all assessment answers

8/10/2019 All Assessment Answers

1/29 NAS Chemistry Teachers Guide 2001 Nelson Thornes Ltd.

Section 7 Solutions to Assessment Questions

SN A


2/29


3/29 NAS Physics Teachers Guide 2001 Nelson Thornes Ltd.

Unit 1

Mechanics and Radioactivity

Solutions to Assessment Questions

Distanced travelled = area under graph between 6 s and 8 s 1

= 1 2 (3.3 m s 1 + 6.0 m s 1 ) 2.0 s = 9.3 m 1 [2]

5 force = mass acceleration 2 [2] Any three from:

different directions

act on different bodies

different types of force

different points of application or different lines of action

act at different times or act for different lengths of time 3 1 [3]

6 rate of change in momentum is proportional to the (resultant) force acting 1

and occurs in the direction of the force 1 [2]

Diagram (see Figure 13.1 on page 28) showing:

how a range of quantifiable forces can be applied to a movable object of constant mass 1

apparatus suitable for measuring the acceleration 1 [2]

statement of how the magnitude of the variable force is known 1

identification of the distance 1

and time measurements that could yield a value for the acceleration 1

repeat acceleration measurements for different forces 1 [4]

how acceleration is determined from the distance and time measurements 1

draw graph of force against acceleration 1

proportionality indicated by a straight line passing through the origin 1 [3]

7 v 2 = u 2 + 2 ax = 0 + (2 9.81 m s 2 2.00 m) 1= 39.2 m 2 s 2 1

v = (39.2 m 2 s 2 ) = 6.26 m s 1 1 [3]Tangent drawn at base of second line 1

Gradient of tangent = (1.85 m 0 m)/(1.10 s 0.72 s) 1

Speed = 4.9 m s 1 {range: 4.4 m s 1 to 5.3 m s 1} 1 [3]

Average force = ( m 2v2 m 2v1 )/ t 1= [0.12 kg (4.9 m s 1 + 6.26 m s 1 )]/(0.08 s) 2= 16.7 N {range: 16.1 N to 17.3 N} 1 [4]



Unit 1


Solutions to Assessment Questions8

Weight

Force fromhand

Normalcontactforce fromfloor

Pullfrom nail

Friction(either

direction)

Any four of the five forces shown, directions and labels must be correct 4 1 [4]No energy transferred by either friction or normal contact force from floor 1

as there is no movement in the direction of the force 1 [2]

Advantage of keeping x small: greater leverage 1

Disadvantage of keeping x small: nail pulled out a shorter distance 1 [2]

9 weight = mg = 45 000 kg 9.81 N kg 1 = 441 000 N 1 [1]Moments about A:

increase in upward force at B 60 m = 441 000 N 40 m 1increase in upward force at B = 441 000 N 40 m/(60 m) = 294 000 N 1 [2]increase in upward force at A = 441 000 N 294 000 N = 147 000 N 1 [1]

10 Energy stored = area under graph or 1

2 force extension 1= 1 2 800 N 0.5 m = 200 J 1 [2]

Kinetic energy = 1 2 mv2 = 200 J 1

v = [(200 J 2)/(0.060 kg)] 1= 82 m s 1 1 [3]

Line added to graph:

with same start and finish points 1

and enclosing a larger area under it 1 [2]

Slight decrease in horizontal velocity 1

due to opposing force of air resistance 1 [2]

Upwards velocity decreases to zero 1

and then increases downwards 1

because of gravity 1 [3]



Unit 1



11 KE before =1

2 22 000 kg (3 m s 1

)2

+1

2 66 000 kg (1 m s 1

)2

1KE after = 1 2 66 000 kg (2 m s 1 )2 1KE before = 132 000 J and KE after = 132 000 J so collision is elastic 1 [3]

KE at midway = 1 2 88 000 kg (1.5 m s 1 )2 1= 99 000 J 1

which is less than the total energy after the collision as some energy has been storedelastically in the compressed buffers 1 [3]

Occurs when the springs are in the act of either compressing or expanding 1 [1]

Impulse = change in momentum 1

= 66 000 kg (2 1) m s 1 [or 22 000 kg (3 0) m s 1] 1= 66 000 N s 1 [3]

The impulse is the same 1

because the momentum exchange is the same 1 [2]

12 Incoming particle is an alpha particle or a helium nucleus 1

Target atoms are gold or silver 1 [2]

Conclusions:

Large spaces between nuclei or nucleus is a very small part of the atom 1

Nucleus is dense 1 [2]

The diameter of a nucleus ( or nucleon or alpha particle) is approximately 10 15 m 1

The diameter of an atom ( or the electron orbits) is approximately 10 11 m 1 [2]

13 Compare and contrast:

1

1

1

1

1 [6]

1

14 Number of electrons = number of protons = 6 1

Number of neutrons = 14 6 = 8 1 [2]14

6C 147N + 10 2 [2]Take background reading 1

Corrected count unaffected when paper used as absorber 1

Corrected count becomes zero when lead used as absorber 1 [3]

Deep inelastic scattering Alpha particle scattering

Both fire projectiles at a target and look at deflections

Both look for substructure/charge distribution

Projectiles are electrons Projectiles are particles

Electrons have been accelerated/ particles from natural source/ have high energy low energy Target is neutrons/protons Target is atoms/gold foil

Discovered quarks Discovered nucleus



Unit 1



152211Na

2210 X +

01 2 [2]

Decay constant = ln 2/( t 1 2 in seconds) = 8.45 10 9 s 1 1

Activity = N = 8.45 10 9 s 1 6.0 1021 1= 5.1 1013 Bq (or s 1 ) 1 [3]

either correct for background or take readings for weeks 1 [1]

16 22086 Rn 21684Po + 42 1 [1] An appropriate set-up for detecting radiation (GM tube or ionisation chamber) 1Measurement of successive count rates 1

at 10 20 s intervals 1 [3]Plot of count rate against time graph 1

How half-life is found from graph 1 [2]

Count rate is so low, it will be difficult to distinguish it either from backgroundradiation or from random fluctuations 1

Very slow change in count rate difficult to detect over normal experimental times 1

Beta-minus absorbed within the milk sample 1 [any 2]

17 Density = mass/volume

Volume = mass/density = 0.197 kg/(19 000 kg m 3 ) 1

= 1.04 10 5 m3 1 [2] Volume of one atom = 1.04 10 5 m3 /(6.0 1023 ) = 1.73 10 29 m3 1Radius 3(1.73 10 29 m3 ) = 2.6 10 10 m 1 [2]Radius of nucleus 10 5 radius of atom so volume of nucleus 10 15 volume of atom 1Density of nucleus 10+15 density of atom= 10 +15 19 000 kg m 3 = 2 1019 kg m 3 1 [2]

Assumption: all the mass of the atom is in the nucleus 1 [1]



Unit 2

Electricity and Thermal Physics


drift speedv

= I /(

nAq )

1

= 8 10 3 A/(7 1023 m 3 3 10 6 m2 1.6 10 19 C) 1= 2.4 10 2 m s 1 1 [3]

Drift speed in the copper wire is much lower 1

but I , A and q are the same for the two specimens 1

so n must be much larger in copper than in the semiconductor material 1 [3]

Vibration of the ionic lattice increases with temperature 1

obstructing the flow of electrons and resulting in a reduced drift speed 1 [2]

2 (a) False 1

Wires in series carry the same current and since I = nAqv 1

If A is less then v will be more 1 [3]

(b) True 1

Power = V 2 /R and both resistors have same V (since in parallel) 1

so if R is less then P will be more 1 [3]

3 Ammeter added anywhere in series 1

Voltmeter added in parallel with the resistance wire 1 [2]

Record (corresponding) values of V and I 1

for different values of V 1

by adjusting R1 1 [3]

Draw a graph of V against I 1

Resistance = gradient 1 [2]

Length 1

diameter ( or cross-sectional area) 1 [2]

Any good reason for an implied yes or no [2]

4 Resistance R = l/A 1

= 1.72 10 8 m 6.00 m/(0.25 10 6 m2 ) 1= 0.41 1 [3]

Current I = nAqv 1

= 1.10 1029 m 3 0.25 10 6 m2 1.60 10 19 C 0.093 10 3 m s 1 1= 0.41 A 1 [3]

Idea that electrons behave as an incompressible fluid 2

so that current flow is immediate throughout circuit 1 [3]

5 Voltage drop across 4 resistor = IR = 0.75 A 4 = 3 V 1 Voltage across parallel combination = 9 V 3 V = 6 V 1

I1 = 6 V/(24 ) = 0.25 A 1 [3]




Unit 2


Solutions to Assessment Questions I

s = 0.75 A 0.25 A = 0.50 A 1

R = 6 V/(0.50 A) = 12 1 [2]

6 E.m.f. of single dry cell allow from 1.30 V to 1.60 V 1

Terminal p.d. allow from 2/3 to full e.m.f. value 1

N cells drawn in series, where N = 9/(e.m.f. value) 1 [3]

Any 3 (1 mark each) from:

keys (metal) have low resistance

since I = E/R (or P = E 2 /R )

small R means large I (or large P )

large I (or P ) causes heating [3]

7 I = I 1 + I 2 [1]

I = V/RT I 1 = V/R1 I 2 = V/R2 1

V/RT = V/R1 + V/R2 1

1 /RT = 1 /R1 + 1 /R2 1 [3]

if R V >> Rlow then 1/ R V



Unit 2



Rate of energy supply =mc T

/ t 1

for saucepan 1.20 kg 900 J kg 1 K 1 0.073 K s 1 = 79 J s 1 1for water 0.5 kg 4200 J kg1 K 1 0.073 K s 1 = 153 J s 1 1total rate of energy supply = 232 J s 1 = 230 W 1 [4]

As the temperature of the saucepan increases 1

an increasing proportion of supplied heat goes to surroundings 1 [2]

11 Pressure = force/area 1

At pedal, small force on small area produces large pressure 1

Pressure transmitted by fluid 1

at brakes, pressure acting on large area produces large force 1 [4]

12 pV = nRT 1 [1]

R = pV/nT

Units of R = N m 2 m3 /(mol 1 K) 1

= kg m s 2 m 2 m3 /(mol 1 K) 1

= kg m 2 s 2 mol 1 K 1 1 [3]

0 K Temperature

Pressure

Correctly labelled axes (allow any temperature) 1

Straight line through the correct origin 1 [2]

13 Energy transferred = 1 2 mv2 1

= 1 2 7.2 10 3 kg (32 m s 1 )2 1

= 3.7 J 1 [3]

Energy for work done by air on disc 1

comes from kinetic energy of air molecules 1but molecular kinetic energy absolute temperature or air temperature falls 1 [3]either disc moves in the direction of the force applied to it or disc gains energy fromthe air 2 [2]



Unit 2



Speed

N u m

b e r o

f m o

l e c u

l e s

A

B

axes suitably labelled 1shape of graph A 1 [2]

peak of graph B to the right of peak of graph A 1

both graphs having roughly equal areas 1 [2]

15 U = 0 1because the temperature of the filament is constant 1 [2]

W = 24 W 2 s = 48 J 1Electrical work done on the filament by the power supply 1 [2]

Q = U W = 0 48 J = 48 J 1Energy given to filament by heating (hence, negative here) 1 [2]only a small proportion of input energy comes out as light 1

the rest heats the surroundings 1 [2]

16 A mechanism in which heat from a higher temperature source 1

flows, in part, to a lower temperature sink 1

while the remainder is converted into useful work 1 [3]

Efficiency = heat transformed into useful work 1

divided by total heat flowing from hot source 1 [2]

A1 lines in spectrum 1

are characteristic of chemical composition 1 [2]

A2 examine the spectrum of the star or sketch graph of spectral distribution 1

Determine spectral type or measure wavelength at which spectrum peaks 1

Find temperature from H-R diagram or use Wiens law to find T 1 [3]

A3 definition of parallax (labelled diagram) 1Measure p for few months against background of distant stars 1



Unit 2



Find distance from p

=r/R

( or

equivalent) 1 [3]

Use of Wiens law ( or colour temperature) 1

to find T ( or spectral class) 1

Luminosity L from Hertzsprung-Russell diagram 1 [3]

measure intensity I at Earth 1

use of I = L /4 R2 1 [2]use of nearby stars 1

to draw H-R diagram 1

H-R diagram then gives other distances 1

and these distances give the structure 1 [4]

A4 axes correctly labelled 1

main sequence correctly drawn and labelled 1

red giants and white dwarfs shown and labelled 1 [3]

A5 luminosity L = intensity 4 R2 1 L = 1.2 10 8 W m 2 4 (4.3 1017 m) 2 1

= 2.8 1028 W 1 [3]4r 2 sT 4 = 2.8 1028 W r 2 = 2.8 1028 W/(4T 4 ) 1r 2 = 2.8 1028 W/[4 5.7 10 8 W m 2 K 4 (5200 K) 4] 1r = (5.3 1019 m2 ) = 7.3 109 m 1 [3]

A6 white dwarfs are very small stars 1

that are also very hot 1 [2]

Sirius B originally more massive than Sirius A 1

so it burnt up faster and has already passed beyond its red giant stage 1 [2]

A7 rise of temperature at centre 1

results in outer layers being blown away 1 [2]

either core mass > 1 M S or star mass > 8 M S [1]

high density or another characteristic 1

because (with more detail) 1 [2]

pulsar action or another characteristic feature 1

with mechanism or more detail 1 [2]

B1 find gradient/read values off graph and divide stress by strain 1using the linear region of the graph 1 [2]

Pa or N m 2 or kg m 1 s 2 [1]



Unit 2



Line B:

steeper than A 1

straight with no curving 1 [2]

Line C:

less steep than A 1

becoming plastic at a lower strain 1 [2]

B2 A to B:

bonds stretched 1bonds recover when force removed 1 [2]

B to C:

planes of atoms 1

slip over each other 1

leaving a permanent strain 1 [3]

Area of triangle = 1 2 base height= 1 2 4 10

3 8 108 Pa 1= 1.6 106 Pa ( or N m 2 or J m 3 ) 1 [2]

Area represents energy per unit volume [1]stored energy = 1.6 106 J m 3 volume of wire [ r 2 h ] 1

= 1.6 106 J m 3 [ (0.2 10 3 m) 2 2 m] 1= 1.6 106 J m 3 2.5 10 7 m3 = 0.40 J 1 [3]

B3 diagram and words convey idea of a number of crystals/grains 1

idea of planes of atoms in different directions 1 [2]

B4 work hardening is repeated stress cycles under elastic limit [1]

any two from:metal becomes harder

metal becomes stronger

metal becomes more brittle

Strain

S t r e s s

/ P a

Material A

B

C



Unit 2



elastic limit increases [2]

reverse by:

heating to a high temperature/just under melting point 1

then gradually cooling 1 [2]

B5 T G: temperature at which the material changes from a glassy/brittle state to arubbery/viscous state [1]

Strain

S t r e s s

< T G> T G

Line A straight line i.e. brittle 1Line B getting shallower then steeper and with a large strain i.e. rubbery 1

Initial gradient of line A greater than that of graph B 1 [3]

Below T G: long tangled chains 1

Above T G: long chains free to untangle and move relative to each other 1 [2]

B6 components of concrete: stone, sand and cement [1]

Concrete is poured onto steel rods that are in tension 1

Concrete sets; tension is removed 1

Steel tries to contract and pulls the concrete into compression 1 [3]

C1 alpha particles emitted all with the same energy 1

Beta particles emitted with a range of energies 1 [2]

Beta decay involves a second particle (antineutrino) 1

which takes lost energy 1

Alpha is the only particle emitted, therefore no lost energy 1 [3]

C2 mass before = [(92 1.0073) + (146 1.0087)] u = 239.9418 u 1m = (239.9418 238.0003) u = 1.9415 u 1Binding energy = 1.9415 u 930 MeV u 1 = 1806 MeV 1Binding energy per nucleon = 1806 MeV/(238 nucleons) = 7.6 MeV 1 [4]



Unit 2


Solutions to Assessment QuestionsC3

electron neutrino is difficult to detect because:it does not react by electromagnetic interaction, hence no ions to detect 1

it does not feel the strong interaction 1

it only interacts via the weak interaction 1

it has very small/zero mass so little/no gravitational interaction 1 [4]

Second generation other members are:

charmed quark 1

muon neutrino 1

anti charmed quark/anti muon neutrino 1 [3]

discovery of the tau [1]

C4 total mass after reaction = [1.0073 + (14 0.1503) + 0.5305 + 1.1974] u = 4.8394 u 1m = [4.8394 (2 1.0073)] u = 2.8248 u 1 [2]Equivalent energy = 2.8248 u 930 MeV u 1 = 2630 MeV 1Kinetic energy of one proton = 1 2 2630 MeV = 1315 MeV 1 [2]Minimum since unlikely that all products will be at rest [1]

C5 conservation of charge/strangeness/charm/baryon number [1]

Conservation of lepton number/lepton muon number

either L (1) = (1) + (+1) + (1)

or L_ (1) = (0) + (0) + (1) [2]

C6 the weak interaction [1]

Strangeness not conserved ( or quarks change flavour) or lifetime of 10 10 s is longerthan expected [1]

W, Z [1]

Large mass of W, Z [1]

C7 diagram 1:

Baryon number not conserved (1 0) 1 means electromagnetic interaction so single strange particle impossible 1 [2]Diagram 2:

lepton number not conserved (0 2) 1 W should be W + /charge not conserved at either vertex 1 [2]

D1 example of medical investigation 1

Further significant detail 1

Suitable value for half-life 1

with reason 1 [4]131

53I 13154 Xe + 10 + [2]



Unit 2



D29942 Mo

99m43 Tc + 1

0

[1]Molybdenum remains attached to alumina 1

Technetium enters the saline solution 1

and is run off when concentration has risen 1 [3]

Time

( I n )

A c

t i v i t y

B C

axes labelled and second peak lower 1

correct shape twice 1

increasing activity of saline shown by B 1

C explained by run off of saline 1 [4]

gamma camera:

source and collimator 1

scintillator and photomultiplier 1

shown in correct sequence 1

with satisfactory reference to output 1 [4]

D3 diagnosis is identifying the cause of the problem 1

Therapy is the treatment 1 [2]

70 kV X-rays distinguish soft tissue from bone 1

since absorption increases with Z 3 1

Megavolt X-rays destroy any tissue 1

since absorption is independent of Z 1 [4]

D4 high frequency sound from probe 1

reflected from an interface within body 1

Echo received by probe and time delay measured 1

to give distance so picture can be constructed 1 [4]

D5 examination of leg 4 mark maximum from:

transmitter placed on leg and detect signal reflected by glass 1



Unit 2



repeat at each of several positions 1

show cross-section of leg on diagram 1

detector linked to monitor 1

details of either A scan or B scan 1 [4]

limit set by wavelength used 1

due to diffraction effects 1 [2]

ultrasonics or X-rays?

one reason for choice 1

with a supporting statement 1

one problem with alternative 1

with a supporting statement 1 [4]



Unit 4

Waves and our Universe


When a body changes direction at a constant speed 1

its velocity (a vector quantity) changes 1

and a change in velocity is an acceleration 1 [3]

Towards the Earth 1 [1]

On the Earth 1

Towards the Moon 1 [2]

2 Angular speed = / t = 2 rad/[(120 60) s] 1

= 8.7 10 4 rad s 1 1 [2]

Free fall: when the gravitational force is the only force acting on the object 1

Without gravity, satellite would continue along a tangent to its orbit 1

Free fall of satellite keeps pulling satellite back onto its circular path 1 [3]

3 v = r 1

v = 16 rad s 1 80 10 2 m = 12.8 m s 1 1 [2]

a = 2r 1

a = (16 rad s 1 )2 80 10 2 m = 205 m s 2 1 [2]

Tension in string has its maximum value when the stone is nearest the ground 1

as it equals centripetal force + weight at this point 1 [2]

4 Oscillatory motion where the force (or acceleration) is proportional to the displacement 1

but in the opposite direction 1 [2]

Using T = 2 ( l/g )

l = T 2 g /4 2 = (2.0 s) 2 9.81 m s 2 /42 1

l = 0.99 m 1 [2]

Amplitude = 3.0 cm 1 [1]

Either attempt to find gradient at Z or vmax = 2 3.0 cm/(8.0 s) 1

v = 2.4 cm s 1 1 [2]

4

3

2

1

0

1

2

3

4

/ c m

s 1

Time/s

4 6 80 42 6 8 10

Negative sine wave 1

Correct scale added 1 [2]


18/29



Unit 4



Vibrations of atoms are parallel to the direction of energy propagation 1 [1]

Y R C R C R C

(i) Any 2 compressions (C) correctly labelled 1

(ii) Any 2 rarefactions (R) correctly labelled 1

(iii) Wavelength correctly labelled 1 [3]

Speed v = f so v = / T (or vT = or T = v / ) 1 [1]

9 A transverse wave 1 which is restricted to vibrate in one plane only 1 [2]

Either using light:

view light through a Polaroid filter 1

rotate Polaroid by 90 goes completely dark if light is plane polarised 1 [2]

Or using microwaves:

microwave transmitter and detector 1

rotate detector (or transmitter) by 90 and reading falls to zero 1 [2]

(or rotate grille of metal bars positioned between transmitter and detector)

(i) False 1 All waves can be diffracted

(ii) False 1

Sound is a pressure wave that transmits energy

(iii) True 1

= v/ f = 331 m s 1 /(436 Hz) = 0.759 m = 75 cm 1 cm

10 Any 5 points from:

level ripple tank

two dippers on same beam or use of a single beamdippers just touch surface of water or plane waves pass through 2 slits

vibrated electrically implied e.g. motor/oscillator/vibrator/signal generator

suitable illumination e.g. light or stroboscope

projected onto paper, screen or production of frozen image 5 1 [5]


20/29

A

C

B

C

B

S 1

S 2

NAS Physics Teachers Guide 2001 Nelson Thornes Ltd.

Unit 4



(i) A correct 1

(ii) B or B correct 1

(iii) C or C correct 1

Both B and C correct 1 [4]

Decrease 1

Increase 1 [2]

11 Any 2 points from:

both waves are the same type (allow either both transverse or both longitudinal)

waves have same frequency or wavelength

waves are travelling in opposite directions 2 1 [2]

Diagram showing a microwave transmitter pointing towards a metal plate 1

Stationary wave from superposition of waves from transmitter and those reflectedfrom metal plate 1

Move probe between transmitter and reflector and note series of maxima and minima 1 [3]

12 (i) True 160 Hz produces 3 loops (3 rd harmonic)

20 Hz produces 1 loop (fundamental)

160 Hz produces 8 loops 1


21/29



Unit 4



Graph B:starts at less than 0.5 V 1

saturates at less than 0.2 A (allow at 0.2 A) 1 [4]

15 2810 eV or 4.5 10 16 J 1 [1]

Use of E = hc / 1

= hc / E = 6.63 10 34 J s 3.0 108 m s 1 /(4.5 10 16 J) 1

= 4.4 10 10 m 1

X-rays or gamma rays 1 [4]

m = 207 9.1 10 31 kg or v = 0.11 3 108 m s 1 1

Momentum p = mv = 1.88 10 28 kg 3.3 107 m s 1 = 6.2 10 21 kg m s 1 1

= h/p = 6.63 10 34 J s/(6.2 10 21 kg m s 1 ) = 1.1 10 13 m 1 [3]

16 The ability of something to exhibit both wave and particle behaviour 1

e.g. light behaves like a wave when it is diffracted 1

and like a particle in the photoelectric effect 1 [3]

Momentum p = mv = 0.06 kg 2 m s 1 = 0.12 kg m s 1 1

= h/p = 6.63 10 34 J s/(0.12 kg m s 1 ) = 5.5 10 33 m 1 [2]

This wavelength is far too small to detect or smaller than that of gamma radiation 1 [1]

17 Measure or f [comparison must be stated or implied] 1

Quote Doppler formula for light waves e.g. / = v/c 1

State how to decide direction e.g. blue shift towards or red shift away 1 [3]

Galactic speed distance or v = Hd with symbols defined 1

[BUT must refer to galaxies to get this mark]

Sketch graph of velocity against distance straight line through origin 1

Hubble constant = gradient of graph 1 [3]

v = ( / ) c = 0.1 3.0 108 m s 1 = 3.0 107 m s 1 1

d = v/H = 3.0 107 m s 1 /(3 10 18 s 1 ) 1

d = 1 1025 m 1 [3]



Unit 5

Fields, Forces and Synthesis

Solutions to Assessment QuestionsSolutions to Assessment Questions

1 Force F = GM J MG / r 2 2 [2] M G r

2 1

M G r 2 = GM J M G / r

2 1

Cancel M G and rearrange to show that r 3 2 = GM J 1 [3]

T G = 7.16 days = 7.16 24 60 60 s = 6.19 105 s 1

G = 2 / T G = 2 /(6.19 10

5 s) = 1.02 10 5 rad s 1 1

M J = r 3 2 / G = (1.07 109 m) 3 (1.02 10 5 rad s 1 )2 /(6.67 10 11 N m 2 kg 2 )

= 1.89 1027 kg 1 [3]

2 g =

GM/r 2 1

v = (2 gr ) = (2GM / r ) 1

Planet has same mass and twice the radius

vp = vE / 2 1

vp = 11 km s 1 / 2 = 8 km s 1 1 [4]

Fuel must supply sufficient kinetic energy ( 1 2 mv 2 ) for the whole space probe 1 [1]

3

Diagram showing only ball A (which can be just a dot) with:

weight acting vertically downwards 1

electrostatic force acting leftwards 1

tension acting upwards to the right 1 [3]

For vertical equilibrium:

T cos 50 = mg 1

mg = 1.5 10 3 kg 9.81 N kg 1 = 1.5 10 2 N 1

T = mg /cos 50 = 1.5 10 2 N/cos 50 = 2.3 10 2 N 1 [3]

For horizontal equilibrium:

F E = T sin 50 = 2.3 10 2 N sin 50 = 1.75 10 2 N 1.8 10 2 N 1 [1]

F E = kq 1 q 2 / r 2 1

1.8 10 2 N = kq 2 /(15.3 10 2 m) 2 1 q = [1.8 10 2 N (15.3 10 2 m) 2 /(8.99 109 N m 2 C 2 )] = 2.1 10 7 C 1 [3]

Electrostatic force is repulsive and the gravitational force is attractive 1

Electrostatic force is much larger than the gravitational force 1 [2]

Weight

Electrostaticrepulsion

Tension



Unit 5



4

Diagram showing a speck of dust (which can be just a dot) with:

electric force upwards 1

gravitational force (or weight) downwards 1 [2]

Electric force = Eq = 100 N C 1 1.6 10 19 C = 1.6 10 17 N 1 Weight = mg = 1.0 10 18 kg 9.8 N kg 1 = 9.8 10 18 N 1Resultant = 1.6 10 17 N 9.8 10 18 N = 6.2 10 18 N 1Resultant force is upwards 1 [4]

5

Diagram showing:

two parallel plates labelled 200 V and 0 V 1

field shape including fringing 1

correct direction 1 [3]

Equipotentials added to diagram:

50 V equipotential 1

100 V equipotential 1 [2]

6 Capacitance = charge displaced/voltage 2 [2]

Current flows in the bell when the capacitor is charging 1

but not when the capacitor is fully charged 1 [2]

Labels added to circuit showing:

voltage across capacitor = 50 V 1

voltage across bell = 0 V 1 [2]

The capacitor discharges through the switch 1

the discharge current flows in the bell 1 [2]

Electric forcepulls dust up

Gravitational forcepulls dust down

200 V

0 V

100 V

50 V



Unit 5



Switch added to circuit in parallel with bell i.e. between pins 3 and 5 1 [1]

With anti-tinkling switch closed, current flows in this switch and not in the bell 1 [1]

7 At the beginning:

Q = CV = 40 000 10 6 F 12.0 V 1

Q = 0.48 C 1

At the end:

Q = CV = 40 000 10 6 F 10.5 V = 0.42 C 1 [3] Q = 0.48 C 0.42 C = 0.06 C 1

I = Q / t = 0.06 C/(10.0 10 3 s) 1

I = 6.0 A 1 [3]

8

Diagram 1

V t = V 1 + V 2 1

V = Q / C so Q / C t = Q / C 1 + Q / C 2 1

Cancel Q to give 1/ C t = 1/ C 1 + 1/ C 2 1 [4]

(a) 1/ C t = [1/(200 F)] + [1/(1000 F)] 1

C t = 167 F 1 [2]

(b) Q = C tV = 167 10 6 F 9 V 1

Q = 1.5 10 3 C 1 [2](c) V 1 = Q / C 1 = (1.5 10

3 C)/(200 10 6 F) 1

V 1 = 7.5 V 1

V 2 = Q / C 2 = (1.5 10 3 C)/(1000 10 6 F) = 1.5 V 1 [3]

9 W =1 2 CV 2 1

W =1 2 3 10 6 F (6 V)2 1

W = 5.4 10 5 J 1 [3]

Q = CV = 3 10 6 F 6 V = 1.8 10 5 C 1

capacitors in parallel, C t = C 1 + C 2 = 3 F + 5 F = 8 F 1W = Q 2 /2C = (1.8 10 5 C)2 /(2 8 10 6 F) 1

W = 2.0 10 5 J 1 [4]

+ Q

+ Q

+ Q

Q

+ Q

Q

V 2

V 1

V t = V 1 + V 2


26/29



Unit 5



14 The generation of an e.m.f. 1 when a conductor moves 1

relative to a magnetic field 1 [3]

Suitable experiment e.g. two coils with one carrying a.c. or bar magnet and solenoid 1

How change of flux linkage is produced 1

Measurement of induced e.m.f. with voltmeter in correct place 1

How to change the rate of flux cutting 1

What is then observed 1 [5]

v = 860 km h 1 = 860 103 m/(60 60 s) = 240 m s 1

Area swept out by aircraft each second = 240 m s 1 54 m = 1.29 104 m2 s 1 1Flux cut each second = B area swept out each second =6.0 10 5 T 1.3 104 m2 s 1 = 0.77 Wb s 1

E.m.f. induced = flux cut each second = 0.77 V 1 [2]

direction of magnetic field or whether plane is flying in the northern orsouthern hemisphere 1 [1]

15 (a) At 30 mm, B = 1 mT

Magnetic flux = BA = 1 10 3 T 16 10 4 m2 1 = 1.6 10 6 Wb ( or T m 2 ) 1

(b) At 10 mm, B = 30 mTMagnetic flux = BA = 30 10 3 T 16 10 4 m2 = 4.8 10 5 Wb 1 [3]Induced e.m.f. = / t 1

t = (4.8 10 5 Wb 1.6 10 6 Wb)/(15 10 6 V) = 3.1 s 1Speed = distance/time = 20 10 3 m/(3.1 s) = 6.5 10 3 m s 1 1 [3]Slow down or decelerate 1 [1]

16 (a) F = Gm 1m 2 / r 2 1

(b) F = kQ 1 Q 2 / r 2 1 [2]

Difference: gravitational only attractive, electric attractive and repulsive 1Similarity: both have an infinite range or obey inverse square laws 1 [2]

17 From line 2 of table:

N Q N 0 Q 0 l 1/ RC line 3 d Q /d t = Q/RC 1

line 4 1/ 1

line 5 RC ln 2 1

line 6 N 0 /e2 1

line 7 Q 0 /8 1 [5]



Unit 5



18 One value for half-life from graph between 33 s and 36 s 1Evidence of at least two values used and averaged 1 [2]

= ln 2/ t 1 2 = ln 2/(34.5 s) = 0.02 s 1 1 [1]

Tangent drawn to curve at N = 3.0 10 20 1 Attempt to find gradient (ignore negative sign) 1

Value between 5.5 1018 Bq and 6.0 1018 Bq 1 [3] A = N

= A/N = 5.75 1018 Bq/(3.0 1020 ) 1 = 0.019 s 1 1 [2]

First method since can take several values and average or not second method asdifficult to draw an accurate tangent 1 [1]

19 Q = CV 1

Q = 30 F 9 V = 270 C 1 [2]Time constant = RC = 20 103 30 10 6 F = 0.6 s 1 [1]

After 0.6 s, charge remaining = Q0 /e

After 1.2 s, charge remaining = Q0 /e2

After 1.8 s, charge remaining = Q0 /e3 1

Charge remaining = 270 C/e3 = 13 C 1 [2]

Graph of charge against time showing charge decreasing 1

Approximately exponentially (a charge intercept and not crossing time-axis) 1

Q 0 = 270 C Q 0.6 s = 99 C Q 1.2 s = 37 C Q 1.8 s = 13 C 1 [3]

20 Total mass before = 2.0136 u + 3.0160 u = 5.0296 u

Total mass after = 4.0026 u +1.0087 u = 5.0113 u

Mass of energy released = 5.0296 u 5.0113 u = 0.0183 u 1

Mass of energy released = 0.0183 u

1.66

10 27

kg u 1

= 3.038

10 29

kg 1

Energy released = c2 m = 3.038 10 29 kg (3.00 108 m s 1 )2 1Energy released = 2.73 10 12 J 1 [4]

C h a r g e

/ C

0

Time/s

0.5 1.5 210

50

100

150

200

250

300


29/29

Unit 5



21 F = mv2 / r p 1

F = Bqv 1

Bqv = mv 2 / r p so q/m = v/Br p 1 [3]

Additions to diagram:

both alpha particles and protons bend downwards 1

with the protons bending the most (smaller radius) 1 [2]

r = mv/Bq 1

m = 4 m p and q = 2 q pso r : r p = 4:2 = 2:1 1 [2]

Further additions to diagram:

neutrons travel straight through magnetic field 1

electrons bend upwards 1

with a very small radius 1 [3]

22 Vacuum 1

Radio frequency cavities or alternating electric field 1

Magnetic field for circular motion 1

Magnetic field for focusing or deflection device 1 [4]

Change in dimensions means that time to complete circuit changes 1

so particles become out of step with the alternating electric field 1 [2]

Alpha particles

Protons

n

e

all assessment answers

Documents