State Newton's second law in terms of momentum.

The resultant force acting on a body is equal to the rate of change of its momentum, and the force acts in the direction of that change of momentum.

Show that F = ma is a special case of the momentum form of Newton's second law.

When the mass is constant, the change of momentum is m(v − u), so F = m(v − u)/t, and since (v − u)/t is the acceleration this reduces to F = ma.

Explain why the momentum form is regarded as the more general statement of Newton's second law.

It holds whether the mass, the velocity or both are changing, whereas F = ma is restricted to constant mass, and multiplying it by the time gives the impulse–momentum equation directly.

Show that the units of rate of change of momentum are the same as the units of force.

Rate of change of momentum has units kg m s⁻¹ divided by s, which is kg m s⁻², and that is exactly the newton.

CCEA A-Level Physics 1210

Physical Quantities, Scalars and Vectors

Topic overview Physical quantities and SI base units Prefixes and standard form Scalars and vectors Resolving vectors and finding resultants Coplanar forces in equilibrium

Moments and Equilibrium

Topic overview Moment of a force Centre of gravity The principle of moments Determining unknown masses using moments

Linear Motion and Dynamics

Topic overview Displacement, velocity and acceleration The equations of motion for uniform acceleration Measuring the acceleration of free fall with light gates Displacement–time and velocity–time graphs Projectile motion Projectile calculations

Newton's Laws of Motion

Topic overview Newton's first law and inertia Newton's second law and F = ma Applying Newton's laws to simple situations Newton's third law Friction as a force opposing motion

Linear Momentum and Impulse

Topic overview Momentum and its calculation Conservation of linear momentum Collisions in one dimension Elastic and inelastic collisions Impulse and the impulse–momentum equation Impacts and reducing the force of a collision Newton's second law in terms of momentum

Work, Energy and Power

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Current, Charge and Potential Difference

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Resistance and Resistivity

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Internal Resistance and Potential Dividers

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Waves

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Refraction

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Quantum Physics

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Wave-Particle Duality

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Astronomy

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Practical Skills and Data Analysis

Topic overview Choosing and using measuring instruments Precision, accuracy and recording measurements Using the cathode ray oscilloscope to measure voltage and frequency Tables, graphs and presenting results Linearising a relationship and using the gradient Absolute and percentage uncertainty Combining uncertainties and uncertainty on a graph Random and systematic error, reliability and refining an experiment

Deformation of Solids

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Thermal Physics

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Uniform Circular Motion

Topic overview Angular velocity, period and frequency Linear speed and angular velocity: v = rω Centripetal acceleration and centripetal force Applying F = mv²/r to motion at constant speed

Simple Harmonic Motion

Topic overview Defining simple harmonic motion and a = −ω²x Displacement, velocity and acceleration: x = A cos ωt The simple pendulum and T = 2π√(l/g)The loaded spiral spring and T = 2π√(m/k)SHM graphs and finding velocity from a gradient Free vibrations, forced vibrations and resonance Damping: light, critical and over-damping

The Nucleus and Radioactive Decay

Topic overview Alpha-particle scattering and the nuclear atom Nuclear radius and nucleon number The density of nuclear matter Alpha, beta and gamma radiation: nature, penetration and range Nuclear equations and changes in nucleon and proton number The random and exponential nature of radioactive decay Activity, the decay constant and A = −λN Exponential decay calculations with A = A₀e^(−λt)Half-life and measuring it

Nuclear Energy, Fission and Fusion

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Gravitational Fields

Topic overview Fields of force and field lines Gravitational field strength Newton's law of universal gravitation The field strength of a point mass Circular motion of planets and satellites Kepler's third law Geostationary satellites

Electric Fields

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Capacitors

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Magnetic Fields

Topic overview Magnetic fields around currents The force on a current-carrying conductor Magnetic flux density The electric motor Magnetic flux and flux linkage Faraday's law of electromagnetic induction Lenz's law The alternating current generator Transformers Power losses and high-voltage transmission

Charged Particles and Accelerators

Topic overview Charged particles in a uniform electric field Charged particles in a magnetic field Circular paths in a magnetic field Comparing electric and magnetic deflection The synchrotron Relativistic mass increase Antimatter and annihilation

Fundamental Particles

Topic overview Fundamental particles The four fundamental forces Classifying particles Quarks and the structure of hadrons Conservation laws Beta decay in the quark model

Linear Momentum and Impulse

Newton's second law in terms of momentum

3 things to know
Unit AS 1

Overview

Newton's second law in terms of momentum and why a firefighter has to brace against a high-pressure hose.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Key term

State the momentum form of Newton's second law as an equation, defining each symbol and its unit.

Key term

Give an example of a situation in which the momentum form applies but F = ma does not.

Numbers

A rocket engine ejects 20 kg of exhaust gas each second at 500 m s⁻¹. Calculate the thrust produced.

Questions from this subtopic

Each one with the model answer underneath.

1
State Newton's second law in terms of momentum.
Model answer
The resultant force acting on a body is equal to the rate of change of its momentum, and the force acts in the direction of that change of momentum.
2
State the momentum form of Newton's second law as an equation, defining each symbol and its unit.
Model answer
The equation is F = (mv − mu)/t, where F is the resultant force in newtons, m the mass in kilograms, u the initial and v the final velocity in m s⁻¹ and t the time in seconds over which the change occurs.
3
Show that F = ma is a special case of the momentum form of Newton's second law.
Model answer
When the mass is constant, the change of momentum is m(v − u), so F = m(v − u)/t, and since (v − u)/t is the acceleration this reduces to F = ma.
4
Give an example of a situation in which the momentum form applies but F = ma does not.
Model answer
A rocket burning fuel loses mass as it accelerates, and a conveyor belt being loaded with sand gains mass, so in both cases the mass is not constant and only the rate of change of momentum gives the correct force.
5
Explain why the momentum form is regarded as the more general statement of Newton's second law.
Model answer
It holds whether the mass, the velocity or both are changing, whereas F = ma is restricted to constant mass, and multiplying it by the time gives the impulse–momentum equation directly.
6
Show that the units of rate of change of momentum are the same as the units of force.
Model answer
Rate of change of momentum has units kg m s⁻¹ divided by s, which is kg m s⁻², and that is exactly the newton.

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