of friction between a car's tyres and the road is 0.75. If the mass of the caris 1300 kg and the acceleration due to gravity is 9.8 m.s−2, the maximum deceleration thatthe car can have while braking is nearest to:

a. 7.35 m.s−2 b. 13.07 m.s−2 c. 5.51 m.s−2 d. 9.80 m.s−2

e. 9.05 m.s−2

Answer: The weight force on the car provides the normal force holding it againstthe road. We can thus calculate the maximum friction force as F = μ N = μmg . Themaximum acceleration or deceleration is then found by writing

20.75 9.80 7.35 .

F ma

F mg a g m s

m m


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A box has a mass of 6.7 kg and the coefficient of friction between the box and a plank onwhich it is resting is 0.47. If the acceleration due to gravity is 9.8 m.s−2 and one end of theplank is lifted, the steepest angle with the horizontal before the box just slips is nearest to:

a. 28.0 °b. 25.2 °

c. 62.0 °d. 4.6 °e. 30.9 ° Answer: The slipping of a box on a slope is discussed in practical 3 and in lecture4. The critical angle for slipping is given by tanθ = μ thus θ = tan−1 μ = tan−10.47 =25.2°. The angle of slip does not depend on the mass of the box, only on thecoefficient of friction.

A locomotive has a mass of 270 tonnes. The railway line is wet and the coefficient offriction between the locomotive's wheels and the wet rails is 0.28. If the acceleration dueto gravity is 9.8 m.2−2, the maximum horizontal tractive force that it can provide before the

wheels slip is nearest to:a. 7.41×102 Nb. 7.41×105 Nc. 2.65×106 Nd. 2.65×102 Ne. 9.45×106 N Answer: The weight force on the locomotive provides the normal force holding itagainst the rails. We can thus calculate the maximum friction force as F = μ N = μmg = 0.28×(270×103)×9.8 = 7.41×105 N


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Advanced level Questions in Dynamics for Part B

A small freight elevator of mass 450 kg and counterweight of mass400 kg are attached by a light flexible cable over a frictionless pulleyas shown.

The acceleration due to gravity is 9.8 m.s−2. If the elevator is allowedto run freely the tension in the cable will be nearest to:

a. 3.92 kNb. 3.69 kNc. 4.67 kNd. 4.41 kNe. 4.15 kN

Answer: Sketch the problem andshow all forces acting on bothmasses. Then write down theequation of motion of bothmasses as shown.

You are given m1 = 450 kg,m2 = 400 kg and g = 9.8 m.s

−2.You now have two equations andtwo unknowns. Solve themsimultaneously to find T .

1

1

2

2

1 2

1 2

1 2 2 1 1 2

1 2 1 2 1 2 1 2

31 2

1 2

1

2

24.15 10

m g T

mass a m

T m g mass a

m

m g T T m g

m m

m m g m T m T m m g

m m g m m g m T m T m m T

m m g T N

m m

Elevator

Counterweight

Elevator

Counterweight

m1 g

m2 g

T T

m2a = T m2 g m1a = m1 g T

m1 g

m2 g

T T

m2a = T m2 g m1a = m1 g T


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A box of mass 15 kg rests on a smooth horizontal surfaceand is connected by a light flexible string to a freelyhanging block of mass 12 kg as shown.

The coefficient of friction between the first block and the

surface is 0.62 and the acceleration due to gravity is 9.8m.s−2. If the blocks are released the system will accelerateand the tension in the string will be nearest to:

a. 162 Nb. 129 Nc. 106 Nd. 26 Ne. 91 N Answer: Sketch theproblem and show allforces acting on both

masses. Then writedown the equation ofmotion of both massesas shown.

You are givenm1 = 15 kg,m2 = 12 kg, μ = 0.62and g = 9.8 m.s−2.You now have twoequations and twounknowns. Solve themsimultaneously to find T .

1

1

2

2

1 2

1 2

2 1 2 1 2 1

2 1 1 2 1 2

1 2

2 1

1

2

1106

T m g mass a

m

m g T mass a

m

T m g m g T

m m

m T m m g m m g m T

m T m T m m g m m g

m m g T N

m m

m1

m2

m1

m2

m1

m2

m1 g

m2 g

T

T

N

F f = m1 g

m1a = T m1 g

m2a = m2 g T

m1

m2

m1

m2

m1 g

m2 g

T

T

N

F f = m1 g

m1a = T m1 g

m2a = m2 g T


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A box is to be dragged across a smooth floor as shown.

The box has a mass of 8.0 kg, the coefficient offriction between the box and the floor is 0.58, theangle between the direction of the force and the

horizontal is 52° and the acceleration due togravity is 9.8 m.s−2. The magnitude of the forcerequired to just move the box is nearest to:

a. 32.4 Nb. 45.5 Nc. 35.8 Nd. 42.4 Ne. 286.7 N

Answer: Sketchthe problem.

Resolve the force F into vertical andhorizontalcomponents.Then sketch thesecomponents plusthe weight forceand the frictionforce acting on thebox.

You are given m = 8 kg, μ = 0.58, θ = 52° and g = 9.8 m.s−2.The nett vertical force downwards (the Normal force) is N = mg − F sinθ The nett horizontal force which will just cause slipping occurs when F cosθ = μ N = μ( mg − F sinθ )make F the subject of this expression to calculate the answer. (42.4 N)

m

F

m

F

m

F

F cos

F sin

F cos

F sin

mg

F f = N

m

F

F cos

F sin

F cos

F sin

mg

F f = N


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A box to be pushed across a smooth floor as shown.

The box has a mass of 8.0 kg, the coefficient of frictionbetween the box and the floor is 0.58, the anglebetween the direction of the force and the horizontal is

52° and the acceleration due to gravity is 9.8 m.s−2

. Themagnitude of the force required to just move the box isnearest to:

a. 263.8 Nb. 45.5 Nc. 57.7 Nd. 286.7 Ne. 42.4 N

Answer: Sketch theproblem. Resolve the force

F into vertical andhorizontal components.Then sketch thesecomponents plus theweight force and the frictionforce acting on the box.

You are given m = 8 kg, μ = 0.58, θ = 52° and g = 9.8 m.s−2.The nett vertical force downwards (the Normal force) is N = mg + F sinθ The nett horizontal force which will just cause slipping occurs when F cosθ = μ N = μ( mg + F sinθ )make F the subject of this expression to calculate the answer. (286.7 N)

m

F

m

F

m

F cos

F sin

F cos

F sin

mg

F f = N

F

m

F cos

F sin

F cos

F sin

mg

F f = N

F


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A box is resting on a slope and is prevented from slipping bya rope as shown.

The mass of the box is 45 kg , the coefficient of frictionbetween the box and the slope is 0.25, the angle made by

the slope with the horizontal is 35° and the acceleration dueto gravity is 9.8 m.s−2. The tension in the string supportingthe box is nearest to:

a. 163 Nb. 343 Nc. 441Nd. 253 Ne. 90 N

Answer: First resolve theweight force mg into

components parallel andperpendicular to theslope. Then sketch thesituation showing allforces acting on the box.Note there are twoforces pulling up theslope, the friction forceand the tension in thestring.

You are given m = 45 kg, μ = 0.25, θ = 35° and g = 9.8 m.s−2.First find the normal force N = mg cosθ = 361.2 NThen calculate the friction force F f = μ N = 90.3 NThe component of the weight force acting down the slope is mg sinθ = 252.9 NThe total force acting up the slope is equal to the total force acting down the slopeT + F f = mg sinθ thus T = 252.9 − 90.3 ≈ 163 N

m

m

mg cos

m

mg

mg sin

mg sin

N

mg cos

F f = N

T

mg cos

m

mg

mg sin

mg sin

N

mg cos

F f = N

T


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Pass level questions on Work and Kinetic Energy.

A projectile of mass 1.65 kg has a kinetic energy of 55 J. The velocity of the projectile isnearest to:

a. 4.1 m.s−1 b. 66.7 m.s−1 c. 8.2 m.s−1 d. 33.3 m.s−1 e. 90.8 m.s−1 Answer: You are given m = 1.65 kg and KE = 55 J

212

2

1

2

2 2 558.2 .

1.65

KE mv

KE v

m

KE v m s

m

A projectile has a velocity of 64 m.s−1 and a kinetic energy of 1700 J. Its mass is nearestto:

a. 0.21 kgb. 0.83 kgc. 4.82 kgd. 1.20 kge. 7.29 kg Answer: You are given v = 64 m.s−1 and KE = 1700 J.

21

2

2 2

2 2 17000.83

64

KE mv

KE m kg

v

If the mass of an object is decreased by a factor of 3 and it velocity is increased by afactor of 4 then the kinetic energy of the object will:

a. increase by a factor of24

1

b. decrease by a factor of3

16

c. decrease by a factor of 49

d. increase by a factor of16

9

e. increase by a factor of16

3

Answer:

210 2

22 21 1

1 02 2

4 164

3 3 3

KE mv

m KE v mv KE


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The work done accelerating a 17 tonne aircraft from rest to 66 m.s−1 for take off isnearest to:

a. 7.41×107 Jb. 3.70×104 Jc. 5.61×105 J

d. 3.70×107

Je. 9.54×103 J Answer: Work done = change in kinetic energy. Since initial kinetic energy is zero

2 3 2 71 12 2 17 10 66 3.70 10W mv J

A freight locomotive exerts a constant tractive force of 50 kN. The total work done haulinga freight train along a level track is 5.50×107 J. The distance the train was hauled isnearest to:

a. 1100 mb. 909 m

c. 33 md. 30 me. 2750 m

Answer: 7

3

cos

5.50 101100

cos 50 10 cos0

W Fd

W d m

F

Note: the angle θ = 0° because the locomotive is exerting the force in the directionof motion.

An old method of storing mechanical energy for a clock was to have a mass on a chain

which could be pulled up and then would gradually descend. If the acceleration due togravity is 9.8 m.s−2 and the maximum distance through which the mass can descend is1.25 m the mass required to store 15.0 J of energy is nearest to:

a. 1.91 kgb. 1.84 kgc. 1.22 kgd. 1.18 kge. 0.82 kg Answer: This is a gravitational potential energy problem

15.0

1.229.8 1.25

PE mgh

PE

m kg gh


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A string is attached to a box and it is draggedacross a floor at a constant speed as illustrated.

The magnitude of the force necessary to dragthe box is 325 N, the horizontal distance it is

dragged is 6.5 m and the total work done is 1750J. The angle theta between the string and thehorizontal is nearest to:

a. 34°b. 56°c. 40°d. 50°e. 75°

Answer:

1

cos

1750cos 0.8284

325 6.5cos 0.8284 34

W Fd

W

Fd

The coil spring in the front suspension of a car has a spring constant of 13600 N.m. Whenthe car's wheel hits a bump the spring is momentarily compressed by 12.5 cm. Theamount of energy temporarily stored in the spring is nearest to:

a. 1.06×106 Jb. 1.06×102 Jc. 8.50×104 Jd. 8.50×102 Je. 1.70×103 J Answer: Make sure you convert values to SI units. Energy stored in a spring

2

2 2 21 12 2

13600 12.5 10 1.06 10 PE kx J

m

F

m

F


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Advanced level Questions on Work, Energy and Power for Part B

A McDonalds large French fries has an energy content of 1900 kJ. A cyclist has anefficiency of 20% converting food energy to mechanical work. The mass of the cyclist plusbicycle is 95 kg. The acceleration due to gravity is 9.8 m.s−2. Ignoring friction, the totalvertical height up a series of hills the cyclist would have to cycle up to use up the energyin a McDonalds large French fries is nearest to:

a. 380 mb. 2041 mc. 102 md. 41 me. 408 m Answer: work done against gravity = efficiency × input energyRemember to convert percentages to fractions divide by 100. I.e. 20% ≡ 0.20. Alsoremember to convert all values to SI units, 1900 kJ = 1900×103 J = 1.9×106 J

30.2 1900 10408

95 9.8

in

in

mgh E

E h m

mg

A hammer head has a mass of 800 g and a velocity of 6.0 m.s−1 just before it strikes thehead of a nail which is then driven 2.5 cm into a piece of timber. The average forceexerted on the nail is nearest to:

a. 1736 Nb. 5760 Nc. 174 Nd. 576 N

e. 96 N Answer: kinetic energy of hammer = work done on nail

212

3 22

2

800 10 6.0576

2 2 2.5 10

mv Fd

mv F N

d

Remember to convert all values to SI units.


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A dart has a mass of 0.025 kg and is pressed against the spring in a toy dart gun. Thespring has a spring constant of 320 N.m and is compressed 55 mm. The dart is firedvertically upwards. The acceleration due to gravity is 9.8 m.s−2. Ignoring air friction themaximum height above the gun reached by the dart will be nearest to:

a. 1.98 m

b. 0.48 mc. 0.51 md. 35.9 me. 7.90 m Answer: potential energy stored in spring = increase in gravitational potential

energy

212

232 320 55 10

1.982 2 0.025 9.8

kx mgh

kxh m

mg

Remember to convert all values into SI units

A box is started sliding across a horizontal surface. The mass of the box is 4.2 kg and it isgiven an initial kinetic energy of 62 J. The coefficient of friction between the box and thesurface is 0.45 and the acceleration due to gravity is 9.8 m.s−2. The distance the box willslide before being brought to rest by friction is nearest to:

a. 3.35 mb. 1.51 mc. 5.43 md. 2.99 me. 6.64 m

Answer: First find the friction force F = μ N = μmg = 0.45×4.2×9.8 = 18.52 NWork done against friction = Initial kinetic energy

623.35

18.52

Fd KE

KE d m

F


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A cyclist plus her bicycle has a mass of 78 kg and is travelling at 5.4 m.s−1 as she comesover the crest of a hill. She rolls down the hill which is 15.0 m high. The acceleration dueto gravity is 9.8 m.s−2. If friction can be ignored the velocity of the cyclist at the bottom ofthe hill will be nearest to.

a. 16.3 m.s−1

b. 17.1 m.s−1

c. 18.0 m.s−1 d. 12.7 m.s−1 e. 22.5 m.s−1 Answer: change in gravitational potential energy = change in kinetic energy

2 21 12 2

2 2 21 1 12 2 2

212

2

1

78 9.8 15.0 78 5.4 12603

2

2 2 1260318.0 .

78

f i

f i

f f

f

f

f

f

mv mv mgh

mv mgh mv J

mv KE

KE

v m

KE v m s

m

A car of mass 1600 kg accelerates from 22 m.s−1 to 28 m.s−1 in a time of 4.0 s. Theaverage power output of the car during this time is nearest to:

a. 240 kWb. 60 kWc. 97 kWd. 157 kW

e. 29 kW

Answer:

2 21 12 2

2 21 12 2

3

1600 22 387200

1600 28 627200

627200 38720060 10 60

4.0

i i

f f

f i

KE mv J

KE mv J

change in energy power

change in time

KE KE W kW

t

A typical stair is 17 cm high. A person of mass 82 kg takes 31 s to walk up a flight of 40stairs. Their power output during this time is nearest to:

a. 1037 Wb. 441 Wc. 176 Wd. 18 We. 170 W Answer: Total height climbed h = 0.17×40 = 6.8 m

82 9.8 6.8176

31

work done against gravity mgh power W

time t


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A ski lift can carry 850 people per hour to the top of a 250 m high hill. Assuming that anaverage person has a mass of 70 kg the power supplied by the motor moving the peopleup the hill is nearest to:

a. 8.40×106 Wb. 1.72×105 W

c. 2.43×106

Wd. 4.13×103 We. 4.05×104 W Answer: Total mass moved per hour m = 850×70 = 59500 kg.1 hour = 60×60 = 3600 s.

459500 9.8 250 4.05 103600

work done against gravity mgh power W

time t

The total mass of a freight train is 4500 tonnes (4.50×106 kg). A section of straight trackhas a grade of 1:110 (i.e. it rises a vertical distance of 1 m for every 110 m horizontal).The power a locomotive supplies to haul this train up the slope at 5.0 m.s−1 is nearest to:

a. 2.00×106 Wb. 4.85×109 Wc. 4.85×106 Wd. 4.01×105 We. 2.48×106 W Answer: First find the angle of the slope then calculate the component of theweight force parallel to the slope the locomotive has to overcome

3

1tan , 0.52086

110

sin 4500 10 9.8 sin 0.52086 400893 F mg N

Now power = force × velocity = 400893×5.0 = 2.00×106 W

A jet aircraft has a mass of 280 tonnes (2.80×105 kg) and can accelerate at 2.0 m.s−2 during takeoff. The power delivered by the engines at a takeoff speed of 79 m.s−1 isnearest to:

a. 4.42×107 Wb. 2.21×107 Wc. 4.42×104 Wd. 2.21×104 We. 8.75×105 W Answer: First find the force delivered by the engines F = ma = (2.80×105) ×2.0 = 5.6×105 NThen calculate the power at takeoff speed

P = Fv = 5.6×105×79 = 4.42×107 W

1.0

L

mg

F = mg sin

1.0

L

1.0

L

mg

F = mg sin


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Pass level questions on Momentum and Impulse for Part B

An empty rail freight wagon of mass 24 tonnes is rolling at 1.4 m.s−1 on a straight leveltrack under a coal loader when 62 tonnes of coal, initially at rest, is dropped into thewagon. The velocity of the loaded wagon will be nearest to:

a. 1.01 m.s−1 b. 1.40 m.s−1 c. 0.54 m.s−1 d. 0.39 m.s−1 e. 0.62 m.s−1 Answer:

1 1 2 2 1 1 2 2i i f f m v m v m v m v

Since the final velocities of the coal and the wagon are the same1 2 f f f v v v we

can write

1 1 2 2 1 2 2

3 311 1 2 2

2 3 31 2

24 10 1.4 62 10 0.00.39 .

24 10 62 10

i i f

i i f

m v m v m m v

m v m vv m s

m m

A dodgem car at an amusement park has a mass of 180 kg and is being driven in astraight line at 1.60 m.s−1 when a second car of mass 140 kg traveling in the samedirection at 2.40 m.s−1 hits it from behind. Immediately after the collision the velocity of therear car is 1.60 m.s−1, the velocity of the front car after the collision will be nearest to:

a. 2.40 m.s−1 b. 2.22 m.s−1 c. 2.00 m.s−1

d. 1.87 m.s−1 e. 1.95 m.s−1 Answer:

1 1 2 2 1 1 2 2

1 1 2 2 2 2 1 1

1 1 2 2 2 2 1

1

1

180 1.6 140 2.4 140 1.62.22 .

180

i i f f

i i f f

i i f

f

m v m v m v m v

m v m v m v m v

m v m v m vv m s

m


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A military tank has a mass of 32 tonnes (32000 kg) and is travelling horizontally at 8.20m.s−1. It fires a shell with a mass of 45 kg straight ahead with a velocity of 760 m.s−1. Thevelocity of the tank immediately after firing the shell is nearest to:

a. 8.20 m.s−1 b. 9.27 m.s−1

c. 7.13 m.s−1

d. 1.07 m.s−1 e. 6.60 m.s−1 Answer: The initial velocity of the tank and the shell are the same so vTi = vSi = vi

132000 45 8.2 45 760 7.13 .32000

T i S i T Tf S Sf

T S i S Sf T Tf

T S i S Sf

Tf

T

m v m v m v m v

m m v m v m v

m m v m vv m s

m

A block of timber with a mass of 4.6 kg is sitting at rest when it is struck by a bullet ofmass 8.5 g which embeds in the timber. Immediately after the collision the recoil velocityof the timber block is 1.60 m.s−1. The velocity of the bullet just before impact is nearest to:

a. 626 m.s−1 b. 105 m.s−1 c. 542 m.s−1 d. 867 m.s−1 e. 136 m.s−1 Answer:

B Bi T Ti B Bf T Tf m v m v m v m v

The initial velocity of the timber vTi = 0. The final velocity of both timber and bullet isthe same so we can write v Bf = vTf = v f . Substituting this into the equation ofconservation of momentum gives:

3 13

8.5 10 4.6 1.60867 .

8.5 10

B Bi B f T f B T f

B T f

Bi

B

m v m v m v m m v

m m vv m s

m


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A cricket ball with a mass of 160 g and a velocity of 30 m.s−1 strikes the bat and is hitdirectly back at the bowler with a velocity of 25 m.s−1. If the cricket ball is in contact withthe bat for a time of 150 μs the average force exerted on the ball by the bat is nearest to:

a. 5.33×103 Nb. 5.87×104 N

c. 5.87×107

Nd. 5.33×106 Ne. 5.87×101 N Answer: Note that the ball changes direction so the velocities before and after theball is hit must have opposite signs. If we set the velocity after the ball is hit as+25 m.s−1 then the velocity before it is hit is −30 m.s−1.impulse = change in momentum

46

0.160 25 305.87 10

150 10

f i

f i

Ft m v m v v

m v v F N

t

A satellite of mass 1850 has a maneuvering rocket motor which can provide 30 N ofthrust. If the rocket motor is fired for 17 s the change in the velocity of the satellite will benearest to:

a. 3.63 m.s−1 b. 2.34 m.s−1 c. 0.28 m.s−1 d. 0.55 m.s−1 e. 4.69 m.s−1 Answer: impulse = change in momentum

130 17 0.28 .1850

Ft m v

Ft v m sm

A bullet of with a velocity of 280 m.s−1 mass 9.4 g hits a timber fence post and embeds inthe post. The bullet takes 290 μs to decelerate during the collision. The average forceexerted on the fence post during this time is nearest to:

a. 9.1 kNb. 29.8 kNc. 3.0 kNd. 11.0 kNe. 2.6 kN Answer: impulse = change in momentum

33

6

9.4 10 2809.1 10 9.1

290 10

Ft m v

m v F N kN

t


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A cricket ball has a mass of 160 g and is travelling at 26 m.s−1 when it is caught by afielder. The average force exerted by the fielder to stop the ball is 180 N. The time it takesto decelerate the ball is nearest to:

a. 29msb. 4 ms

c. 42 msd. 34 mse. 23 ms Answer: impulse = change in momentum

0.160 260.023 23

180

Ft m v

m vt s ms

F


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Examples of Multiple Choice Questions for Part C

Pass level questions on DC electricity, resistance and Ohm’s Law.

The current drawn by a 18 Ω resistor when it is connected to a 12 V power supply isnearest to:

a. 1.500 Ab. 0.667 Ac. 0.150 Ad. 0.080 Ae. 0.128 A Answer : Using Ohm’s law write

12 0.667 A18

V IR

V I R

A light bulb is rated as 12 V, 35 W. The current drawn by the light bulb when it is switchedon is nearest to:

a. 0.41 Ab. 0.34 Ac. 2.92 Ad. 2.43 Ae. 3.43 A Answer : From the expression for electrical power

352.92 A

12

P VI

P I

V

A small radio is rated as 3.2 W and has a 9.0 V battery. The effective resistance of theradio is nearest to:

a. 1.1 Ω b. 2.8 Ω c. 25.3 Ω d. 7.9 Ω

e. 28.8 Ω Answer : Starting with the expression for power2

2 29.025.3

3.2

V P

R

V R

P


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A laptop computer dissipates 40 W when drawing a current of 3.3 A from its powersupply. The effective resistance of the computer is nearest to:

a. 132 Ω b. 12.1 Ω c. 0.27 Ω

d. 147 Ω e. 3.7 Ω Answer : Starting with the expression for power

2

2 2

403.7

3.3

P I R

P R

I

A the voltage across an unknown resistor is measured and found to be 6.2 V when thecurrent through it is 3.1 mA. The value of the resistance is nearest to:

a. 2.0 kΩ

b. 2.0 Ω c. 2.0 mΩ d. 5.0 kΩ e. 5.0 Ω Answer : Remember that 1 mA = 10−3 A. Then using Ohm’s law write

3

3

6.22.0 10 2.0 k

3.1 10

V IR

V R

I

The resistivity of nichrome is 1.00×10−6 Ω.m. The length of 0.18 mm diameter nichrome

wire needed to make a 72 Ω heater element is nearest to:a. 40.0 mb. 2.83 mc. 7.33 md. 1.83 me. 0.55 m Answer : The cross sectional area of a wire is A = π r 2. Remember that the radius ishalf the diameter and to convert all units to SI units. Start with the expression forresistance of a long wire and make the length the subject

22 3

6

0.18 1072

221.83 m

1.00 10

l R

A

d R

RAl


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The resistivity of nichrome is 1.00×10−6 Ω.m. A heater element for an appliance needs tobe manufactured from a 1.8 m length of nichrome wire and have a resistance of 72 Ω.The diameter of the nichrome wire should be nearest to:

a. 0.45 mmb. 5.60 mm

c. 0.36 mmd. 2.24 mme. 0.18 mm

Answer : The cross sectional area of a wire is A = π r 2 so r

.

Remember that the diameter is double the radius and to convert all units to SIunits. Start with the expression for resistance of a long wire and make the area thesubject. Then calculate the diameter

68

84

1.00 10 1.82.50 10

72

2.50 102 2 2 1.8 10 m 0.18 mm

l R

A

l A

R

Ad r

Copper has a resistivity of 1.68×10−8 Ω.m. The resistance of a 25 m long cable containinga 1.20 mm diameter copper conductor is nearest to:

a. 371 mΩ b. 3.71 Ω c. 93 mΩ d. 930 mΩ e. 9.3 Ω Answer : Remember that the radius is half the diameter and to convert all units toSI units. Use the expression for resistance of a long wire.

8

2231

2

1.68 10 250.371 371 m

1.20 10

l l R

A r


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The resistance of a 0.42 mm diameter 18.0 m length of an alloy being trialed as aconductor is measured and found to be 530 mΩ. The resistivity of the alloy is nearest to:

a. 4.08×10−9 Ω.mb. 1.63×10−8 Ω.mc. 4.08×10−6 Ω.m

d. 1.63×10−5

Ω.me. 1.24×10−5 Ω.m Answer : Remember that the diameter is double the radius and to convert all unitsto SI units. Start with the expression for resistance of a long wire and make theresistivity the subject.

2

3 3122 9

530 10 0.42 104.08 10 .m

18

l R

A

RA R r

l l

Three resistors with values 33 kΩ, 82 kΩ and 15 kΩ are connected in parallel. Theeffective parallel resistance is nearest to:

a. 109 kΩ b. 130 kΩ c. 9.16 kΩ d. 43.3 kΩ e. 7.69 kΩ Answer : When adding resistances in parallel remember that we add thereciprocals of the resistances. Do not forget to take the reciprocal of the final result. Also, because only resistances are involved we can work in any units as long as all

resistances are in the same unit. In this case we work in kΩ

1 2 3

1 1 1 1 1 1 10.10916

33 82 15

19.16 k

0.10916

P

P

R R R R

R

Three resistors with values 33 kΩ, 82 kΩ and 15 kΩ are connected in series. Theeffective series resistance is nearest to:

a. 43.3 kΩ b. 9.16 kΩ

c. 109 kΩ d. 130 kΩ e. 7.69 kΩ Answer : When adding resistors in series we can work in any units as long as allresistances are in the same unit. In this case we work in kΩ

1 2 3 33 82 15 130 k S R R R R


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Four resistors with values 33 Ω, 82 Ω, 15 Ω and 47 Ω are connected in parallel. Theeffective parallel resistance is nearest to:

a. 5.8 Ω b. 177 Ω c. 130 Ω

d. 13.0 Ω e. 7.7 Ω Answer : When adding resistances in parallel remember that we add thereciprocals of the resistances. Do not forget to take the reciprocal of the final result.

1 2 3 4

1 1 1 1 1 1 1 1 10.130441

33 82 15 47

17.7

0.130441

P

P

R R R R R

R

Four resistors with values 33 Ω, 82 Ω, 15 Ω and 47 Ω are connected in series. The

effective parallel resistance is nearest to:a. 5.8 Ω b. 7.7 Ω c. 130 Ω d. 13.0 Ω e. 177 Ω Answer : Adding resistors in series is straightforward

1 2 3 4 33 82 15 47 177S R R R R R

The internal resistance of a 1.50 V battery whose terminal voltage drops to 1.45 V when acurrent of 1.2 A is drawn from the battery is nearest to:

a. 24.00 Ω b. 1.250 Ω c. 1.208 Ω d. 0.042 Ω e. 0.800 Ω Answer : The internal resistance

voltage drop inside batteryInternal resistance

total current

1.50 1.450.042

1.2 s out

s

V V R

I

A 6.00 V battery has an internal resistance of 0.050 Ω. When the battery delivers acurrent of 3.2 A the terminal voltage of the battery will be nearest to:

a. 6.00 Vb. 0.16 Vc. 5.95 Vd. 5.84 Ve. 1.88 V Answer : The voltage drop across the internal resistance is

3.2 0.050 0.16 sV IR V

The output voltage is thus 6.00 0.16 5.84out S V V V V


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The internal resistance of a 1.20 V battery whose terminal voltage drops to 1.12 V when itis connected to a 1.00 Ω resistor is nearest to:

a. 1.071 Ω b. 0.071 Ω c. 1.000 Ω

d. 14.00 Ω e. 0.933 Ω Answer : First find the current driven through the load by the output voltage.

1.121.12 A

1.00out

L

V I

R

The internal resistance is now given by the internal voltage drop over the current.1.20 1.12

0.0711.12

S out S

V V R

I

A 6.00 V battery has an internal resistance of 0.050 Ω. When the battery is connected to

a load resistance of 1.400 Ω the current delivered by the battery will be nearest to:a. 4.444 Ab. 1.200 Ac. 4.268 Ad. 4.138 Ae. 4.000 A Answer : The total resistance of the circuit is just the sum of the source resistanceplus the load resistance so

6.004.138 A

0.050 1.400S S

Tot S L

V V I

R R R


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Advanced level questions on electric fields, electrical potential and forces involving charges

The electrical potential a distance 20 cm from a point charge is 21000 V. The forceexerted on a dust particle at this point which contains 8 excess electrons will be nearestto:

a. 1.34x10−13 Nb. 2.69x10−14 Nc. 1.34x10−15 Nd. 5.38x10−15 Ne. 5.38x10−13 N Answer : This is a two step problem. First use the potential and distance tocalculate the point charge then use coulombs law to find the magnitude of the forcebetween the point charge and the dust particle.Firstly given V = 21000 V and r = 0.20 m in SI units.

7

9

21000 0.20, 4.667 10 C

9.0 10

kQ Vr V Q

r k

Second calculate the force on the dust particle given the charge on the particle isq = 8e = 8×1.60×10−19 = 1.280×10−18 C

7 189 13

2 2

4.667 10 1.280 109.0 10 1.34 10 N

0.20

Qq F k

r

A dust particle with 22 excess electrons is subjected to an electric force of 2.56x10−13 Nwhen it is between two parallel metal plated separated by a distance of 8.0 mm. Thevoltage applied to the plates will be nearest to:

a. 73 V

b. 12800 Vc. 582 Vd. 704 Ve. 2.8 V Answer : This is a two step problem. First calculate the strength of the electric fieldand then use this value to determine the voltage applied to the plates.Given q = 22e = 22×1.60×10−19 = 3.520×10−18 C

134 1

18

2.56 10, 7.273 10 V.m

3.520 10

F F qE E

q

Now for parallel plated with a separation of d = 0.0080 m

4

, 7.273 10 0.0080 582 VV

E V Ed d


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Two parallel plates are separated by a distance of 2.4 mm. A potential difference of 105 Vis applied to the plates. The magnitude of the force on a small oil drop with a chargeequivalent to 6 excess electrons is nearest to:

a. 1.05x10−14 Nb. 4.20x10−14 N

c. 4.20x10−17

Nd. 1.05x10−17 Ne. 2.42x10−16 N Answer : This is a two step problem. First calculate the electric field strengthbetween the plates. The separation between the plates is d = 0.0024 m so

1105 43750 V.m0.0024

V E

d

Secondly given the charge is q = 6e = 6×1.60×10−19 = 9.60×10−19 C19 149.60 10 43750 4.20 10 N F qE

300828 physics 1 examples of quiz questions and past exam papers with solutions 2016

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