The key idea
Newton's laws stated precisely, then applied to connected bodies and collisions. AS raises the bar in one way above all: the second law is now about momentum, not just F = ma.
Understand it · 01
Draw forces on one chosen object
A free-body diagram includes every external force acting on the selected object. It does not include forces that the object exerts on something else. Label weight, contact forces, tension, thrust and drag where appropriate, then resolve their resultant along useful axes. A force arrow shows the direction of a force, not necessarily the direction of motion. An object moving to the right can have a resultant force to the left while it slows down.
Newton’s second law relates resultant force to the rate of change of momentum. For constant mass this becomes ΣF = ma. The equality applies component by component. If two connected objects are treated as a single system, their mutual forces are internal and cancel from the system equation. To find the tension between them, select one object separately and write a second force equation for that object.
Understand it · 02
Keep action–reaction pairs on different bodies
Newton’s third-law forces are equal in magnitude, opposite in direction, of the same interaction type, and act on different objects. Weight and the normal contact force on a resting book are therefore not a third-law pair: both act on the book. The partner of Earth’s gravitational force on the book is the book’s gravitational force on Earth. Equal interaction forces can produce very different accelerations because the interacting masses can be very different.
During a short collision, the contact forces can be much larger than other external forces. If the net external impulse is negligible, total momentum of the chosen system is conserved. This does not imply that kinetic energy is conserved. In a completely inelastic collision the objects move together afterwards, while some kinetic energy becomes internal energy, sound or deformation. Always use signed velocities when constructing the momentum balance.
Understand it · 03
Explain changing acceleration through the forces
For an object falling through a fluid, weight is approximately constant over a small height, while drag usually increases with speed. Upthrust also acts upward. Initially the downward resultant is relatively large; as speed increases, drag grows and the resultant decreases. At terminal speed the forces balance and acceleration is zero. The object continues moving: zero resultant force implies constant velocity, rather than requiring the object to stop.
In a trolley experiment, measure acceleration as the resultant driving force changes while keeping total accelerated mass constant. Friction must be included or estimated; using only the hanging weight as the driving force can be wrong if the hanging mass is accelerating. Repeating trials reduces scatter but cannot correct a wrongly modelled force. A graph of acceleration against resultant force should be a straight line through the origin under the constant-mass model.
Key definitions
- Newton's first law
- An object remains at rest or continues at constant velocity unless acted on by a resultant force.
- Newton's second law
- For a closed system, the resultant external force equals its rate of change of momentum. Favg = Δp/Δt gives the average over an interval and reduces to F = ma for constant mass.
- Newton's third law
- When body A exerts a force on body B, body B exerts a force on body A that is equal in magnitude, opposite in direction, and of the same type.
- Elastic collision
- A collision in which both momentum and total kinetic energy are conserved. In a perfectly elastic collision the relative speed of approach equals the relative speed of separation.
Equations, units & conditions
Second law (average)
Favg = Δp ÷ Δt
Units: N
Use the external resultant on a defined closed system.
Second law (constant mass)
F = m × a
Units: N
Momentum
p = m × v
Units: kg m/s
Worked examples, step by step
Example 1
A 0.40 kg ball moving at 5.0 m/s strikes a wall and rebounds at 3.0 m/s. The contact lasts 0.20 s. Calculate the average force on the ball.
Follow the method
- Take the initial direction as positive: u = +5.0, v = −3.0 m/s.
- Δp = m(v − u) = 0.40 × (−3.0 − 5.0) = −3.2 kg m/s.
- F = Δp ÷ t = −3.2 ÷ 0.20.
F = −16 N (16 N acting away from the wall)
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Resultant force first
A 750 kg vehicle has a forward driving force of 1800 N and total resistance of 450 N. Find its acceleration.
Show solution and self-check: question 1
- Resultant force = 1800 − 450 = 1350 N forwards.
- a = F/m = 1350/750.
1.8 m s⁻² forwards.
Self-check · one mark per point
- Subtract resistance from driving force.
- Use resultant force divided by mass.
- State magnitude and direction.
Stretch · 4 suggested marks
2. Objects join after a collision
A 0.30 kg trolley moving at 2.0 m s⁻¹ hits a stationary 0.20 kg trolley and they join. Neglect external impulse. Find their common velocity and the kinetic energy lost.
Show solution and self-check: question 2
- Momentum: 0.30 × 2.0 = (0.30 + 0.20)v, so v = 1.2 m s⁻¹.
- Initial KE = ½ × 0.30 × 2.0² = 0.60 J.
- Final KE = ½ × 0.50 × 1.2² = 0.36 J; loss = 0.24 J.
1.2 m s⁻¹ in the original direction; 0.24 J lost.
Self-check · one mark per point
- Conserve system momentum.
- Obtain the common velocity.
- Calculate both kinetic energies.
- Find a positive loss of 0.24 J.
Common mistakes & how to avoid them
- Third-law 'pairs' that act on the same object. Weight and the normal contact force on a book are NOT a third-law pair — they act on the same body and are different types. The pair of the book's weight is the book pulling the Earth up.
- Testing elasticity with momentum. System momentum is conserved when the external impulse is negligible; the test for an elastic collision is whether total kinetic energy is conserved. Compute KE before and after.
- Connected bodies treated with the wrong mass: for the whole system use the total mass; for one body, include the tension. Mixing the two gives self-consistent nonsense.
- Use resultant force, not just one force. For a closed system, external force changes total momentum. A changing-mass subsystem, such as a rocket alone, also requires accounting for momentum carried across its boundary.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
What makes a valid Newton's third law pair?
A Newton’s third-law pair consists of equal and opposite forces of the same interaction type acting on two different objects. The forces act simultaneously, but they do not cancel in a free-body diagram of either object separately.
Name both bodies and reverse the interaction sentence: Earth attracts a book, and the book attracts Earth. Weight and the normal reaction on the book are not partners because both act on the book. The book’s normal force on the table is the partner of the table’s normal force on the book. Equal interaction forces also do not imply equal acceleration; the masses can be very different. When the two objects are combined into one system, the interaction forces become internal and cancel in the system’s total force equation.
Is momentum conserved in inelastic collisions?
System momentum is conserved in an inelastic collision when the net external impulse is negligible. Kinetic energy is not conserved: some transfers into deformation, internal energy or sound. Momentum conservation alone therefore cannot establish that a collision is elastic.
The chosen system must include both colliding objects. If a ball hits a wall, the ball’s momentum alone changes; the momentum balance would also involve the wall and Earth. For two isolated trolleys that join, use the combined mass with their final common velocity. Work with signed velocities so a rebound contributes in the opposite direction. After finding the motion from momentum, compare total kinetic energy before and after to quantify the energy transfer. A small external force can be neglected only when its impulse over the collision time is small enough.
Joining trolleys
A 0.40 kg trolley at +3.0 m s⁻¹ joins a 0.20 kg trolley at −1.0 m s⁻¹. External impulse is negligible. Find their final velocity.
- Initial momentum = 0.40 × 3.0 + 0.20 × (−1.0) = 1.0 kg m s⁻¹.
- Combined mass = 0.60 kg.
- v = 1.0/0.60.
+1.67 m s⁻¹, in the initial direction of the heavier trolley.
Why is F = Δp/Δt more fundamental than F = ma?
The momentum statement connects external force with the rate of change of total momentum of a defined closed system. F = ma follows for constant mass, while Δp/Δt gives an average force over a finite interval.
This distinction matters in impacts because force often rises and falls rapidly. A measured total momentum change divided by contact time gives the average, not the peak. It also matters when defining the system: for a rocket alone, mass and momentum cross the boundary, so simply differentiating the rocket’s mv and calling it the external force is incomplete. One can instead include the ejected material in a closed system or explicitly account for momentum flow. At International AS, constant-mass force equations are central; impulse and more developed collision analysis extend into Unit 4.
How do I find impulse and average force from a force–time graph?
The signed area under a resultant force–time graph is impulse, equal to the change in momentum. For a triangular pulse, use half base times height; dividing the area by its duration gives average force.
The graph’s highest point is the peak force, which can be much larger than the average. A longer stopping time reduces the average force for the same momentum change, explaining the role of padding and deformation zones. Preserve the sign of any sections on opposite sides of the axis. If the graph shows only a contact force while another force has a significant impulse, include that additional contribution before equating the result to total momentum change. The area has units newton seconds, equivalent to kilogram metres per second.
Unit 4: triangular force pulse
A resultant force pulse rises linearly from zero to 120 N and falls to zero over 0.050 s. Find impulse and average force.
- Impulse = ½ × 0.050 × 120 = 3.0 N s.
- Average force = impulse/duration = 3.0/0.050.
Impulse 3.0 N s; average force 60 N.
Does an object’s weight change when it reaches terminal speed?
At terminal speed, drag plus upthrust balances weight, making the resultant force and acceleration zero. Weight has not decreased to zero; the object continues moving because balanced forces maintain constant velocity.
An object released in a fluid initially has relatively little drag because its speed is small. Its speed then increases, raising drag until the forces balance. Changing the fluid or opening a parachute changes the relationship between drag and speed, so a different terminal speed can result. Upthrust is sometimes negligible in air but can be important in a liquid; state the approximation rather than deleting it without thought. Terminal speed is not an intrinsic value belonging only to the falling object, because the medium and orientation affect the forces.
Which mass should I use for connected objects, and where does tension go?
For the whole connected system, use the total accelerated mass and exclude internal tensions from the external resultant. To find a particular tension, isolate one object and include the tension as a force on that object.
A common mistake is to use the driving force on the complete system but divide by only one of its masses. Another is to add the connecting tension as an extra external force even though its equal opposite partner is internal to the selected system. Draw separate diagrams and label each equation with the object or system it describes. A light taut inextensible connection often makes the acceleration magnitudes equal, while a smooth light pulley supports the equal-tension model. Friction or a pulley with significant inertia requires a more careful force model.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- State and apply Newton's three laws of motion
- Use F = ma for single bodies and connected systems
- Define linear momentum; state force as rate of change of momentum
- Apply conservation of momentum in one dimension; two-dimensional collisions are an A2 Unit 4 extension
- Distinguish elastic collisions (KE conserved) from inelastic (KE not conserved)
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Dynamics. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/dynamics
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