Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
Energy Concepts
Conservation of energy
- 3 things to know
- Component 1
Overview
Principle of conservation of energy, three mechanical energy stores considered in this topic and closed system in energy calculations.
What you need to know
Answer each one before you reveal it.
State the principle of conservation of energy.
Name the three mechanical energy stores
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the principle of conservation of energy.
Model answer
Energy can be transferred from one form to another but it cannot be created or destroyed, so the total energy of a closed system stays constant.
- 2
Name the three mechanical energy stores considered in this topic.
Model answer
The three are gravitational potential energy, elastic potential energy and kinetic energy.
- 3
A ball of mass 0.50 kg is dropped from rest through 5.00 m. Calculate its speed just before it lands, ignoring air resistance. Take g as 9.81 m s⁻².
Model answer
Equating mgh to ½mv² gives v = √(2 × 9.81 × 5.00) = 9.90 m s⁻¹ downwards. Substituting back, ½ × 0.50 × 9.90² = 24.5 J, which equals 0.50 × 9.81 × 5.00 J.
- 4
Why does the mass cancel when the landing speed of a freely falling body is found by energy conservation?
Model answer
Both the potential energy lost and the kinetic energy gained are proportional to the mass, so it appears on both sides of the equation and divides out.
- 5
A roller coaster car starts from rest and descends 20.0 m. Calculate its speed at the bottom if friction is neglected. Take g as 9.81 m s⁻².
Model answer
Using v = √(2gh) gives √(2 × 9.81 × 20.0) = 19.8 m s⁻¹. Substituting back, 19.8²/(2 × 9.81) = 20.0 m.
- 6
A pendulum bob is released from a height of 0.20 m above its lowest point. Calculate its maximum speed. Take g as 9.81 m s⁻².
Model answer
The speed at the lowest point is √(2 × 9.81 × 0.20) = 2.0 m s⁻¹. Substituting back, 1.98²/(2 × 9.81) = 0.20 m.
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