Newton's laws, momentum and force
Key idea: H2 Physics lessons on motion graphs, uniformly accelerated motion, momentum and Newton’s laws.
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The core idea
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Big question: How do Newton's laws connect an interaction to a change in motion?
Inertia is the tendency of mass to resist acceleration. For one chosen object, the vector resultant of all external forces equals ma. Newton's third-law forces instead act on two different interacting bodies: they are equal in magnitude, opposite in direction and of the same type. Keep these ideas separate when analysing connected bodies, lifts or changing forces.
Mass describes resistance to acceleration
Inertia is not a force stored inside an object. It is the tendency to resist a change in velocity, measured by mass. At the same resultant force, a larger mass has a smaller acceleration; at the same acceleration, it requires a larger resultant force.
Momentum combines this inertia with motion. Because velocity is a vector, momentum depends on the chosen reference frame and sign convention. Mass remains a positive scalar even when momentum is negative.
Check your understanding: Can a stationary truck have less momentum than a moving tennis ball?
Yes. In the chosen frame the stationary truck has zero momentum, while the moving ball has non-zero momentum.
Start with the system and the resultant
Newton's first law says a body remains at rest or at constant velocity unless a resultant external force acts. Newton's second law says the rate of change of momentum is proportional to, and in the direction of, the resultant force; in SI units the proportionality becomes equality.
Newton’s first and second laws describe one chosen body or system. Add only forces acting on it, find their vector resultant, then connect that resultant to acceleration or rate of momentum change. Velocity may point elsewhere while the object slows or turns.
For constant mass, F = ma follows from F = dp/dt. This shortcut is powerful, but writing the resultant component equation first prevents a single force from being mistaken for the net force.
Check your understanding: A car moves east while its resultant force is west. What happens?
Its acceleration is west, so its eastward speed initially decreases; force need not point with velocity.
Third-law partners belong to different diagrams
A third-law pair comes from one interaction: A pushes B and B simultaneously pushes A with equal magnitude and opposite direction. Because the forces act on different bodies, they never cancel within the free-body diagram of either one.
Check your understanding: What is the partner of Earth’s gravitational force on a falling ball?
The gravitational force exerted by the ball on Earth, equal in magnitude and opposite in direction.
Key ideas to keep
- Balanced forces on one body are not a third-law pair.
- Constant velocity requires zero resultant force, not zero individual forces.
- Use the same positive direction for force components, acceleration and any linked-body constraints.
See the reasoning
Worked example
Compare unequal vehicles with equal momentum magnitudes
Question: A 900 kg car moves east at 20 m s⁻¹. A 1200 kg car moves west at 15 m s⁻¹. Compare their momenta using east as positive.
Step 1: Attach signs to velocity
Why: Direction enters through velocity, not through mass.
Working: v₁ = +20 m s⁻¹ and v₂ = −15 m s⁻¹.
Step 2: Calculate each momentum
Why: Linear momentum is the product of mass and signed velocity.
Working: p₁ = 900(+20) = +1.80 × 10⁴ kg m s⁻¹; p₂ = 1200(−15) = −1.80 × 10⁴ kg m s⁻¹.
Step 3: Interpret rather than merely compare numbers
Why: A vector comparison needs magnitude and direction.
Working: |p₁| = |p₂|, while their signs are opposite.
Answer: The cars have equal momentum magnitudes, 1.80 × 10⁴ kg m s⁻¹, in opposite directions.
Check: The heavier car needs the smaller speed to produce the same momentum magnitude.
Apparent weight in an accelerating lift
Question: A 60 kg passenger accelerates upward at 2.0 m s⁻². Find the normal contact force from the lift and identify its third-law partner. Use g = 9.81 m s⁻².
Step 1: Draw forces on the passenger
Why: The equation must contain only forces acting on the chosen body.
Working: Normal force N upward; weight mg downward.
Step 2: Apply Newton’s second law
Why: Upward acceleration means the upward force must exceed weight.
Working: N − mg = ma, so N = m(g + a) = 60(9.81 + 2.0) = 708.6 N.
Step 3: Name the third-law partner
Why: The partner must be the same interaction acting on the other body.
Working: The passenger exerts 708.6 N downward on the lift.
Answer: The lift exerts about 709 N upward on the passenger; the passenger exerts 709 N downward on the lift.
Check: N is greater than the 589 N weight, which is required for upward acceleration.
Another worked model
Model 1
A 0.20 kg ball moving at 15 m s⁻¹ east rebounds at 10 m s⁻¹ west in 0.050 s. Find the average force on the ball.
Check the worked solution
Take east positive. Δp = 0.20(−10 − 15) = −5.0 kg m s⁻¹. F = Δp/Δt = −5.0/0.050 = −100 N, so the average force is 100 N west.
Model 2
A 5.0 kg crate is pulled right by 22 N while resistance is 7.0 N left. Explain its motion using Newton’s laws and find its acceleration.
Check the worked solution
The resultant force is 22 − 7 = 15 N right. Newton’s second law gives a = F/m = 15/5.0 = 3.0 m s⁻² right. The pull and resistance are not a third-law pair because both act on the crate.
Use a hint if needed
Practise with support
Question 1
A 900 kg vehicle experiences a 2700 N driving force and 900 N resistance. Find its acceleration.
Hint: Add forces with signs before applying F = ma.
Check your answer
Resultant force = 2700 − 900 = 1800 N forward. a = F/m = 1800/900 = 2.0 m s⁻² forward.
Question 2
A swimmer pushes water backward. State the interaction pair and explain why the swimmer accelerates forward.
Hint: Put each force on its own receiving body.
Check your answer
The swimmer exerts a backward force on the water; the water exerts an equal forward force on the swimmer. The swimmer accelerates according to the resultant force on the swimmer, not by adding the force acting on the water.
Now work without the hint
Practise independently
Question 1
Define inertia and momentum. Compare the momenta of a 1500 kg car at 12 m s⁻¹ and a 3000 kg truck at 6.0 m s⁻¹.
Check your answer
Inertia is resistance to change of velocity and is measured by mass. Momentum p = mv. Each vehicle has momentum 18 000 kg m s⁻¹ in its direction of motion.
Question 2
A resultant force varies so that an object’s momentum increases from 6 to 30 kg m s⁻¹ in 4.0 s. State the governing law and find the average force. If its constant mass is 3.0 kg, find the average acceleration.
Check your answer
Newton’s second law is F = dp/dt. Average force = (30 − 6)/4.0 = 6.0 N. For constant mass, F = ma, so average acceleration = 6.0/3.0 = 2.0 m s⁻².
Avoid these traps
Common mistakes
Common mistake
An object needs a forward resultant force to keep moving at constant velocity.
What is wrong with this reasoning?
Show better thinking
Inertia means constant velocity persists when resultant force is zero. A resultant force produces acceleration or, more generally, a rate of change of momentum.
Common mistake
Newton’s-third-law forces cancel because they act on the same object.
What is wrong with this reasoning?
Show better thinking
An interaction pair is equal and opposite but acts on two different bodies. Only forces on the same chosen body are combined to find its resultant.
Write for the examiner
Exam guidance
Name the body before writing ΣF = ma; this prevents mixing forces from different free-body diagrams.
Exam-style practice [5 marks]
Trolley A has mass 0.40 kg and velocity +3.0 m s⁻¹. Trolley B has mass 0.60 kg and velocity −2.0 m s⁻¹. Find each momentum and the total momentum. Explain what the result does—and does not—tell you about their motion.
Plan before you answer
- Use one sign convention for both trolleys.
- Add signed momenta.
- Distinguish total momentum from individual motion.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
pA = 0.40(+3.0) = +1.2 kg m s⁻¹ and pB = 0.60(−2.0) = −1.2 kg m s⁻¹. Total momentum is zero. The trolleys are not stationary; their equal and opposite momenta cancel in the system total for this reference frame.
Exam-style practice [7 marks]
A 20 kg crate is pulled right by 90 N while friction is 30 N left. (a) Find its acceleration. (b) The pull is removed while the crate is still moving right. Find its acceleration and explain the subsequent motion using Newton’s laws. (c) State the third-law partner of the friction on the crate.
Plan before you answer
- Write a horizontal resultant equation for each stage.
- Keep velocity and acceleration directions separate.
- Name the other body in the third-law interaction.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
Initially Fres = 90 − 30 = 60 N right, so a = 60/20 = 3.0 m s⁻² right. After the pull is removed, friction is the 30 N leftward resultant, so a = 30/20 = 1.5 m s⁻² left. The crate continues moving right at first but slows because acceleration is opposite velocity. The third-law partner is the friction force exerted by the crate on the floor to the right.
Come back in three days
Check what stayed with you
Recall question 1
Define linear momentum.
Check the answer
The product of mass and velocity, p = mv.
Recall question 2
What does a negative momentum mean?
Check the answer
It points opposite the chosen positive direction.
Recall question 3
Is inertia a force?
Check the answer
No. It is a body’s resistance to a change in velocity and is measured by mass.
Recall question 4
State Newton’s first law in terms of resultant external force.
Check the answer
With zero resultant external force, a body remains at rest or continues at constant velocity.
Recall question 5
When does F = ma follow directly from F = dp/dt?
Check the answer
When mass is constant.
Recall question 6
Why do third-law forces not cancel on one free-body diagram?
Check the answer
They act on different bodies.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. This topic connects descriptions of motion to the forces that cause acceleration. Read graph gradients and areas from their axes, and draw forces only on the chosen object before applying Newton's laws.
- GCE A-Level H2 PhysicsTopic 3(f) / Topic 3(g) / Topic 3(h) / Topic 3(i) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027