CCEA A-Level Physics 1210
Physical Quantities, Scalars and Vectors
Moments and Equilibrium
Linear Motion and Dynamics
Newton's Laws of Motion
Linear Momentum and Impulse
Work, Energy and Power
Current, Charge and Potential Difference
Resistance and Resistivity
Internal Resistance and Potential Dividers
Waves
Refraction
Lenses
Superposition, Interference and Diffraction
Quantum Physics
Wave-Particle Duality
Astronomy
Practical Skills and Data Analysis
Deformation of Solids
Thermal Physics
Uniform Circular Motion
Simple Harmonic Motion
The Nucleus and Radioactive Decay
Nuclear Energy, Fission and Fusion
Gravitational Fields
Electric Fields
Capacitors
Magnetic Fields
Charged Particles and Accelerators
Newton's Laws of Motion
Applying Newton's laws to simple situations
- 3 things to know
- Unit AS 1
Overview
Recall questions on the general method for a mechanics problem, worked examples on a person on scales in an accelerating lift, a trolley pushing a second trolley, a car towing a trailer, a block on a smooth slope, a rocket at lift-off, and the development of terminal velocity.
What you need to know
Answer each one before you reveal it.
A 1000 kg car tows a 500 kg trailer with a driving force of 4500 N and negligible resistance. Calculate the acceleration.
Explain the development of terminal velocity for a falling object using Newton's laws.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Outline the method used to apply Newton's laws to a mechanics problem.
Model answer
Choose the single body to be considered, identify every force acting on that body, choose one direction as positive, add the forces along that direction to find the resultant, and then apply F = ma to find the unknown quantity.
- 2
A person of mass 70 kg stands on scales in a lift accelerating upwards at 1.5 m s⁻². Calculate the contact force from the floor, taking g as 9.81 m s⁻².
Model answer
The resultant must be upwards, so N − mg = ma giving N = 70 × (9.81 + 1.5) = 792 N; checking back, (792 − 687) ÷ 70 = 1.50 m s⁻², which is the acceleration given.
- 3
A person of mass 70 kg stands on scales in a lift accelerating downwards at 1.5 m s⁻². Calculate the contact force from the floor, taking g as 9.81 m s⁻².
Model answer
The resultant must be downwards, so mg − N = ma giving N = 70 × (9.81 − 1.5) = 582 N; checking back, (687 − 582) ÷ 70 = 1.50 m s⁻², which is the acceleration given.
- 4
A person of mass 70 kg stands on scales in a lift travelling upwards at constant velocity. Calculate the contact force from the floor, taking g as 9.81 m s⁻².
Model answer
Constant velocity means zero resultant force, so the contact force equals the weight, N = 70 × 9.81 = 687 N; checking back, 687 − 687 = 0 N, so there is indeed no acceleration.
- 5
Explain why a person feels heavier during the moment a lift starts to move upwards.
Model answer
The sensation of weight comes from the contact force of the floor, and while the lift accelerates upwards that force must exceed the person's weight in order to produce the upward resultant force, so it feels greater than usual.
- 6
A 10 N horizontal force is applied to a 3.0 kg trolley which is in contact with, and pushes, a 2.0 kg trolley on a frictionless surface. Calculate their common acceleration.
Model answer
The two move together, so a = F ÷ (m₁ + m₂) = 10 ÷ 5.0 = 2.0 m s⁻² in the direction of the applied force; checking back, 5.0 × 2.0 = 10 N, which is the applied force given.
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