Revisit Newton's laws with full mathematical rigour, connect force to impulse and momentum change, and apply conservation of momentum to collisions and explosions.
AQA A-Level Physics · Unit 4: Mechanics
📜Newton's laws State all three laws and apply them to real situations
💨Momentum Define momentum p = mv and understand it as a vector
⚡Impulse Relate impulse FΔt to change in momentum Δp
🔢F = Δp/Δt Use the rate-of-change form of Newton's second law
🎱Conservation of momentum Apply conservation of momentum to collisions in 1D
💥Explosions Analyse explosions where total initial momentum is zero
Newton's Three Laws of Motion
First Law: An object remains at rest, or moves with constant velocity, unless acted upon by a resultant external force. (Objects resist changes to their motion — this resistance is called inertia.)
Second Law: The resultant force on an object is equal to the rate of change of its momentum. F = Δp/Δt. For constant mass this reduces to F = ma.
Third Law: When object A exerts a force on object B, object B exerts an equal and opposite force on object A. These forces act on different objects and are of the same type.
Newton's Second Law in its most general form is F = Δp/Δt rather than F = ma. This is important when mass changes (e.g. rockets expelling fuel, rain falling on a trolley). The familiar F = ma is a special case valid only when mass is constant.
Newton's Third Law pairs must always be between the same two objects and of the same type of force. For example: the Earth pulls the Moon (gravity) and the Moon pulls the Earth with the same gravitational force. Note that these forces act on different objects — they do not cancel each other.
Newton's Third Law pairs always act on different objects. You cannot use them to cancel forces in a free body diagram of a single object.
Momentum and Impulse
Momentum: p = mv (kg m s⁻¹ or N s)
Momentum is a vector quantity — it has both magnitude and direction. For motion along a line, use + and − to denote direction.
Impulse = FΔt = Δp = mv − mu
Symbol
Quantity
Unit
p
Momentum
kg m s⁻¹
F
Force (resultant)
N
Δt
Time interval
s
Δp
Change in momentum (impulse)
N s
The impulse-momentum theorem tells us that a large force acting for a short time produces the same change in momentum as a small force acting for a long time. This is why:
Crash barriers and car crumple zones increase collision time → reduce peak force
Catching a cricket ball with recoiling hands → increases Δt → reduces F on hands
Airbags extend the time over which the driver decelerates → reduce force on driver
On a force-time graph, the area under the curve equals the impulse (= Δp).
Conservation of Momentum
Principle of Conservation of Momentum: The total momentum of a closed system (no external resultant force) remains constant before and after any interaction.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
This follows directly from Newton's Third Law: during a collision, the forces each object exerts on the other are equal and opposite (Third Law), so the total change in momentum of the system is zero.
Types of collisions:
Elastic collision: Both momentum and kinetic energy are conserved. (Rarely achieved in practice — billiard balls approach this.)
Inelastic collision: Momentum is conserved but kinetic energy is not — some KE is converted to heat, sound, or deformation. This is the common case.
Perfectly inelastic collision: Objects stick together after collision. Momentum conserved; maximum KE lost.
Momentum is always conserved in a closed system. Kinetic energy is only conserved in elastic collisions.
Explosions
An explosion is the reverse of a perfectly inelastic collision. Objects that are initially stationary (total momentum = 0) fly apart after the explosion.
Since momentum is conserved and the total initial momentum is zero:
0 = m₁v₁ + m₂v₂
m₁v₁ = −m₂v₂
This means the two fragments fly off in opposite directions, with momenta equal in magnitude and opposite in direction.
Examples of explosions in physics:
A gun firing a bullet: the bullet goes forward, the gun recoils backward
A rocket: exhaust gases ejected backward, rocket accelerates forward
Radioactive α-decay: the α-particle and daughter nucleus fly apart
Two ice skaters pushing off each other from rest
In an explosion from rest: m₁v₁ = −m₂v₂. The lighter fragment always has the greater speed; the heavier fragment has the smaller speed.
Kinetic energy is gained in an explosion (it comes from stored chemical or nuclear energy). So explosions are not elastic — they are the reverse: energy is input, not lost.
A 0.15 kg cricket ball moving at 30 m s⁻¹ is caught and brought to rest in 0.05 s. Calculate the average force on the ball during catching.
1Initial momentum: p₁ = mu = 0.15 × 30 = 4.5 kg m s⁻¹
2Final momentum: p₂ = 0 (at rest)
3Change in momentum: Δp = 0 − 4.5 = −4.5 kg m s⁻¹
4Force = Δp/Δt = −4.5 / 0.05 = −90 N (negative means opposing motion)
Average force on ball = 90 N (opposing the ball's motion)
A 2000 kg car moving at 15 m s⁻¹ collides with a stationary 1500 kg car. They stick together. Find their common velocity after the collision.
1Before: total momentum = 2000 × 15 + 1500 × 0 = 30 000 kg m s⁻¹
2After: both move together at velocity v, total mass = 3500 kg
3Conservation: 30 000 = 3500 × v
4v = 30 000 / 3500 = 8.57 m s⁻¹
Common velocity = 8.57 m s⁻¹ (in original direction)
A rifle of mass 3.0 kg fires a bullet of mass 0.012 kg at 400 m s⁻¹. Find the recoil velocity of the rifle.
Rifle recoils at 1.6 m s⁻¹ in the opposite direction to the bullet.
A 0.5 kg ball hits a wall at 8 m s⁻¹ and bounces back at 6 m s⁻¹. The collision lasts 0.02 s. Calculate the average force exerted by the wall on the ball.
1Take towards the wall as positive: u = +8 m s⁻¹, v = −6 m s⁻¹
2Δp = m(v − u) = 0.5 × (−6 − 8) = 0.5 × (−14) = −7 N s
3F = Δp/Δt = −7/0.02 = −350 N
Force on ball = 350 N, directed away from the wall.
Q1. Newton's Second Law in its most general form is F = ma. True or false?
Q2. A 60 kg person falls and hits the ground, stopping in 0.1 s from a speed of 5 m s⁻¹. Calculate the average force the ground exerts on the person.
Q3. Which of the following is an example of a Newton's Third Law pair?
Q4. A 1200 kg car travelling at 20 m s⁻¹ collides with a stationary 800 kg car. After the collision the 1200 kg car moves at 8 m s⁻¹ in the original direction. Find the velocity of the 800 kg car after the collision.
Q5. A force-time graph shows a triangular pulse: force rises from 0 to 500 N over 0.04 s, then falls back to 0. What is the impulse delivered?
Challenge Q1. A rocket of initial mass 500 kg (including 200 kg of fuel) ejects gas at 800 m s⁻¹ relative to the rocket. If it burns all its fuel in 40 s, estimate the average thrust force produced.
Challenge Q2. Ball A (mass 0.3 kg, velocity +4 m s⁻¹) collides elastically with ball B (mass 0.3 kg, at rest). Determine the velocities after the collision and verify kinetic energy is conserved.
Challenge Q3. A stationary nucleus of mass number 226 undergoes alpha decay, emitting an alpha particle (mass number 4) at 1.5 × 10⁷ m s⁻¹. Calculate the recoil speed of the daughter nucleus.