Oxford Revise AQA GCSE Physics

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

chemical store

(associated with/of) the food/in her muscles

1

1

AO2

01.2

1

1

AO2

02.1

created or destroyed

1

AO1

02.2

there is no net changed to the total energy

1

AO1

02.3

is not

1

AO1

02.4

energy is transferred out of the system

because it no longer has kinetic/potential

energy/mechanical energy

accept energy is wasted/dissipated

accept change to either kinetic or potential energy

1

AO2

03.1

the height of the ball when she drops it

the mass of the ball

the rebound height of the ball after it first bounces

measure heights with a ruler/video analysis

measure mass of ball with a digital balance

1

AO1

03.2

the ball is moving quickly, so the height of the bounce is difficult to measure accurately/precisely

use a video camera to video the experiment

1

AO2

03.3

plan:

• use gravitational potential energy = mgh to calculate the initial gravitational potential energy
• use initial height, mass and g
• use the same equation to calculate the final gravitational potential energy
• use height after first bounce, mass and g
• subtract the final gravitational potential energy from the initial gravitational potential energy to find the energy

1

1

1

1

1

03.4

yes because there is energy

wasted/dissipated/transferred to the surroundings

or

no because the ball is not doing anything useful in terms of energy

justification must match answer to be awarded the marks

1

1

AO3

04.1

kinetic energy = 0.5 × mass × (speed)²

1

AO1

04.2

kinetic energy = 0.5 × 40 × (10) ²

= 2000 J

1

1

AO2

04.3

elastic potential energy = 0.5 × spring constant × extension²

2000 = 0.5 × 20 000 × e²

1

1

1

AO2

04.4

actual compression is less

because some energy from the kinetic energy store is transferred by sound/to the thermal energy store of the surroundings

1

1

1

AO3

05.1

energy in the elastic potential energy store is transferred to

the kinetic energy store

energy is transferred due to work done by forces

1

AO1

05.2

1

1

AO2

05.3

video analysis

accept any sensible suggestion

1

AO2

05.4

less energy is stored than predicted because the extension is less than 5 cm

or

the spring constant is less than 10⁵ N/kg

1

AO2

05.5

the thermal energy of the surroundings

1

AO1

06.1

the streamlined shape reduces the energy transferred to the surroundings/dissipated

1

AO2

06.2

kinetic energy = 0.5 × mass × (speed)²

1

AO1

06.3

kinetic energy = 0.5 × 700 000 × (90)²

= 2 835 000 000

= 2 835 000 kJ

= 28 400 00 kJ to three significant figures

AO2

06.4

2 840 000 kJ

accept 2 835 000 000 (J)

1

AO2

07.1

energy is transferred from the gravitational potential energy store to the kinetic energy store

1

1

AO1

07.2

work done by (gravitational) forces

1

AO2

07.3

the gravitational potential energy depends on mass, gravitational field strength and height

the mass and height are the same

if the gravitational field strength is less then there is less energy in the gravitational potential energy store

so less energy is transferred to the kinetic store

so the hammer ends up going slower on the Moon than the Earth

accept g on Earth is bigger, so gravitational potential energy is bigger, so kinetic energy is bigger, so speed is bigger

1

1

1

1

1

AO3

08.1

gravitational potential energy = mass × gravitational field strength × height

1

AO1

08.2

height = 4 floors = 4 × 3 = 12 m

gpe = mgh

= 1220 × 9.8 × 12 = 143 472 J

= 145 000 J

1

1

AO2

08.3

1

AO1

08.4

280 kJ = 280 000 J

accept 51(%) with no working for three marks

1

AO1

08.5

measure the mass of an object in the lift/and the lift

measure the number of floors it moves up

in a certain measured time.

either:

calculate the gravitational potential energy as before

calculate the energy transferred using power × time

calculate efficiency using

or

calculate the gravitational potential energy as before

calculate efficiency using

1

1

1

1

1

1

AO2

08.6

one from:

• difficult to measure time exactly
• floors may be different heights
• difficult to measure distance travelled early

accept any sensible suggestion

1

AO3

09.1

1

AO1

09.2

= 2 (seconds)

1

AO2

09.3

the second truck is less powerful than the first truck

1

AO2

09.4

gravitational potential energy store

1

AO1

10.1

energy is transferred from the gravitational potential energy store

to the kinetic energy store

1

1

AO1

10.2

light gate

1

AO1

10.3

time = 820 × 10^–3 s

speed = 1.3 m/s

distance = speed × time

= 1.3 × 0.82

= 1.07m

assuming the speed of the ball is constant.

no, it will not hit the target

convert to s

1

1

1

1

1

AO1

10.4

raise the height of the ramp

to increase the energy in the gravitational potential energy store and kinetic store

so the ball is moving faster at B

and travels further in the same time

1

1

1

1

AO3

11.1

energy that is no longer useful/stored in less useful ways

1

AO1

11.2

1

AO1

11.3

= 0.6

accept 0.6 with no working for two marks

60% scores one mark

1

1

AO2

11.4

car B

it has a lower efficiency, so wastes more energy

1

1

AO3

12.1

the trolley is moving too fast

light gates/motion sensor

1

1

AO1

12.2

gpe = mgh

Also accept:

gpe = mass × gravitational field strength × height

1

AO1

12.3

gpe = 0.25 × 9.8 × 0.12

= 0.298 J

1

1

AO2

4.1.1.2

12.4

no, the student is incorrect

because the energy has been dissipated/wasted as the trolley moves down the ramp

and transferred to a thermal energy store

no mark without correct reason

1

AO2

4.1.2.1

13.1

the barrier does not behave like a spring/does not behave elastically

1

AO3

13.2

car B:

kinetic energy before = 0.5 × 1000 × 50²

= 1 250 000 J

kinetic energy after = 0.5 × 1000 × 25²

= 312 500 J

energy transferred = 1 250 000 – 312 500

= 937 500

= 9.4 × 10⁵ J to two significant figures

1

1

1

1

1

1

AO2

13.3

car B transfers less energy to the surroundings

because it rebounds/does not stop

1

1

AO3

13.4

the kinetic energy depends on speed squared

so if the speed is reduced to 50%, the kinetic energy will be reduced to 25%

1

1

AO3

14.1

petrol/chemical energy (store)

1

AO2

14.2

kinetic energy store

1

AO2

14.3

work done by

force of friction/drag

1

1

AO2

14.4

motorcycle: the efficiency decreases with speed

at a decreasing rate

car: the efficiency decreases with speed

at a constant rate

1

1

1

1

AO2

14.5

evidence of tangent drawn to the curve at 30 mph

correct changes in efficiency/speed

the tangent should be drawn as a straight line between 60 on the y-axis and 60 on the x-axis, which touches the motorcycle curve at 30 mph.

rate = 60 % per 60 mph

1

AO3

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

temperature change = 50°C – 20°C

= 30°C

1

AO2

01.2

energy transferred = mass × specific heat capacity × change in temperature

= 1.2 × 900 × 30

= 32 400 J

1

1

AO2

01.3

energy is wasted/transferred to the thermal energy store of surroundings.

1

AO2

02.1

use loft insulation

1

AO1

02.2

the thicker the layer of bricks the slower the energy is transferred/the rate of energy transfer is less/less energy is transferred per second

1

AO1

02.3

a low thermal conductivity

thermal conductivity is related to the rate of energy transfer /low thermal conductivity means a low rate of energy transfer

1

AO1

03.1

swimming pool:

energy transferred = mass × specific heat capacity × change in temperature

28.8 × 10⁶ = mass × 4200 × (28 – 18)

paddling pool:

energy transferred = mass × specific heat capacity × change in temperature

88 000 × 10⁶ = mass × 4200 × (25 – 18)

1

AO1

03.2

there is more energy in the thermal energy store of the swimming pool

the paddling pool is at a higher temperature.

accept other reasonable suggestions

1

AO3

04

Level 3: Correct calculation of specific heat capacity, definition of specific heat capacity and power used to explain effect of doubling specific heat capacity.

5-6

AO2

Level 2: Calculation of specific heat capacity with conversion errors/comment about heating liquid B but without reference to definition of specific heat capacity.

3-4

Level 1: Comment about the difference in heating liquid B/attempt at calculating specific heat capacity with errors.

1-2

No relevant content.

0

Indicative content

• energy transferred = mass × specific heat capacity × change in temperature
• convert 1.8 kJ to 1800 J and 10 g to 0.01 kg
• substituting, 1800 = 0.01 × specific heat capacity × 50
• \(\rm{rearranging,\ specific\ heat\ capacity\ } =
  {\ }\frac{1800}{{{\rm{(0}}{\rm{.01\ }} \times {\rm{\ 50)}}}}
  \)
• solving, specific heat capacity = 3600 J/kg °C
• liquid B has twice the specific heat capacity, so it would take twice as much energy to raise the temperature (of the same mass by 50°C)
• (if he uses the same heater/a heater with the same power) it would take twice as long.

05.1

mass, speed

both needed for the mark

1

AO1

05.2

Level 3: Clear and coherent description of energy stores and transfers involved in both situations. Difference clearly stated with reason.

5-6

AO3

Level 2: Beginning or end stores described and transfers involved in both situations. Difference stated but no reason given.

3-4

Level 1: One or more relevant stores stated, with no description of transfer mechanism, or difference.

1-2

No relevant comment.

0

Indicative content

accelerating car:

• more energy in the chemical store (petrol) at the beginning
• as it accelerates there is an increasing amount transferred to
• the kinetic energy store
• the thermal energy store of the car
• the surroundings

motorway car:

• more energy in the chemical store (petrol) at the beginning
• this is transferred to the kinetic energy store
• to keep it at a constant level, energy is passed on to
• the thermal energy store of the car
• the surroundings

both cases:

• energy is transferred by
• mechanical working/force of the engine
• friction

05.3

oil provides lubrication

so less energy is transferred to the thermal energy store of the surroundings/less energy is dissipated

1

1

AO2

06.1

the type of insulation/material of can/starting temperature of water/colour of can/lid or no lid

1

AO2

06.2

72 in the third column/temperature after 15 minutes for 2cm

misreading the thermometer/starting temp of water was higher/any sensible suggestion

1

1

AO3

06.3

sensible suggestion with benefit

repeat the experiment more than once and calculate average /makes outliers easier to see/improves accuracy

use a bigger range of thicknesses/trend in data is easier to see

one mark for suggestion

one mark for benefit only if it clearly links to suggestion

2

AO3

07.1

crosses should be plotted along the following points:

(0,20), (2,35), (4,45), (6,50), (8,52) with a curved line of best fit that passes through all points

one mark for correct variable on x-axis and y-axis

one mark for appropriate scale on the x-axis and y-axis

one mark for three of four points of data plotted correctly

two marks for all data points plotted correctly

one mark for drawing a line of best fit

5

AO2

07.2

the rate of temperature increase decreases over time

the slope of the line decreases/the line becomes less steep/gradient decreases

1

1

AO2.2

07.3

energy transferred = mass × specific heat capacity × change in temperature

15 800 = 1 × specific heat capacity × (52 – 20)

allow 494 with no substitution for two marks

1

1

1

AO2

07.4

iron

allow error carried forward for incorrect calculation from question 07.3

1

AO3

08.1

type of material

1

AO2

08.2

suitable variable e.g., temperature drop over 10 minutes

1

AO2

08.3

suitable method e.g.,

wrap a can with a certain thickness of insulating material

fill the can with water at a certain temperature

put a lid on the can

record the temperature drop after ten minutes

repeat experiment three times, find average temperature drop

repeat whole experiment using a different insulating material each time

best insulator is the one with the smallest temperature drop

rank materials in order of smallest to largest temperature drop.

four marks are available for describing the method in detail, one mark is available for describing the repetition of measurements

one mark is available for describing how to rank them

1

1

1

AO2

08.4

any sensible suggestion e.g.,

thickness of insulation

starting temperature of water

1

AO3

09.1

the temperature decreased

thermal energy was transferred from the inside of the room to the outside of the room

1

1

AO2

09.2

the thermal conductivity of the walls

temperature difference across the walls

1

1

AO1

09.3

line drawn:

• same initial temperature
• shallower gradient

explanation:

• the line is above the first line
• because energy is transferred more slowly through the thicker walls
• the temperature at any time during the night will be higher

1

1

1

1

1

AO2

10.1

D

1

AO3

10.2

assuming (two from):

• the thickness of the walls is the same
• the same amount of energy is transferred
• temperature difference across walls is the same

so energy travels 2.5 times more slowly through material B

time to cool down will be 2.5 times longer

2 hours × 2.5 = 5 hours

one mark for each correct assumption up to a maximum of two marks

2

1

1

AO3

10.3

if the thermal conductivity decreases, the rate of energy transfer decreases

so the time it takes the temperature to go down will increase

1

AO3

11.1

there is a hazard/risk of injury/the teacher did a risk assessment that suggested that they not do it

1

AO2

11.2

so that the temperatures could be compared/the energy transferred per second was the same

1

AO2

11.3

for both liquids:

as the time increases the temperature increases

at an increasing rate

and

the oil increases at a higher rate than the water/temperature of oil stays steady for a longer time

1

1

1

AO2

11.4

assuming the energy supplied to each liquid is the same

assuming the mass of each liquid is the same

the temperature change of the oil is higher at the end

so the oil has a lower specific heat capacity

1

1

1

AO3

12.1

energy to raise to melting point= mass × specific heat capacity × change in temperature

= 0.005 × 371 × (29.8 – 20)

= 18(.2) J

1

1

1

AO2

12.2

energy is proportional to mass

energy = 3 × 18.2 = 54(.6) J

1

1

AO2

12.3

it would be about 3 times less

if the specific heat capacity is three times bigger, it takes three times the energy to produce the same temperature rise

or words to that effect

1

1

AO2

13.1

gravitational potential energy = mass × gravitational field strength × height

1

AO1

13.2

estimate mass of human = 50 kg – 100 kg

gravitational potential energy = 75 × 9.8 × 0.5

= 367.5 J

1

1

1

AO2

13.3

energy stored in a spring = 0.5 × spring constant × extension²

assuming all the energy stored in the muscles/tendons is transferred to the gravitational potential energy store

367.5 = 0.5 × spring constant × 0.01²

1

1

1

1

AO2

13.4

gravitational potential energy is proportional to mass if the height is the same

mass is 2000 times smaller, so gravitational potential energy is 2000 times smaller

spring constant is proportional to energy if the extension is the same

spring constant is 2000 times smaller

1

1

1

1

AO3

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

temperature change = 20 °C – 4 °C = 16 °C

energy transferred = mass × specific heat capacity × change in temperature

= 0.5 × 3930 × 16

= 31 440 J/3.1 × 10⁴ J

1

1

AO2

01.2

to keep the inside of the refrigerator cool/at 4 °C

by slowing the transfer of thermal energy from the outside to the inside of the refrigerator

1

1

AO2

01.3

a low thermal conductivity

so that the energy transfers very slowly across the material

1

1

AO3

01.4

a refrigerator with a high efficiency rating has insulation material with a low thermal conductivity

because less electricity is used so less energy is wasted

accept reverse argument

1

AO3

02.1

the data are not continuous

the names are categoric

1

1

AO2

02.2

one in 1990

three in 2017

accept three times as many in 2017 than in 1990 for 2 marks

1

1

AO2

02.3

total fossil fuels in 1990 = 230 + 20 + 0 = 250

total fossil fuels in 2017 = 20 + 10 + 140 = 170

change = –80 (TWh)

1

1

1

AO2

02.4

coal

plausible reason e.g., coal more expensive/less available/too polluting/causes global warming or greenhouse gases

1

AO2

03.1

the cost of production of solar cells/photovoltaic cells is very high

1

AO1

03.2

the cheapest method is coal

which produces the highest mass of CO₂ per unit

CO₂ is a greenhouse gas/contributes to climate change/global warming

1

1

1

AO2

AO3

4.1.3

03.3

two from:

• there are other considerations such as nuclear fuel produces radioactive waste
• nuclear accidents cause radioactive material to be released
• which could have a significant impact on the environment

one for each correct answer up to a maximum of two marks

2

AO3

03.4

(biomass involves) growing plants

plants take in CO₂ from the atmosphere

which would lower the concentration of CO₂/reduce the greenhouse effect/effects of climate change/idea of carbon neutral

1

1

1

AO2

04.1

points should be plotted at:

(0,0), (2,0), (4,0.1), (6, 1.2), (8, 2.4), (10, 3.6), (12, 3.6), (14, 3.6)

one mark for correct variable on the x-axis and y-axis

one mark for appropriate scale on the x-axis and y-axis

one mark for three or four points of data plotted correctly

two marks for all data points plotted correctly

one mark for drawing a line of best fit

5

AO2

04.2

for small wind speeds the output is zero

as the wind speed increases, the output power increases

for a wind speed over 10 m/s, the power output doesn’t change/is constant

1

1

1

AO2

04.3

advantage – no greenhouse gases produced while it is in use/renewable resources/can be used in remote places

disadvantage – wind speed is variable/wind doesn’t always blow/needs a large space/noisy

1

1

AO1

05.1

independent – number of sheets of transparent film

dependent – energy per second

1

1

AO2

05.2

three from:

• the distance of the lamp from the solar cell
• the angle of the solar cell
• the type/thickness of transparent film
• the type/area of solar cell

one mark for each correct answer up to a maximum of three marks

3

AO3

05.3

statement one: do not incorporate 2.11 into the calculation of the mean/change mean to 3.65/repeat test

statement two: add units to columns 2/3 and change mean to mean energy per second (J/s)

statement one: change all the measurements to the same number of significant figures/three significant figures.

1

1

1

AO3

05.4

1

1

AO2

06.1

a renewable resource can be replenished as it is used, but a non-renewable resource cannot

1

AO1

4.1.3

06.2

one mark for each correct column

3

AO1

06.3

non-renewable resources are very reliable/can produce a more/steady supply of electricity

1

AO2

4.1.3

07.1

wind/wave/hydroelectric/geothermal/solar/biofuel

1

AO1

4.1.3

07.2

in 1990, the total kWh for these resources was 225 million kWh, out of a total of 250 million kWh

in 2015 there were 190 million kWh out of 225 million kWh

The percentage has decreased/so has the overall energy use

accept 230 million kWh, giving 92%

1

AO2

07.3

change in energy use in 5 years = 230 million kWh – 250 million kWh = -20 million kWh

current use = 230 million kWh hours, half of this is 115 million kWh

accept 4 with no working for one calculation mark

accept 29 with no working for one calculation mark

1

1

1

1

AO3

07.4

sensible suggestions:

• the energy use might halve in this time because people use more energy efficient devices/want to save money
• the energy use might not halve in this time because this is an estimate based on past data/you cannot be sure that the downward trend will continue / world population is increasing

1

1

AO3

08.1

annual energy required by the village = 7000 × 10⁶ × 60 min × 60 sec

= 2.52 × 10¹³ J (2.5 × 10¹³ to two significant figures)

1

1

AO2

08.2

energy generated by one turbine per year = 33 000 W × 60 sec × 60 min × 24 h × 365 days

1

1

1

1

AO2

08.3

25 × 1 million = £25 million

7000 MWh = 7 000 000 kWh

total cost = 7 000 000 × £0.50

= £3.5 million

biofuel is cheaper

1

1

1

1

AO3

08.4

Level 3: Both resources evaluated, with at least one advantage and disadvantage of both given.

5-6

AO3

Level 2: Both resources evaluated, but an advantage or disadvantage missing for one resource.

3-4

Level 1: Only one resource evaluated, or only advantages or disadvantages given.

1-2

No relevant comment.

0

Indicative content:

• both resources are renewable
• biofuels are reliable
• biofuels could be carbon neutral/carbon dioxide released by burning fuel (theoretically should) equal the carbon dioxide absorbed from the atmosphere by the living matter
• however, carbon dioxide also produced during the process to make and transport biofuels
• biofuel would contribute to climate change by producing CO₂.
• wind turbines can be noisy
• wind turbines are not reliable, as only produce electricity when it is windy
• wind power does not contribute to climate change/ no pollutant gases

09.1

two from:

• oil
• coal
• gas

one mark for each correct answer up to a maximum of two marks

2

AO1

09.2

suitable resource e.g., hydroelectric, tidal power

correct description e.g., water in a lake moves down a hill/mountain

through a generator that produces electricity

1

1

AO1

09.3

two comments e.g.,

tides happen regularly/twice a day

water can be released from a lake on demand

or

the height of tides varies

rainfall to fill the lake is variable

one mark for each correct comment up to a maximum of two marks

2

AO3

09.4

carbon dioxide is a greenhouse gas

it contributes to climate change/causes global warming

1

1

AO1

09.5

power station affects habitats of wildlife

accept any suitable comment

1

AO2

10.1

gravitational potential energy = mass × gravitational field strength × height

1

AO1

10.2

60 × 9.8 × 10

= 5880 J

1

1

AO2

10.3

(extension =) 10 – 3.2 = 6.8

5880 J = 0.5 × k × 6.8²

1

1

1

1

AO1

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

the potential difference of the mains electricity in the UK is: about 230 V

the frequency of mains electricity in the UK is: 50 Hz

the mains supply in the UK produces a current that is: alternating

1

1

1

AO1

01.2

earth or neutral

earth; neutral in any order

live; neutral in any order

1

2

2

AO1

01.3

if the casing on an appliance becomes live, the earth wire conducts the current safely to earth

do not accept ‘for safety’

do not accept ‘to protect the user’

1

AO1

02.1

power = potential difference × current or P = I x V

= 6 V × 1.5 A

(= 9 W)

1

1

AO1

02.2

\)

1

AO1

02.3

\)

accept 270 with no working for three marks

1

AO2

02.4

both devices transfer the same amount of energy

1

AO2

03.1

a fault can be caused by the live wire touching the case of the fan

the earth wire is connected to the case

providing a (low resistance) path to ‘earth’

so the current flows through the earth wire and not through the person touching the case

a large current flows, so the fuse melts so the current stops flowing

1

1

1

1

1

AO1

03.2

make the case of the fan out of plastic/non-conducting material

if the live wire touches the case the current will not travel through the case to the person

1

1

AO1

03.3

power = potential difference × current

= 230 × 4.5

= 1035 W

= 1000 W (to two significant figures)

accept 1000 W with no working for two marks

1

1

AO1

03.4

or

1

1

1

1

or

AO1

04.1

a transformer changes the potential difference/steps a potential difference up or down

1

AO1

04.2

Level 3: Detailed explanation of why the National Grid uses a higher potential difference. Calculations of the current in each wire and of power loss in each wire

5-6

AO3

Level 2: Explanation of why the National Grid uses a higher potential difference.

Calculation of the current in each wire or an attempt at calculation of power loss in each wire

3-4

Level 1: Explanation of why the National Grid uses a higher potential difference. Calculation of the current in each wire or an attempt at calculation of power loss in each wire.

1-2

No relevant comment.

0

Indicative content:

• transmitting power at a higher potential difference means that the current is smaller
• the wires have a resistance, so they will get hot
• so there is less energy transferred to the thermal store of the surroundings
• \(
  begin{array}{lr}
  rm{at a power of 80 times 10^6 W and a potential difference of 400 000V, the current in the wire is: current } =
  frac{rm{power}}{rm{potential difference}}=frac{80 times 10^6}{400 000}= 200 rm{A}
  end{array}
  \)
• power loss is: P = I² x R = 200² x 4 = 1.6×10⁵ W
• \(
  begin{array}{lr}
  rm{at a power of 80 times 10^6 W and a potential difference of 4000 V the current in the wire is: current} =
  frac{rm{power}}{rm{potential difference}}= frac{80 times 10^6}{4000}= 20000 rm{A}
  end{array}
  \)

power loss is: P = I² x R = (20 000)² x 4 = 1.6×10⁹ W

05.1

power = current × potential difference

accept symbol equation P = V × I

1

AO1

05.2

potential difference of the mains = 230 V

power = 2 kW = 2000 W

2000 = current × 230

accept 8.2 with no working for the four calculation marks

1

1

1

1

1

1

AO1

05.3

potential difference = current × resistance

accept V = IR

1

AO1

05.4

230 = 8.69 × resistance

\)

= 26.47 Ω

= 26 Ω (to two significant figures)

or

power = current² × resistance

2000 = (8.69)² × resistance

= 26.48 Ω

= 26 Ω (to two significant figures)

allow 26.4 = 26 Ω if 8.7A used

allow 26.4 = 26 Ω if 8.7A used

1

AO2

05.5

energy transferred by kettle = power time

accept E = Pt

1

AO1

05.6

E = 2000 × 2 × 60

= 240 000 J

for toaster:

240 000 = 1200 × time

= 200 seconds

= 3 minutes 20 seconds.

1

1

05.7

yes

they both transfer energy from a chemical/nuclear energy store (in the power station) to the thermal energy store (of the surroundings)

by an electric current

no marks for ‘yes’

1

1

AO1

06.1

2000 W means 2000 joules of energy are transferred per second/unit time

1

AO1

06.2

230 V means 230 joules of energy are transferred by each coulomb of charge (that flows in the circuit)

1

AO1

06.3

energy = power × time

accept symbol equation E = P × t

1

AO1

06.4

E = 2000 x 5 x 60

= 600 000 J

accept 600 000 with no working for two marks

1

1

AO2

06.5

energy = charge × potential difference

accept symbol equation E = Q × V

1

AO1

4.2.4.2

06.6

accept 2609 with no working for three marks

1

AO2

4.2.4.2

07.1

1

evidence of metal rod as independent variable in the first column

evidence of repeat readings

evidence of calculation of mean

1

AO2

07.2

two from:

• distance of a nail from end of rod
• Bunsen burner, type/air hole position/position on rod
• amount of wax on nail
• size/material/mass of nail
• initial temperature of nail

one mark for each correct point up to a maximum of two marks

2

AO3

07.3

the control variables are difficult to control

leading to a big uncertainty in the data produced

1

1

AO3

07.4

use a different method for working out when the end of the rod has gotten hot e.g., thermal paint, temperature sensor attached to rod

1

AO3

08.1

230 V for all appliances

this is the potential difference of the mains

1

1

AO1

08.2

thermal (energy store)

1

AO1

08.3

iron→ hairdryer → toaster

1

1

AO2

08.4

2000 = 8.7² resistance

= 26 Ω (to two significant figures)

1

AO2

09.1

the energy from the Sun will not run out (in the immediate future)

1

AO1

09.2

4 × 10²⁶ W

1

AO2

09.3

energy per year = power × time

= 500 × 3.1 × 10⁷

= 1.55 × 10¹⁰ J

1

1

AO2

09.4

1

AO2

10.1

(E_e = 0.5ke²)

= 0.5 × 500 × 0.01²

= 0.025 J

accept 0.025 (J) with no working

1

1

AO2

10.2

at the top of the first bounce there is more energy in the gravitational potential energy store

at the top of the second bounce the energy has been transferred to a gravitational potential energy store and the thermal energy store of the surroundings

there is less energy in the gravitational potential energy store, so the second bounce is not so high

1

1

1

AO1

10.3

energy is transferred by forces/mechanically

and by heating

1

1

AO1

11.1

the National Grid

1

AO1

11.2

both transformers change the potential difference

the transformer near the power station/transformer 1 is a step-up transformer/increases the potential difference

the transformer near the house/transformer 2 is a step-down transformer/decreases the potential difference

1

1

1

AO1

11.3

(the energy is transferred at a high potential difference so) the current is small

so the energy/power/heat lost is small

1

1

AO1

4.2.4.3

12.1

power = potential difference current

9000 = 230 current

1

1

AO1

12.2

power = potential difference × current

2000 = 230 × current

the current in the oven is over 4 times bigger

1

1

1

AO1

12.3

the current is very large, so the heating effect is very big

the wire needs to be thicker so that there is less resistance and so less heating in the wire/the wire does not melt

1

AO2

12.4

there is an earth wire connected to the casing of an appliance

through which current flows if the casing becomes live/connected to the live wire

1

1

AO1

13.1

as the light intensity increases, the resistance decreases

at a decreasing rate

1

1

AO2

13.2

the resistance of B also decreases with increased light intensity

but after light intensity reaches 30 lux, resistance of B is constant

whereas the resistance of A continues to go down

1

1

AO3

13.3

when the light level is 13 lux, the resistance of A is about 40 kΩ

either:

the total resistance = (40 × 10³) + (100 × 10³) = 140 × 10³ Ω

so potential difference across the light dependent resistor

V = IR = (4.28 × 10^–5) × (40 × 10³) = 1.7 V

so the potential difference across the resistor V_out = 6 – 1.7 = 4.3 V

or:

so the potential difference across the resistor V_out = 6 – 1.7 = 4.3 V

1

either

1

1

1

or

1

1

1

AO1

13.4

light dependent resistor A

for light dependent resistor B, the lowest value of the resistance of light dependent resistor B is 40 kΩ

so the lowest potential difference across the light dependent resistor is 1.7 V

so the maximum potential difference of V_out is 4.3 V (because they add up to 6 V)

to get a potential difference that is higher than 5 V, you need the potential difference across the light dependent resistor to be lower (than 1 V)

so you need a light dependent resistor that has a resistance that can go lower than 40 kΩ

1

AO1

14

Level 3: Calculations of power generated in each bulb in each circuit.

The link between power and brightness is explicit and the statement about brightness is correct.

synoptic question involving ideas from the previous chapter

5-6

AO1

Level 2: Calculations of currents in circuits and a comment linking current and resistance to power. Statements about brightness that follow from previous reasoning.

3-4

Level 1: Recognition that power depends on current and potential difference/resistance. A general statement that bulbs are brighter in parallel/less bright in series

1-2

No relevant comment.

0

Indicative content:

• the brightness of the bulb depends on the power generated in it, P = I²R
• in series, the current through each bulb is the same
• \(
  rm{current} = frac {rm{potential difference}}{rm{total resistance}}=frac{12}{15}= 0.8 rm{A}
  \)
• the 10 Ω lamp will have a power of 0.8² × 10 = 6.4 W
• the 5 Ω lamp will have a power of 0.8² × 5 = 3.2 W
• the 10 Ω lamp will be brighter.
• in parallel the potential difference across each bulb is the same, 12 V
• \(
  begin{array}{lr}
  rm{current through the 5 Omega lamp} =
  frac{rm{potential difference}}{rm{total resistance}}\
  = frac{12}{5}
  = 2.4 rm{A}
  end{array}
  \)
• current through the 10 Ω lamp will be 1.2 A (as the resistance is double, current will be half)
• the 10 Ω lamp will have a power of 1.2² × 10 = 14.4 W
• the 5 Ω lamp will have a power of 2.4² × 5 = 28.8 W
• the 5 Ω lamp will be brighter

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

A should be in the uppermost circle

V should be in the right hand circle

The resistor is between the cell and the voltmeter

The variable resistor is the lower-most circuit component

one mark for ‘A’ in correct place

one mark for ‘V’ in correct place

one mark for resistor labelled correctly

one mark for variable resistor labelled correctly

1

1

1

1

AO1

01.2

the rate of flow of charge/charge flowing per second

1

AO1

01.3

1

AO1

01.4

accept 20 with no working for two calculation marks

1

1

1

AO2

02.1

circuits A and C are parallel circuits

no marks if more than one box ticked

1

AO1

02.2

A, C

B, D

B, D

in all cases, both letters required for the mark

letters can be in either order

1

AO2

02.3

no

the bulbs will be the same brightness

because they are in a series circuit

1

1

1

AO2

03

Level 3: Correct diagrams with description of measurements (total current and potential difference) to be taken in each circuit. Rearrangement of equation to give an equation for resistance. Correct statement about relative magnitudes of equivalent resistances.

5-6

AO1

Level 2: Diagrams or description of measurements (total current and potential difference) to be taken in each circuit lacking one or two details

Evidence of use of equation involving current, potential difference, and resistance. Correct statement about relative magnitudes of equivalent resistances.

3-4

Level 1: One correct diagram. Either potential difference or current measurement mentioned. Little or no evidence of use of equation. Little or no statement about the relative magnitudes of the equivalent resistances.

1-2

No relevant comment.

0

Indicative content:

• two correct circuits drawn with an ammeter in each circuit in an appropriate place to measure the total current, with or without a voltmeter across the battery
• one circuit should be a parallel circuit with a bulb in each circuit, with an ammeter in the circuit closest to the cell
• the other circuit should be a single circuit with two bulbs and one ammeter
• you need to measure the total current in each circuit, and the potential difference of the supply.
• you use the equation potential difference = current × resistance
• \(textrm{rearrange to give resistance }=frac{rm{potential difference}}{rm{current}}textrm{ to calculate the equivalent resistance of each circuit}
  \)

the equivalent resistance of the series circuit is bigger than the equivalent resistance of the parallel circuit

04.1

as the temperature increases, the resistance decreases

1

AO1

04.2

the reading on the voltmeter will not change

it is connected directly across the battery/there is only one component (other than the battery)

1

AO2

04.3

the potential differences across resistors in a series circuit are in the same proportion as the size of the resistances

\(textrm{there is 8 V out of the total 12 V across the thermistor, which is}frac{2}{3}
\)
of the total

so the resistance of the thermistor is 2 × 10 = 20 kΩ

this happens when the resistance is very cold

so the resistance of the thermistor is 3.3 kΩ

this happens when the resistance is very hot

1

1

AO2

05.1

potential difference = 6 – 4

= 2 V

accept 2 V with no working for two marks

1

1

AO2

4.2.2

05.2

potential difference = current × resistance or V = IR

accept 20 with no working shown for three marks

1

1

1

AO1

05.3

or

so resistance of resistor = 30 – 20 = 10 Ω

1

AO2

05.4

decrease

adding a resistor in series increases the total resistance of the circuit so less current will flow

1

1

AO2

06.1

component A

1

AO2

06.2

evidence of reading current and potential difference from the graph e.g., current = 0.5 A, potential difference = 5.0 V

accept any correct pair of readings

accept ten with no working shown for three marks

1

1

1

AO1

06.3

as the potential difference increases the resistance of component A stays the same

the ratio of all the potential difference and current readings stays the same = 10 (Ω)

as the potential difference increases, the resistance of component B increases

do not accept any answer involving gradients of the lines

1

1

1

1

AO2

06.4

at 3.0 V, total current = 0.3 A + 0.4 A

= 0.7 A

1

1

AO2

06.5

accept 4.3 with no working shown for two marks

1

07.1

negative

1

AO2

4.2.5.2

07.2

an arrow in a direction away from the sphere

1

AO2

4.2.5.2

07.3

(the force) decreases/gets smaller

1

AO2

07.4

charge = current × time or Q = It

student must show working for full marks

1

AO1

08.1

electron(s)

1

AO1

08.2

negatively charged particles (electrons) are transferred from the jumper to the balloon

which leave a (net) negative charge on the balloon

1

1

AO2

08.3

yes

because opposite charges attract

and an attractive force keeps the balloon on the wall

or

no

the wall is neutral

but when the balloon is close to the wall it repels the electrons which leaves a net positive charge on the surface

all marks for this question are awarded for explanations, rather than for the ‘yes’/’no’ answer

1

1

AO2

4.2.5.2

08.4

the balloons will repel

they have the same charge on them/both are negatively charged

like/similar charges repel

1

1

1

AO1

09.1

charge = current × time or Q = It

accept 40 with no working shown for three marks

accept 0.04 (not converting from mA) for one mark

1

1

1

AO1

09.2

graph should show a smooth curve with resistance increasing sharply initially before beginning to gradually plateau

One mark for a graph labelled mass on x-axis, resistance on y-axis

one mark for a curved shape with decreasing gradient

2

AO3

09.3

as the mass of salt increases the current increases

the potential difference is constant (6 V)

the current increases at a decreasing rate, so the resistance will decrease at a decreasing rate

from 25 g to 30 g the resistance decreases from 400 Ω to 240 Ω

from 75 g to 80 g the resistance decreases from 150 Ω to 146 Ω

1

AO2

10.1

graph should show that resistance initially drops sharply, but begins to gradually plateau as light intensity increases

the curve should be a smooth arch showing a negative correlation

one mark for plotting light intensity on x-axis and resistance on y-axis

one mark for correct units

one mark for correct shape of graph

3

AO2

10.2

sensible suggestion e.g., street lights, security lights

1

AO2

10.3

description of how it is used

appropriate circuit diagram

explanation of why it is needed

for example:

turning the lights on in a house when it gets dark outside

connect up the light dependent resistor in a circuit with an LDR, a resistor and a battery

circuit diagram with labelled components

use the output potential difference across the LDR or the resistor to switch on the lights

as the light level changes, the changing resistance produces a changing potential difference

which can be used to turn on the lights when the potential difference reaches a certain level

2

1

2

AO2

11.1

points should be plotted at the following co-ordinates:

(5, 1.5), (10, 3.8), (15, 4.6), (20, 5.9), (25, 7.8)

line of best fit should be straight

three or four points plotted correctly for one mark

all points plotted correctly for two marks

one mark for appropriate scales on correctly labelled axes

one mark for plotting an appropriate line of best fit

4

AO2

11.2

independent variable = length

dependent variable = resistance

control variable = type of metal/diameter of wire/temperature of wire

1

1

AO3

11.3

6.8 (Ω)

accept values between 6.5 and 7.3 (Ω)

1

AO2

11.4

take repeat measurements and calculate/plot the average/mean of repeat measurements

1

AO3

11.5

both students A and B are correct

the line is straight (so is linear)

and foes through (0,0) (so it is directly proportional)

1

1

1

AO3

12.1

energy in the kinetic energy store

is transferred to a gravitational potential energy store as the ball moves upwards.

as the ball moves through the air some energy is transferred to the thermal energy store of the surroundings

as the ball moves down energy is transferred from a gravitational potential energy store to a kinetic energy store

some of this energy is transferred to the thermal energy store of the hand

1

1

1

AO2

12.2

gravitational potential energy = mass × gravity × (change in) height

allow gpe = mgh

1

AO1

12.3

0.4 = 0.1 × 9.8 × change in height

accept 0.4 with no working shown for three marks

1

1

1

AO2

13.1

accept 40% with no working shown for three marks

accept 0.4 for 1 mark

1

AO1

13.2

efficiency of vehicle B is 2 × 40% = 80% or 0.8

accept 7500000 J with no working for three marks

1

1

1

AO2

13.3

accept 833 with no working shown for two marks

1

1

1

1

AO1

14.1

A – the graph for a resistor with a small resistance

B – the graph for a resistor with a large resistance

C – the graph for a filament lamp

one mark for two lines correct

two marks for all lines correct

2

AO1

14.2

while potential difference is negative, current should be 0

once potential difference becomes positive, current should increase sharply

one mark for current with positive potential difference

one mark for approximately zero current with negative potential difference

2

AO1

14.3

none of them will light up

light emitting diode X is in the reverse direction/cannot allow current from positive to negative terminals of the battery.

1

1

AO2

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

volume before = 72 cm³

volume after = 97 cm³

difference in volume = volume of clay

= 97 cm³ – 72 cm³

=15 cm³

1

1

AO1

01.2

1

1

AO1

01.3

digital balance

1

AO2

01.4

1

AO1

01.5

accept 1.56 or 1.6 with no working shown for two marks

1

1

AO2

01.6

measure the length of each side (in cm)/measure the length, breadth and height

cube the answer/multiply the length, breadth and height

1

1

AO1

4.3.1.1

02.1

vibrating

potential

moving fast

kinetic

1

1

1

1

AO1

02.2

the internal energy changes from mainly potential

to mainly kinetic/more kinetic

do not accept answers involving solids/liquids/gases

1

1

AO1

02.3

the particles in a gas are in random motion

1

AO1

4.3.2.1

03.1

20 °C

1

AO2

03.2

70 °C

1

AO2

03.3

energy transferred = power × time

allow E = P × t

1

AO1

03.4

energy = 1000 × 2 × 60

= 120 000 J

energy transferred = mass × specific latent heat of vaporisation

allow E = mL

1

1

1

AO2

03.5

an overestimate

(the energy transferred to vaporise the water is lower because) some energy is transferred to the thermal energy store of the surroundings

1

1

AO2

03.6

a line that starts at 20 °C

steeper than the original line

becomes horizontal at 70 °C at about 2 minutes

1

1

1

AO2

04.1

energy = power × time

allow E = P × t

1

AO1

04.2

2 kW = 2000 W

2 minutes = 120 seconds

energy = 2000 × 120

= 240 000 J

1

1

1

AO1

AO2

4.1.1.4

4.3.2.3

04.3

change in mass = 1.276 – 1.180 = 0.096 kg

energy transferred = specific latent heat × change in mass

240 000 J = specific latent heat of vaporisation × 0.096 kg

allow 2500 with no substitution for two marks

allow E = mL

1

1

1

1

1

AO1

AO2

04.4

the energy transferred was heating the material of the kettle/air around it as well as vaporising the water

a lower mass of water vaporised than should have been the case

so the value calculated was bigger than the textbook value

1

1

1

AO2

05.1

the gas has been compressed/is exerting a big pressure on the inside of the container

1

AO1

05.2

the student has done work on the gas (so has transferred energy)

so has increased the internal energy of the system (which produces an increase in temperature)

1

1

AO1/2

05.3

when the student does work, the particles move more quickly/energy is transferred to the particles

the average speed/kinetic energy of the particles increases, so the temperature increases

1

1

AO1

4.3.3.3

06.1

(when the student is in the shower) water evaporates to make water vapour

the water vapour condenses when it comes in contact with the colder mirror

and energy is transferred to the mirror

1

1

1

AO1

06.2

for converting all kJ to J, cm to m

1

1

1

1

1

1

AO1

AO2

4.3.1.1

4.3.2.3

07.1

the particles in a solid are arranges in a regular pattern/array

the particles in a gas are moving in all/random direction

the particles in a solid are vibrating about a fixed position

the particles in a gas are moving quickly

1

AO2

07.2

the particles collide with the walls of the container

each particle exerts a force on the wall

pressure is force per unit area

the total force of all the collisions of the particles per unit area is the pressure

1

AO3

07.3

pressure on y-axis, temperature on x-axis

straight line with positive gradient

that intercepts y-axis above zero (does not need to be extrapolated)

1

AO1

08.1

the steam is at a high temperature (and in the gas state)

steam transfers latent heat energy to the milk (steam cools/changes to a liquid state and the temperature of the milk rises)

1

1

AO1

08.2

convert 242 g to 0.242 kg, and 3.93 kJ/kg °C to 3930 J/kg °C

temperature difference = 70 – 20 = 50°C

energy required = mass × specific heat capacity × change in temperature

= 0.242 × 3930 × 50

= 47 553 J (
\(
\approx
\)
48 000 J so about 48 kJ)

allow E = m c 

allow 47533 with no substitution for three marks

1

1

AO1

08.3

convert 2260 kJ/kg to 2260 000 J/kg

energy = mass × specific latent heat of vaporisation

allow E= mL

1

AO2

08.4

Assume that all the energy transferred to the milk to heat it comes from the change of state of the steam.

1

AO2

09.1

B

D

the temperature isn’t changing/doesn’t change

even though the substance is being heated

1

AO1

4.3.1.2

09.2

solid

it changes state twice/goes from solid to liquid, then liquid to gas

1

1

AO1

09.3

A

1

AO1

4.3.1.2

09.4

C

E

1

1

AO1

4.3.1.2

10.1

volume = (5)³

= 125 cm³

1

1

AO2

10.2

\)

1

AO1

10.3

sublimation

1

AO1

10.4

if the process is reversed, the material recovers its original properties

1

A01

10.5

the internal energy of the gas is bigger than the internal energy of the solid

the particles have more kinetic energy/are moving faster

1

AO2

11.1

one mark for correct symbol

one mark for correct label

2

AO1

11.2

(when they put the test tube into iced water) the temperature of the water decreases

the resistance of the thermistor increases

the potential difference across the thermistor increases

V_out decreases

because the total potential difference across the thermistor and the resistor = 12 V at all times

1

1

1

1

1

AO1

11.3

find the potential difference across the resistor by measuring V_out

find potential difference across thermistor by subtracting V_out from 12 V

use
\(
\begin{array}{lr}
\frac{\rm{potential\ difference\ across\ the\ resistor}}{\rm{potential\ difference\ across\ the\ thermistor}}\\
= \frac{\rm{resistance\ of\ resistor}}{\rm{resistance\ of\ thermistor}}
\end{array}
\)
to find the resistance of the thermistor

use a graph/table of the resistance of the thermistor at different temperatures to work out the temperature of the water.

1

AO1

11.4

sensible suggestions e.g., the human body continually generates thermal energy, but the water in the test tube does not

1

AO3

12

Level 3: Well organised answer with descriptions of reasons for calculations. Equations for power, efficiency and gravitational potential energy used

5-6

AO1/1

Level 2: Some relevant calculations, but descriptions lacking detail or missing, or parts of calculations/conversions incorrect.

3-4

Level 1: Some relevant calculations completed, but unit conversions may be missing, and no explanation of method.

1-2

No relevant comment.

0

Indicative content:

• calculate energy needed to be transferred to bulb:
• energy = power × time

  = 0.24 × 60

  = 14.4 J

• energy transferred to motor must be greater than this because the generator is only 90% efficiency
• \(
  \rm{efficiency} = \frac{\rm{energy\ out}}{\rm{energy\ in}}\times\ 100
  \)
• \(
  90 = \frac{14.4}{\rm{energy\ in}}\times\ 100
  \)
• \(
  \rm{energy in} = \frac{14.4}{90}\times\ 100
  \)
• = 16 J
• this energy is transferred by the falling mass.
• gravitational potential energy = mass × gravity × height
• 16 = 0.3 × 9.8 × height
• height = 5.4 m

13.1

voltmeter

1

AO2

4.2.1.3

13.2

potential difference = current × resistance

allow V = IR

1

AO1

13.3

potential difference = 15 × 10

= 1.5 V

allow 1.5 with no substitution for two marks

1

1

AO1

13.4

charge = current × time

allow Q = It

1

AO1

13.5

charge = 0.15 × 60

= 9

C/coulombs

allow nine with no substitution for two marks

1

1

AO1

13.6

it would increase

the potential difference has increased (but the clock resistance is the same)

1

1

AO1

14.1

before the air is pumped out the pressure of the air in the marshmallow pockets = the pressure of the air outside the marshmallow

the particles of the air (inside and outside the marshmallow) collide with the (surface/material of the) marshmallow,

producing a force, force = pressure × area

initially the forces are the same

when the teacher pumps the air out, the force/pressure exerted by the particles of the air on the outside is smaller than the force exerted by the particles inside, (so the marshmallow expands)

1

1

1

AO2

14.2

initially: pressure × volume = 101 × 10³ × 8 × 10^–6

= 0.808 (Pa m³)

= final pressure × final volume

final pressure =
\(
\frac{{{0}{\rm{.808}}}}{{{2}{\rm{.7\ }} \times {\rm{\ 1}}{{0}^{{\rm{\ -\ 5}}}}}}
\)

= 2.99 × 10⁴ Pa

= 29.9 kPa.

change in pressure = 101 – 29.9

= 71.1 kPa (71 100 Pa)

1

1

1

1

1

1

AO1

4.3.3.2

Question

Answers

Extra information

Mark

AO / Specification reference

01.1

10

1

AO2

4.4.1.2

01.2

the number of electrons is the same as the number of protons

electrons and protons have opposite charges/charge cancels/atoms are neutral so have equal positive and negative charges

1

1

AO1

AO2

4.4.1.1

01.3

A and B

1

AO1

01.4

isotopes have the same number of protons

but different numbers of neutrons

1

1

AO1

02.1

particle B

there are the same number of those particles in each nucleus

1

1

AO1

4.4.1.2

02.2

particle A is a neutron

1

AO1

02.3

the element is lithium

it has an atomic number of three

1

1

AO1

02.4

one mark for correct top number/atomic mass in each case

one mark for correct bottom number/atomic number in each case

one mark for correct symbol (credit once only)

5

AO1

02.5

the same

the charge comes from/depends on the number of protons, which is the same

1

1

AO1

03.1

10^-10 m

1

AO1

03.2

an orbit/distance from the nucleus where there are atoms

1

AO1

03.3

when hydrogen absorbs electromagnetic radiation, the electron will move up a level

e.g., from level one to level two or three

when the electron moves back down it emits electromagnetic radiation

e.g., when it moves from three to two, or from two to one

1

1

1

1

AO1

04.1

looking at the atomic/bottom number/seven

subtracting the atomic/bottom number from the mass/top number

looking at the atomic/bottom number/seven/it is the same as the number of protons

1

1

AO1

04.2

no, there are atoms where the mass/top number/atomic mass is not double the atomic/bottom number/atomic number

suitable example e.g., Be, Na

1

1

AO1

AO1

4.4.1.2

05

Level 3: Description of the how the nuclear model came to be proposed, and the limitations of it. Comment about the refinement of the model when the proton and neutron were discovered.

5-6

AO1

Level 2: Description of the nuclear model and some discussion about how the structure of the nucleus wasn’t worked out until later.

3-4

Level 1: Some comment about the development of the model in terms of the alpha scattering experiment and/or discovery of particles in the nucleus.

1-2

No relevant comment.

0

Indicative content:

• the discovery of the electron was before the nuclear model was proposed
• Rutherford knew that electrons were negatively charged and that atoms were neutral
• so he knew that there had to be positive charge in the nucleus
• the results of the alpha particle experiment showed that there was a massive, small positively charged object at the centre of the atom
• and that most of the atom was empty space
• his model included a positive massive nucleus but did not include the names or number of particles in it

the model was refined when the proton and neutron were discovered and named

06.1

some energy from the gravitational potential energy store at the start is transferred to the thermal energy store of the surroundings when the ball bounces

so there is less energy in the gravitational potential energy store at the end of the bounce

1

AO1

06.2

points should be plotted at:

(20, 16), (40, 30), (60, 49), (80, 55), (100, 70)

one mark for correct variables on the x and y-axes

one mark for appropriate scales on the x and y-axes with labels and units

one mark for plotting three or four points of data correctly (within ± 0.5 squares)

two marks for plotting all points of data correctly

one mark for drawing an acceptable line of best fit

5

AO2

AO3

06.3

the anomalous results is for the 60 cm drop/49 cm bounce

1

AO3

06.4

yes/bounce height is proportional to the drop height

the line of best fit is a straight line through (0, 0)/origin

or as drop height doubles, the bounce height doubles

1

1

AO3

07.1

1

AO2

4.4.1.3

07.2

percentage going through = 100 – 0.14 = 99.86 %

1

AO2

4.4.1.3

07.3

the majority of particles went through the foil

so the part of the atom/nucleus deflecting alpha particles was very small.

1

1

AO3

4.4.1.3

07.4

positive (or 2+)

1

AO1

07.5

If the nucleus is negative the alpha particles would be attracted to the nucleus and there would be no particles repelled and no deflections greater than 90°

If it is positive, they would be repelled

and be deflected by small or larger angles

1

1

1

1

AO1

08.1

A, D, E, F

two marks for three or four correct

one mark for one or two correct

2

AO1

4.4.1.1

08.2

B

C

1

1

AO1

4.3.3.2

08.3

E

F

1

1

AO1

4.4.1.1

08.4

positive (2+)

1

AO3

4.4.1.2

08.5

the nucleus is smaller by a factor of

1

1

1

1

AO1

4.4.1.1

09.1

the plum pudding model is a model with a positively charged mass

with negatively charged electrons

embedded in it

1

1

1

AO1

09.2

the plum pudding model replaced the Dalton model, which was that atoms were tiny spheres that could not be divided

1

AO1

09.3

most alpha particles went straight through a gold foil

but some came back/were deflected through more than 90°

1

1

AO1

10.1

if the current in the circuit becomes too large the wire inside a fuse melts

which breaks the circuit (and helps to prevent injury)

1

1

AO1

10.2

it takes a certain amount of energy to raise the temperature of the metal sufficiently to melt the fuse wire

the energy required depends on the mass of the wire/thinner wires can only carry smaller currents

if the wire needs to melt at a higher current the mass needs to be bigger, so the wire needs to be thicker

1

1

AO1

10.3

mass = density × volume

1

AO1

10.4

0.5 cm = 5 × 10^-3 m

volume = length × area

= 5 × 10^-3 × 1×10^–6

= 5 × 10^-9

mass = 7000 × 5 × 10^-9

= 3.5 × 10^–5 kg

1

1

1

1

1

AO1

10.5

energy = mass × specific heat capacity × change in temperature

= 3.5 × 10^–5 × 230 × (687 – 20)

=5.37 J (5.4 J)

energy = mass × specific latent heat

= 3.5 × 10^–5 × 300 000

= 10.5 J

Total energy = 5.37 + 10.5 = 15.87 = 15.9 J

1

AO2

10.6

energy transferred = power × time

1

AO1

10.7

power = current² × resistance

1

AO1

10.8

15.87 = current² × 1.8 × 0.5

or

allow E = I²Rt

accept 4.2 with no working shown for two marks

1

1

1

or

1

1

1

AO2

11.1

one mark for battery, ammeter and component in series

one mark for voltmeter in parallel with the component

one mark for battery, ammeter and voltmeter correctly labelled

3

AO2

11.2

(connect up the circuit, and) measure the potential difference and current record the measurements

change the potential difference of the battery/add a variable resistor, and take another set of measurements

repeat for different values of potential difference

calculate the ratio of potential difference to current/plot a graph of potential difference against current.

if the ratio/gradient is constant and straight line goes through origin is constant, the component is ohmic

1

1

1

1

AO1

11.3

1

AO1

11.4

one mark for drawing the graph in third quadrant

one mark for curved lines

one mark for correct curvature

3

AO3

11.5

potential difference = current × resistance

1

AO1

4.2.1.3

11.6

AO1

4.2.1.3

11.7

the resistance changes by a factor of two (is not constant)/not a straight line through the origin

so the component is non-ohmic

1

AO3

12.1

greater than

greater than

same as

1

1

1

AO1

12.2

greater than

1

4.3.2.1

12.3

the specific heat capacity for hot water is the same as for cold water

1

4.3.2.2

12.4

the average speed and specific heat capacity will stay the same

because these quantities do not depend on the number of particles in the beaker.

the kinetic energy, potential energy and internal energy will be halved

because the number of particles in the beaker has halved.

1

1

AO1

13

Level 3: Clear and coherent description and comparison of fission and fusion.

5-6

AO1

Level 2: Processes of fission and fusion described but not compared.

3-4

Level 1: A relevant part or description of fission or fusion with no comparison.

1-2

No relevant comment.

0

Indicative content:

• fission is the splitting of large unstable nuclei
• into smaller nuclei
• fusion is the joining together of smaller nuclei
• to make larger nuclei
• fission processes can be started and maintained easily
• fission occurs at lower temperatures / fusion occurs at higher temperatures
• fission is a process that happens in nuclear power stations but there are no nuclear fusion power stations
• fission causes a chain reaction/fusion does not
• both release energy (more energy released in fusion)

14.1

a positively charged mass

with negatively charged electrons embedded in it

1

1

AO1

14.2

two correct points e.g.,

the electron was smaller/less massive than the atom

it showed that there were particles that were smaller than the atom

it was negatively charged showing that there was positive and negative charge in the atom

one mark for each correct point up to a maximum of two marks

2

AO1

14.3

all of the alpha particles would go through or all the alpha particles would come back

1

AO1

14.4

some of the alpha particle came back, but most went through

which could not be explained by the current model, so the model had to change

1

AO2

14.5

correct suggestion e.g.,

the scientists were using the results to develop an alternative model

the results were being checked by other scientists

1

AO2

Answers

Extra information

Mark

AO / Specification reference

01.1

beta and gamma radiation would not be stopped by the smoke

but alpha particles are

1

1

AO2

4.4.2.1

01.2

the atomic mass is 241 – 4

= 237

the atomic number is 95 – 2

= 93

1

AO2

01.3

the atomic number has changed/the number of protons has changed

every element has a different atomic number/no of protons

1

1

AO2

02.1

5600

1

AO3

02.2

evidence of using graph to find the time when the activity is 5.0 count/min

= 8000 years

1

AO2

02.3

no

made from wood from a tree that died about 8000 years ago (so is too young)

1

1

AO3

4.4.2.3

03.1

the activity of a sample is measured in Becquerels (Bq)

the activity of a sample is the number of decays recorded per second

1

1

AO1

4.4.2.1

03.2

Geiger counter/Geiger Muller tube/GM tube

1

AO1

03.3

one mark for each correct column

2

AO1

4.4.2.1

4.4.2.2

03.4

radioactive decay is a random process

as is throwing a die

or

you never know which nucleus will decay next

or know which face on the die will appear next

1

1

or

1

1

AO2

4.4.2.1

04.1

one mark for one or two correct

two marks for all correct

2

AO1

04.2

no

the radiation that is most ionising (is alpha)

(which) has the smallest range in air

1

1

AO2

04.3

the aluminium absorbs alpha and beta radiation (so the activity goes down)

gamma is not stopped by the aluminium (so you can still detect the gamma radiation)

1

1

AO2

4.4.2.1

05.1

1

1

1

AO2

05.2

1

AO2

05.3

sheep (eat the grass and) become contaminated

the radioactive material inside them decays/emits radiation/could cause cancers

1

1

AO3

4.4.2.2

06.1

the matches/heads of matches are like the (unstable) nuclei

the flames are like the neutrons that move between the unstable nuclei

when you light one match it sets off the rest of the matches, just like when one neutron is absorbed by an unstable nucleus it can start a chain reaction

1

1

1

AO2

4.4.4.1

06.2

the strengths – any two from:

• when you light one match the rest light
• the matches need to be close enough together, just like the nuclei in a bomb/reactor
• if the matches are too far apart the reaction stops, just like if there are not enough nuclei

the weaknesses – any two from:

• the nuclei undergo fission, they do not burn
• particles (neutrons) are transferred between the nuclei, not flame/energy
• the matches continue to burn but nuclei undergo fission once.

one for each correct answer, up to a maximum of two for each of the strengths and weaknesses

4

AO1

07.1

food/drink

1

AO1

07.2

space/the Sun

1

AO1

07.3

any sensible suggestion e.g., more medical x-rays/air travel/depends where you live

1

AO2

07.4

(% from radon gas = 48%)

dose = 0.48 × 2.7

= 1.3 (1.296) mSv

1

1

AO2

07.5

a bar chart/pie chart

the data are categoric/not continuous

1

1

AO2

08.1

beta

1

AO2

08.2

beta particle

high speed/fast moving electron

no marks for charge of -1/no mass

1

1

AO2

4.4.2.2

08.3

neutron converted into proton

neutrons and protons have the same mass

1

1

AO2

4.4.2.1

4.4.2.2

08.4

83 protons before X emitted, 84 after (as a neutron changes to a proton)

the neutron has no charge, and a proton has a charge of +1 (so the charge increases by 1)

1

1

AO2

4.4.2.2

09

Level 3: Correct choice, with clear reasons to do with half-life/radiation transmission through tissue. Clear consequences with discussion of contamination /ionisation and their effects

5-6

AO3

Level 2: Correct choice with reason based on half-life. Some discussion of problems with longer half-life isotopes.

3-4

Level 1: Correct choice with little or no reasoning.

1-2

No relevant comment.

0

Indicative content:

• for a tracer you need an isotope with a short half-life (6 hours) so that it doesn’t emit too much radiation while inside the body and
• which emits gamma radiation that can be detected by the camera
• gamma radiation can travel through the body, but beta cannot
• so technetium-99 is the correct choice
• technetium-99 emits the wrong type of radiation to be detected by the camera
• technetium-95 and technetium-99 have longer half lives
• so the material would continue to cause damage due to contamination for a long time after the injection

10.1

line of best fit should be curved

1

AO2

10.2

two points chosen from count rate and time worked out from them, e.g. 60

evidence of use of graph

two marks for halving initial count rate and reading time

one mark for use of two values

one mark for correctly reading times and subtracting, e.g., 90 – 30

ignore any units written

2

1

AO3

4.4.2.3

10.3

no

radioactive decay is random, so there will be a scatter of points on the graph

1

1

AO2

4.4.2.3

10.4

they have not put unit labels on the axes of the graph

1

AO2

4.4.2.3

11.1

contamination means the presence of radioactive material inside of n the human body

irradiation is the process of exposing the strawberries to radiation

1

1

AO1

11.2

you are not taking radioactive material into your body (so there is little or no increase in risk due to ionizing radiation)

accept the radiation passes through the strawberry, it does not stay inside it, so you are not ingesting any radioactive material

1

AO3

11.3

scientists do experiments and collect data/draw conclusions

their results are checked by other scientists in a process called ‘peer review’

appropriate reason, e.g.,

• the regulations deal with radioactive material which can be harmful
• the hazards of radioactive material to the human body are significant
• people could be harmed if the data is not correct

1

1

1

AO2

12.1

temperature is proportional to time until the temperature reaches 97 °C

energy transferred by the heater is increasing the internal energy by increasing the kinetic energy of the particles

when the temperature reaches 98.5 °C the energy transferred by the heater increases the internal energy by increasing the potential energy of the particles

the temperature is constant but the liquid evaporates

1

1

1

1

AO3

12.2

energy is proportional to change in temperature while the liquid is heating

above 100 °C the increase in energy produces no further change in temperature

1

1

AO3

4.1.1.3

12.3

change in thermal energy = mass × specific heat capacity × temperature change

1

1

1

1

AO1

AO2

4.1.1.3

12.4

largest possible range reduces percentage error

accept average the error over larger value

1

AO3

4.1.1.3

13.1

resistor

1

AO1

4.2.2

13.2

in circuit one, the voltmeter reads 3 V

in circuit two, the voltmeter reads 6 V

in circuit one, the potential difference is split between the components/resistors

in circuit two, the voltmeter is connected directly across the battery

1

1

1

1

AO2

4.2.2

13.3

there would be no change to the readings

the potential difference would still be split in half if the bulbs are identical in circuit one

the potential difference is still connected across the battery in circuit one

1

1

AO3

4.2.2

14.1

solar energy is renewable

because it will not run out

1

1

AO1

14.2

the system of transformers and cables linking power stations to consumers

1

AO1

4.2.4.3

14.3

09:35 to 17:15

1

AO2