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)
Mathematical Requirements (Appendix C)
Rearranging and solving physics equations
- 1 thing to know
- All three components
Overview
Recall questions on making any symbol the subject of the equations physics uses, substituting with consistent units, solving an equation containing a square or a square root, handling an equation with the unknown in more than one place, and checking by substituting back.
What you need to know
Answer each one before you reveal it.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Rearrange E = mc² to make the mass the subject.
Model answer
Dividing both sides by c² gives m = E/c². The mass is then in kg when E is in J and c in m s⁻¹.
- 2
An energy release of 3.0 × 10¹⁴ J is measured. Calculate the corresponding mass defect.
Model answer
m = E/c² = 3.0 × 10¹⁴/(3.00 × 10⁸)² = 3.0 × 10¹⁴/9.0 × 10¹⁶ = 3.3 × 10⁻³ kg. Working back, 3.33 × 10⁻³ × 9.0 × 10¹⁶ = 3.0 × 10¹⁴ J.
- 3
Rearrange λ = h/(mv) to make the speed the subject.
Model answer
Multiplying by v and dividing by mλ gives v = h/(mλ). The speed is in m s⁻¹ when h is in J s, m in kg and λ in m.
- 4
An electron has a de Broglie wavelength of 1.0 × 10⁻¹⁰ m. Calculate its speed, taking its mass as 9.11 × 10⁻³¹ kg.
Model answer
v = h/(mλ) = 6.63 × 10⁻³⁴/(9.11 × 10⁻³¹ × 1.0 × 10⁻¹⁰) = 6.63 × 10⁻³⁴/9.11 × 10⁻⁴¹ = 7.3 × 10⁶ m s⁻¹. Working back, h/(9.11 × 10⁻³¹ × 7.28 × 10⁶) = 1.0 × 10⁻¹⁰ m.
- 5
Rearrange T = 2π√(l/g) to make g the subject.
Model answer
Dividing by 2π and squaring gives T²/4π² = l/g, so g = 4π²l/T². The result is in N kg⁻¹ when l is in m and T in s.
- 6
A pendulum of length 1.00 m has a period of 2.00 s. Calculate the value of g it gives.
Model answer
g = 4π²l/T² = 4π² × 1.00/4.00 = 9.87 N kg⁻¹. Working back, T = 2π√(1.00/9.87) = 2π × 0.3183 = 2.00 s.
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