3.8.1Forces and Motion

Newton’s Three Laws of Motion

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3.8.1 Newton's Three Laws of Motion — OCR A Level Physics A notes: State and apply Newton's three laws, including the momentum form of the second law and third-law pairs.Newton's first law: a body remains at rest or continues to move with constant velocity unless acted on by a re…

Newton's second law
Newton's third law
Worked example 1
Worked example 2
Worked example 3
Resultant force

Section 1 of 9First law

Newton's first lawa body remains at rest or continues to move with constant velocity unless acted on by a resultant external force.

1Inertia

The tendency of a body to keep its state of motion is inertia, and the mass of a body is its measure.

The law explains why passengers lurch forward when a bus brakes suddenly (their bodies tend to continue at constant velocity) and why a spacecraft far from any star keeps moving with no engines.

Note that "constant velocity" includes zero: a book on a table is acted on by forces that balance, so the resultant is zero.

Friction or drag is the reason things on Earth seem to need a force to keep moving at constant speed; in fact the driving force balances the resistive force.

Section 2 of 9Second law

Newton's second lawthe resultant force acting on a body is directly proportional to the rate of change of its momentum, and the change takes place in the direction of the force.

In equation form F = Δp/Δt, with p = mv the momentum in kg m s⁻¹.

F = ma

For a body of constant mass this reduces to F = ma: acceleration is proportional to the resultant force and inversely proportional to mass,

and has the direction of the resultant force.

The newton is defined as the force that gives a mass of 1 kg an acceleration of 1 m s⁻².

The momentum form is needed when mass changes, as for a rocket expelling fuel or a conveyor belt carrying falling sand.

Impulse

It also gives the idea of impulse: FΔt = Δp, the change in momentum.

What the symbols mean: F = force (N) · Δt = time the force acts (s) · Δp = change in momentum (kg m s⁻¹); FΔt is the impulse (N s)

The area under a force–time graph equals the impulse.

This explains safety features: crumple zones, airbags and padded surfaces increase the time Δt over which momentum changes,

reducing the average force F for the same Δp.

In a ball bouncing off a wall, the momentum changes from mv to −mv, so Δp = −2mv for an elastic bounce with equal speeds.

Section 3 of 9Third law

Newton's third lawwhen two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction.

Different

The forces are of the same type (both gravitational, both contact, and so on), act along the same line, and crucially act on different bodies.

They exist at the same time and are therefore called an action–reaction pair.

Examples

the Earth pulls the Moon with gravitational force and the Moon pulls the Earth with a force of the same size;
- a swimmer pushes water backward and water pushes the swimmer forward;
- a rocket pushes gas backward and the gas pushes the rocket forward.

Third law

A common pitfall is to think the forces cancel.

They cannot cancel because each acts on a different object.

To see if a pair is third-law, check: same type, same size, opposite direction, different objects.

Section 4 of 9Conservation of momentum

Combining the second and third laws shows that in an interaction between two bodies the forces are equal and opposite for the same time, so the impulses are equal and opposite and the total momentum is constant if no external force acts.

Conservation of momentum

This is the principle of conservation of momentum: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.

It applies to collisions, explosions and recoil.

For a gun of mass M firing a bullet of mass m at velocity v, the recoil velocity of the gun is V = −mv/M, much smaller because M is much bigger.

Section 5 of 9Using the laws

Typical problem steps

(1) draw a free-body diagram for each object;
- (2) apply F = ma separately along each direction, remembering that with connected objects the tension is internal to the system;
- (3) use third-law pairs to link forces on different bodies.

Using the laws

For two blocks of masses m₁ and m₂ pushed along a smooth surface by a force F, the system accelerates at a = F/(m₁ + m₂) and the contact force between them is m₂a on the second block.

In a lift, the reading of a balance equals the reaction force on the person, R = m(g + a).

Section 6 of 9Conceptual notes

Newton's laws hold in inertial frames, those not accelerating.

In an accelerating lift or a turning car, objects appear to experience forces that are not caused by any interaction; these are effects of the frame and not true forces.

The laws apply to the motion of the centre of mass of any system.

For objects moving close to the speed of light, or at atomic scales, they must be replaced by relativity and quantum mechanics,

but for everyday engineering they are extremely accurate.

Check each answer for the direction of vector quantities: a negative answer for a force or velocity simply means opposite to the direction you chose as positive.

Section 7 of 9Worked examples

A 1200 kg car accelerates from rest to 24 m s⁻¹ in 8.0 s. Find the resultant force and the impulse.
A 0.15 kg ball hits a wall at 20 m s⁻¹ and rebounds at 16 m s⁻¹. The contact lasts 0.010 s. Find the average force on the ball.
Two trolleys of mass 2.0 kg and 3.0 kg are pushed apart by a spring on a smooth track. The 2.0 kg trolley moves off at 1.5 m s⁻¹. Find the velocity of the other.

Section 8 of 9Key terms

Term	Meaning
Inertia	Resistance of a body to changes in its velocity; measured by mass.
Momentum	Mass × velocity, kg m s⁻¹.
Impulse	Force × time, equal to the change in momentum.
Resultant force	Vector sum of the forces on a body.
Newton	Force giving 1 kg an acceleration of 1 m s⁻².
Third-law pair	Equal and opposite forces of the same type on two different bodies.

Section 9 of 9Exam technique

Treating the equal and opposite forces of the third law as cancelling; they act on different objects.

Thinking a moving object needs a resultant force to keep going at constant velocity.

Calling the weight and the normal reaction on a book a third-law pair; they act on the same body and differ in type.

Forgetting momentum is a vector, so rebounds change the sign of velocity.

Using F = ma when mass changes; use F = Δp/Δt.

Physics 0625 & 9702 SUVAT Mastery Graph Interpretation

Constant Acceleration Equations (SUVAT) & Slopes

v = u + at | s = ut + ½at² | v² = u² + 2as | s = ½(u + v)t

Displacement-Time Graph: Gradient = Velocity (curve = accelerating).
Velocity-Time Graph: Gradient = Acceleration; Area under curve = Distance travelled / Displacement.
Free Fall in vacuum: a = g ≈ 9.81 m/s² downwards (air resistance negligible).

Section 1 of 9First law

Section 2 of 9Second law

Section 3 of 9Third law

Section 4 of 9Conservation of momentum

Section 5 of 9Using the laws

Section 6 of 9Conceptual notes

Section 7 of 9Worked examples

A 1200 kg car accelerates from rest to 24 m s−1 in 8.0 s. Find the resultant force and the impulse.

A 0.15 kg ball hits a wall at 20 m s−1 and rebounds at 16 m s−1. The contact lasts 0.010 s. Find the average force on the ball.

Two trolleys of mass 2.0 kg and 3.0 kg are pushed apart by a spring on a smooth track. The 2.0 kg trolley moves off at 1.5 m s−1. Find the velocity of the other.

Section 8 of 9Key terms

Section 9 of 9Exam technique

Marked practiceQuizzes for this lesson

Constant Acceleration Equations (SUVAT) & Slopes

A 1200 kg car accelerates from rest to 24 m s⁻¹ in 8.0 s. Find the resultant force and the impulse.

A 0.15 kg ball hits a wall at 20 m s⁻¹ and rebounds at 16 m s⁻¹. The contact lasts 0.010 s. Find the average force on the ball.

Two trolleys of mass 2.0 kg and 3.0 kg are pushed apart by a spring on a smooth track. The 2.0 kg trolley moves off at 1.5 m s⁻¹. Find the velocity of the other.