1. This document contains sample questions and answers about energy transfers and transformations. Questions cover topics like energy transfers in a fan heater, kinetic and potential energy changes in falling objects, and calculating work, power and efficiency in simple machines. 2. The answers provide calculations and explanations for how energy is transferred or transformed in each situation. For example, kinetic energy of a moving trolley is explained to be lost to friction rather than increasing its speed. 3. Key energy terms like kinetic energy, gravitational potential energy, work, and efficiency are applied to quantitatively analyze examples involving energy changes in physical systems.

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This resource sheet may have been changed from the original. 1
P1 Summary questionsStudent Book answers
Question
number
Answer Marks Guidance
1 a energy transferred by electric current to heater and motor
in heater, increases store of thermal energy so becomes hot
in fan motor, transferred to air as kinetic energy
air warmed by heater heats objects and material it comes into contact
with
heater warms fan heater case and vibrating parts of heater create
sound waves
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1 b i energy transferred where wanted in way wanted
in fan heater, transferred by current to heater transferred heat air
creating convection currents and keeping motor turning to make air
move
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1
1 b ii energy transferred that is not useful
in fan heater, kinetic energy from vibrations of motor transferred as
sound waves
and by heating fan heater case
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2 a i loses gravitational potential energy and gains kinetic energy
some energy lost due to air resistance
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2 a ii loses more gravitational potential energy and all kinetic energy
energy lost mostly transferred as elastic energy into rope
some energy may be lost due to air resistance
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2 b maximum kinetic energy when rope taut
if 100% efficiency, maximum kinetic energy = 12 000 J
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1 Must have unit to gain mark.
3 a i 7000 kW × 300 s = 210 MJ 1 Must have unit to gain mark.
3 a ii 23 m/s × 300 s = 6900 m 1 Must have unit to gain mark.
3 a iii
6900
000000210
= 30 435
≈ 30 000 N
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2. © Oxford University Press 2016 www.oxfordsecondary.co.uk/acknowledgements
This resource sheet may have been changed from the original. 2
P1 Summary questionsStudent Book answers
3 a iv a = zero so resultant force = zero
as driving force and resistive force must be equal and opposite in
direction
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3 b train gains gravitational potential energy as it travels up incline
rate it transfers energy to surroundings unchanged as speed same and
resistive forces unchanged
so output engine power needs to be greater as energy must be
transferred to train as gravitational potential energy as well as to
surroundings
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4 a mg = 150 N
EP = 150 N × 1.2 m = 180 J
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1
4 b W = 11 N × 20 m
= 220 J
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1
4 c trolley did not gain kinetic energy as speed constant
trolley gained 180 J gravitational potential energy which is 40 J < work
done by student
resistive forces such as friction at trolley wheels → 40 J to surroundings
as waste energy
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5 a energy initially stored in stretched cord as elastic potential energy →
arrow as kinetic energy when cord released
as arrow gains height its kinetic energy decreases and gravitational
potential energy increases
at maximum height arrow has maximum gravitational potential energy
and minimum kinetic energy
as it travels through air, some energy transferred to air due to air
resistance
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5 b i 0.5 × 0.015 kg × (25 m/s)2
= 4.7 J
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1
5 b ii 0.015 kg × 10 N/kg × 20 m
= 3.0 J
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3. © Oxford University Press 2016 www.oxfordsecondary.co.uk/acknowledgements
This resource sheet may have been changed from the original. 3
P1 Summary questionsStudent Book answers
5 b iii 4.7 J − 3.0 J
= 1.7 J
EK = ½mv2
= 1.7 J so ½ × 0.015 kg × v2
= 1.7 J
v2
=
1.7 𝐽𝐽
0.5 × 0.015 kg
= 227 m2
/s2
v = 15 m/s
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6 a EK = 0.5 × 75 kg × (60 m/s)2
= 135 kJ
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6 b i 0.5 × 75 kg × (5 m/s)2
= 940 J
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1
6 b ii 75 kg × 10 N/kg × 900 m
= 675 kJ
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6 c W = (GPE + KE) at start – KE after = 675 kJ + 135 kJ − 940 J
= 809 kJ
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7 a i mgh = 500 N × 3.0 m
= 1500 J
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7 a ii energy supplied = 600 W × 25 s = 15 000 J
energy wasted = 15 000 J − 1500 J = 13 500 J
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7 b 1500 J/15 000 J
= 10%
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