Oxford Revise AQA GCSE Physics | Chapter P1 answers

P1: Energy stores and transfers

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AO / Specification reference

01.1

chemical store

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

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AO2

4.1.1.1

01.2

\({\rm{elastic\ strain\ energy\ }} = \frac{1}{2}{\rm{k}}{\rm{e}^2}\)

= 0.5 × 20 × (0.2)2

= 0.4 (J)

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AO2

4.1.1.2

02.1

created or destroyed

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AO1

4.1.2.1

02.2

there is no net changed to the total energy

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AO1

4.1.2.1

02.3

is not

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AO1

4.1.2.1

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

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AO2

4.1.2.1

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

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AO1

4.1.1.2

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

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AO2

4.1.1.2

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

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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

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AO3

4.1.2.1

04.1

kinetic energy = 0.5 × mass × (speed)2

allow \({{\rm{E}}_{\rm{k}}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^{2}}\)

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AO1

4.1.1.2

04.2

kinetic energy = 0.5 × 40 × (10) 2

= 2000 J

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AO2

4.1.1.2

04.3

elastic potential energy = 0.5 × spring constant × extension2

2000 = 0.5 × 20 000 × e2

\({{\rm{e}}^{2}} = \frac{2000}{{{0}{\rm{.5}} \times {2000}}}\)

e2 = 0.2

e = 0.45 m

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AO2

4.1.1.2

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

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AO3

4.1.1.1

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

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AO1

AO2

4.1.1.1

4.1.1.2

05.2

elastic potential energy = \(\frac{1}{2}\) × spring constant × extension2

= 0.5 × 105 × (0.05)2

= 125 J

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AO2

4.1.1.2

05.3

video analysis

accept any sensible suggestion

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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

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AO2

AO3

4.1.1.2

05.5

the thermal energy of the surroundings

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AO1

4.1.1.1

06.1

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

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AO2

4.1.2.1

06.2

kinetic energy = 0.5 × mass × (speed)2

accept \({{\rm{E}}_{\rm{k}}} = \frac{1}{2}{\rm{m}}{{\rm{v}}^{2}}\)

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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

4.1.1.2

06.4

2 840 000 kJ

accept 2 835 000 000 (J)

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AO2

4.1.2.1

07.1

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

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AO1

AO2

4.1.1.1

07.2

work done by (gravitational) forces

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AO2

4.1.1.1

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

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AO3

4.1.1.1

4.1.1.2

08.1

gravitational potential energy = mass × gravitational field strength × height

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AO1

08.2

height = 4 floors = 4 × 3 = 12 m

gpe = mgh

= 1220 × 9.8 × 12 = 143 472 J

= 145 000 J

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AO2

4.1.1.2

08.3

\({\rm{efficiency\ }} = \frac{{{\rm{useful\ output\ energy\ transfer}}}}{{{\rm{total\ input\ energy\ transfer}}}}\)

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AO1

4.1.2.2

08.4

280 kJ = 280 000 J

efficiency = 143 472 × \(\frac{100}{{{\rm{280\ 000}}}}\)

= 51(.2)(%)

accept 51(%) with no working for three marks

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AO1

AO2

4.1.2.2

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

\({\rm{efficiency\ }} = \frac{{{\rm{useful\ output\ energy\ transfer}}}}{{{\rm{total\ input\ energy\ transfer}}}}\)

or

calculate the gravitational potential energy as before

calculate the useful power using power = \(\frac{{{\rm{energy}}}}{{{\rm{time}}}}\)

calculate efficiency using

\({\rm{efficiency\ }} = \frac{{{\rm{useful\ output\ energy\ transfer}}}}{{{\rm{total\ input\ energy\ transfer}}}}\)

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AO2

4.1.2.2

08.6

one from:

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

accept any sensible suggestion

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AO3

09.1

\(\textrm{power} = \frac{{{\rm{energy\ transferred}}}}{{{\rm{time}}}}\)

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AO1

4.1.1.4

09.2

\(15\ 000 = \frac{{{\rm{30\ 000}}}}{{{\rm{time}}}}\) \(\textrm{time} = \frac{30\ 000}{15\ 000}\)

= 2 (seconds)

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AO2

4.1.1.4

09.3

the second truck is less powerful than the first truck

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AO2

4.1.1.4

09.4

gravitational potential energy store

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AO1

4.1.1.1

10.1

energy is transferred from the gravitational potential energy store

to the kinetic energy store

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AO1

AO2

4.1.1.1

4.1.1.2

10.2

light gate

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AO1

4.5.6.1.1

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

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AO1

AO2

AO3

4.5.6.1.2

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

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AO3

4.1.1.1

4.1.1.2

4.5.6.1.2

11.1

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

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AO1

4.1.2.1

11.2

\({\textrm{efficiency}} = \frac{{{\rm{useful\ output\ energy\ transfer}}}}{{{\rm{total\ input\ energy\ transfer}}}}\)

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AO1

11.3

\(\textrm{efficiency} = \frac{12}{20}\)

= 0.6

accept 0.6 with no working for two marks

60% scores one mark

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AO2

4.1.2.2

11.4

car B

it has a lower efficiency, so wastes more energy

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AO3

4.1.2.2

12.1

the trolley is moving too fast

light gates/motion sensor

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AO1

12.2

gpe = mgh

Also accept:

gpe = mass × gravitational field strength × height

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AO1

12.3

gpe = 0.25 × 9.8 × 0.12

= 0.298 J

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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

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AO2

4.1.2.1

13.1

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

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AO3

4.1.1.2

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

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AO2

4.1.1.2

13.3

car B transfers less energy to the surroundings

because it rebounds/does not stop

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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%

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AO3

4.1.1.2

14.1

petrol/chemical energy (store)

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AO2

4.1.1.1

14.2

kinetic energy store

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AO2

4.1.1.1

14.3

work done by

force of friction/drag

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AO2

4.1.1.1

14.4

motorcycle: the efficiency decreases with speed

at a decreasing rate

car: the efficiency decreases with speed

at a constant rate

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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

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AO3

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