What is the difference between force and momentum in Physics?

In Physics, force is a vector quantity that causes an object to accelerate, defined by Newton's second law as F = ma, where F is force, m is mass, and a is acceleration. Momentum, on the other hand, is the product of an object's mass and velocity, represented as p = mv. While force is a measure of the interaction that changes an object's motion, momentum is a measure of the quantity of motion an object has. Both concepts are crucial in the study of Mechanics and Materials, particularly in the Edexcel International A Levels curriculum.

How is momentum conserved in a closed system?

Momentum is conserved in a closed system due to the law of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it. This principle is fundamental in Mechanics and Materials, as it explains how momentum is transferred between objects during collisions or interactions. For example, in an elastic collision, the total momentum before and after the collision remains the same, which is a key concept in the Edexcel International A Levels Physics syllabus.

What role does impulse play in changing momentum?

Impulse is the change in momentum of an object when a force is applied over a period of time. It is calculated as the product of the force and the time duration over which it acts, represented by the equation Impulse = FΔt. This concept is crucial in understanding how forces affect motion, as it directly relates to the change in an object's momentum. In the context of Edexcel International A Levels, understanding impulse helps students analyze real-world scenarios, such as car crashes or sports, where forces are applied over short time intervals.

How do you calculate the resultant force acting on an object?

The resultant force acting on an object is the vector sum of all individual forces acting on it. To calculate it, you must consider both the magnitude and direction of each force. In a two-dimensional scenario, this involves breaking down forces into their components and summing them up. The resultant force determines the object's acceleration according to Newton's second law (F = ma). This calculation is a fundamental skill in the Mechanics and Materials section of the Edexcel International A Levels Physics curriculum.

Why is understanding forces and momentum important in Physics?

Understanding forces and momentum is essential in Physics because they are fundamental concepts that describe how objects interact and move. Forces explain why objects start moving, stop, or change direction, while momentum provides insight into the motion's quantity and persistence. These concepts are integral to solving problems in Mechanics and Materials, making them crucial for students studying Edexcel International A Levels Physics. Mastery of these topics enables students to analyze and predict the behavior of physical systems in various contexts, from everyday phenomena to complex engineering applications.

Why is "Using a single applied force in $\Sigma F = ma$ instead of the resultant force" a common mistake in Forces and Momentum?

Why it happens: Students see one force in the question (e.g. the driving force) and substitute it directly, ignoring friction or drag. How to avoid it: Always find the **resultant force first** by combining all the forces (subtract opposing forces). Newton's second law uses the **net** force, never one force in isolation. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Quoting weight and normal contact force as a Newton's third-law pair" a common mistake in Forces and Momentum?

Why it happens: Both are equal and opposite when an object rests on a surface, so they look like a pair. How to avoid it: A third-law pair acts on **two different objects** and is the **same type** of force. Weight (gravitational) and normal force (contact) both act on the **same** object, so they only **balance** by Newton's first law — they are not a pair. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Ignoring direction (signs) when calculating change of momentum or applying conservation" a common mistake in Forces and Momentum?

Why it happens: Momentum is treated as a scalar, so velocities in opposite directions are added with the same sign. How to avoid it: Momentum is a **vector**. Choose a positive direction and give opposite velocities **opposite signs**. When a ball rebounds, v = v - (-u), so the speeds **add** — e.g. 15 - (-12) = 27, not 3. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Keeping the negative sign when squaring a velocity in $\tfrac{1}{2}mv^2$" a common mistake in Forces and Momentum?

Why it happens: Carrying the vector sign across from a momentum calculation into a kinetic-energy calculation. How to avoid it: Kinetic energy uses v^2, so the **sign disappears**: (-2.0)^2 = +4.0. KE is a scalar — direction is irrelevant once the velocity is squared. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Saying a falling object 'decelerates' or 'slows down' as it approaches terminal velocity" a common mistake in Forces and Momentum?

Why it happens: Confusing decreasing acceleration with deceleration once drag grows large. How to avoid it: The object keeps **speeding up** — its **acceleration decreases** towards zero. At terminal velocity the resultant force is **zero**, so the velocity is **constant** (a flat line on a v–t graph), not falling. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Confusing mass (kg) with weight (N) or omitting units" a common mistake in Forces and Momentum?

Why it happens: The everyday word 'weight' is used loosely for mass, and g is sometimes forgotten. How to avoid it: **Mass is in kilograms; weight is a force in newtons**, W = mg. Always multiply by g = 9.81\ m s^-2 to convert a mass to a weight, and label every numerical answer with its correct unit for the A marks. Source: Edexcel IAL Physics Unit 1 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Forces and Momentum

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Short Notes - Forces and Momentum

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Detailed Study Notes

Detailed notes on Mechanics and Materials for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Forces and Momentum — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 1)

Newton's three laws, $\Sigma F = ma$ , weight $W = mg$ and free-body diagrams; momentum $p = mv$ , impulse $F\Delta t = \Delta p$ , conservation of linear momentum (elastic vs inelastic collisions), and how drag produces terminal velocity. Master the M/A working discipline Edexcel rewards.

At a glance

Newton's 1st law: zero resultant force → object stays at rest or moves at constant velocity.
Newton's 2nd law: $\Sigma F = ma$ — always use the resultant force, never one applied force in isolation.
Newton's 3rd law pairs: equal magnitude, opposite direction, same type of force, on two different objects. Weight and normal reaction are NOT a pair.
Weight $W = mg$ (a force, in newtons) — use $g = 9.81\ \text{m s}^{-2}$ . Mass (kg) ≠ weight (N).
Momentum $p = mv$ is a vector ( $\text{kg m s}^{-1}$ ); assign signs by direction.
Impulse $F\Delta t = \Delta p = mv - mu$ — extend $\Delta t$ to cut the force (crumple zones, airbags).
Conservation of momentum: with no external resultant force, total momentum before = total momentum after. Elastic also conserves KE; inelastic does not.
Terminal velocity: drag rises with speed until drag = weight → resultant force zero → constant velocity (a flat v–t line).

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Recall and use Newton's first, second and third laws of motion, including $\Sigma F = ma$ for the resultant force.
Draw and interpret free-body force diagrams, labelling each force with magnitude and direction.
Identify Newton's third-law pairs and distinguish them from balanced forces acting on one object.
Calculate weight using $W = mg$ and use $g = 9.81\ \text{m s}^{-2}$ .
Define and calculate linear momentum $p = mv$ and use impulse $F\Delta t = \Delta p$ .
Apply the principle of conservation of linear momentum to collisions and explosions in one dimension, and distinguish elastic from inelastic collisions.
Explain how the motion of a falling object changes until it reaches terminal velocity.

Newton's three laws of motion

The three statements that govern every force question in Unit 1.

Newton's laws describe how forces change motion. State them precisely — examiners credit exact wording.

Newton's first law. An object stays at rest, or continues to move at constant velocity, unless acted on by a resultant (net) force. If the forces are balanced (resultant zero), velocity does not change.

Newton's second law. The resultant force on an object equals the rate of change of its momentum. For constant mass this gives the form you use most:

$\Sigma F = ma$

The acceleration is in the same direction as the resultant force. The single biggest exam error is using one applied force here — you must use the resultant of all forces.

Newton's third law. When object A exerts a force on object B, object B exerts an equal and opposite force, of the same type, on A. The two forces of a third-law pair always:

are equal in magnitude, opposite in direction;
are the same type of force (both gravitational, or both contact, etc.);
act on two different objects.

1st law: zero resultant force → rest or constant velocity.
2nd law: $\Sigma F = ma$ (use the RESULTANT force).
3rd law pair: equal, opposite, same type, different objects.

Free-body diagrams and weight

Show every force on ONE object — magnitude and direction.

A free-body diagram isolates a single object and draws every external force acting on it as a labelled arrow (length ~ magnitude, head pointing the right way). It shows only the forces on that object — never the forces it exerts on others.

The weight of an object is the gravitational force on it:

$W = mg$

where $g = 9.81\ \text{m s}^{-2}$ on Earth. Weight is a force in newtons and acts vertically downwards from the centre of mass — do not confuse it with mass (in kilograms). For a $75\ \text{kg}$ person, $W = 75 \times 9.81 = 736\ \text{N}$ .

On a slope, the normal contact force is perpendicular to the surface (not vertical), and the weight is resolved into a component down the slope ( $mg\sin\theta$ ) and into the slope ( $mg\cos\theta$ ).

Free-body diagram: weight acts vertically down (W = mg), the normal contact force is perpendicular to the slope, and friction acts up the slope. The normal force is NOT vertical — that is the classic slip.

Free-body diagram = all forces ON one object, labelled with magnitude + direction.
$W = mg$ , a downward force in newtons ( $g = 9.81\ \text{m s}^{-2}$ ).
On a slope: normal force ⟂ surface; weight resolves into $mg\sin\theta$ and $mg\cos\theta$ .

See the full worked example for forces and momentum →

Momentum and impulse

$p = mv$ is a vector; impulse links force, time and momentum change.

Linear momentum is mass × velocity:

$p = mv$

It is a vector (unit $\text{kg m s}^{-1}$ ), pointing in the direction of the velocity. A $1200\ \text{kg}$ car at $25\ \text{m s}^{-1}$ has $p = 1200 \times 25 = 3.0 \times 10^{4}\ \text{kg m s}^{-1}$ .

Impulse is the product of a force and the time for which it acts. Starting from Newton's second law ( $F = \Delta p / \Delta t$ ) and rearranging gives the impulse–momentum equation:

$F\Delta t = \Delta p = mv - mu$

So impulse equals the change in momentum. Two big consequences:

Direction matters. Choose a positive direction and give opposite velocities opposite signs. If a ball moving at $12\ \text{m s}^{-1}$ rebounds at $15\ \text{m s}^{-1}$ , the change is $15 - (-12) = 27\ \text{m s}^{-1}$ , not $3$ — the speeds add on a reversal.
Extending the time reduces the force. Because $\Delta p$ is fixed by the mass and velocity change, increasing $\Delta t$ decreases the average force $F$ . This is why crumple zones, airbags and crash mats save lives: they lengthen the collision time, lowering the force on the body.

$p = mv$ — a vector in $\text{kg m s}^{-1}$ ; assign signs by direction.
$F\Delta t = \Delta p = mv - mu$ — impulse = change in momentum.
Reversal → speeds add: $v - (-u) = v + u$ .
Bigger $\Delta t$ → smaller force (crumple zones, airbags).

See the full worked example for forces and momentum →

Conservation of momentum and collisions

Total momentum before = total momentum after; elastic vs inelastic.

Principle of conservation of linear momentum. In a closed system with no external resultant force, the total momentum before an interaction equals the total momentum after it.

For two bodies, always write out both states with consistent signs:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Why it holds (a favourite 6-marker): during the collision the two objects push on each other with equal and opposite forces (Newton's 3rd law) for the same time, so they exchange equal and opposite impulses. By Newton's 2nd law ( $F\Delta t = \Delta p$ ), one body gains exactly the momentum the other loses, so the total is unchanged.

Elastic vs inelastic:

Elastic collision — both momentum and total kinetic energy are conserved. (KE before = KE after.)
Inelastic collision — momentum is conserved but kinetic energy is not; some KE becomes heat, sound or deformation. A perfectly inelastic collision is one where the objects stick together and move off with one common velocity.

To test elasticity: conserve momentum to find the velocities, then compare $\sum\tfrac{1}{2}mv^2$ before and after. Remember KE uses $v^2$ , so the sign disappears — $(-2.0)^2 = +4.0$ .

Always draw both states. Total momentum before (12 000 kg m s⁻¹) equals total momentum after. Because the trucks couple and move as one, this is a perfectly inelastic collision — KE is lost to heat and sound.

No external resultant force → total momentum before = after.
Write both states: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$ (keep signs).
Elastic: momentum AND KE conserved. Inelastic: only momentum.
Perfectly inelastic = objects stick together (one common velocity).

See the full worked example for forces and momentum →

Drag and terminal velocity

Drag grows with speed until it balances weight — then velocity is constant.

When an object falls through a fluid (air, oil, water) it experiences drag — a resistive force acting opposite to its motion whose magnitude increases with speed. Trace the motion of a falling object using $\Sigma F = ma$ :

At release the speed is zero, so drag is zero. The only force is weight, giving a large downward resultant and an acceleration close to $g$ .
As it speeds up, the drag increases, so the resultant force decreases. The object still accelerates, but the acceleration gets smaller — the speed rises ever more gently.
Eventually drag grows until it equals the weight. Now the forces balance, the resultant force is zero, the acceleration is zero, and (Newton's 1st law) the object falls at a constant maximum speed — the terminal velocity.

At terminal velocity $W = F_{drag}$ , so $mg =$ drag. The object is not decelerating — it has simply stopped accelerating. On a velocity–time graph this is a curve that starts steep (gradient ≈ $g$ ), flattens as the gradient falls, and finally becomes a horizontal line at the terminal velocity.

Gradient = acceleration. It starts at ≈ g, decreases as drag grows, and reaches zero when drag = weight — the graph levels off at the terminal velocity. The line flattens; it never dips down.

Drag acts opposite to motion and increases with speed.
Falling: resultant force (and acceleration) decreases as drag grows.
Terminal velocity: drag = weight → resultant zero → constant velocity.
On a v–t graph: steep start, flattening curve, horizontal line at terminal velocity.

See the full worked example for forces and momentum →

Quick recap

Newton's 1st law: zero resultant force → constant velocity. 2nd law: $\Sigma F = ma$ (resultant!). 3rd law: equal, opposite, same type, different objects.
Weight $W = mg$ is a downward force in newtons; on a slope the normal force is perpendicular to the surface.
Momentum $p = mv$ is a vector; assign signs by direction — a reversal makes the speeds add.
Impulse $F\Delta t = \Delta p = mv - mu$ ; extending $\Delta t$ reduces the force (crumple zones, airbags).
Conservation of momentum: no external resultant force → total before = total after; write both states with signs.
Elastic collisions conserve KE as well as momentum; inelastic collisions conserve only momentum (KE → heat/sound).
Terminal velocity: drag rises with speed until drag = weight → resultant force zero → constant velocity.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Newton's 1st law: an object stays at rest or moves at constant velocity unless acted on by a resultant force.
Newton's 2nd law: $\Sigma F = ma$ — use the RESULTANT force.
Newton's 3rd-law pair: equal magnitude, opposite direction, same type of force, on two different objects.
Weight $W = mg$ ; $g = 9.81\ \text{m s}^{-2}$ . Mass (kg) is not weight (N).
Momentum $p = mv$ (vector, $\text{kg m s}^{-1}$ ).
Impulse $F\Delta t = \Delta p = mv - mu$ .
Conservation of momentum: with no external resultant force, total momentum before = total momentum after.
Elastic = momentum AND kinetic energy conserved; inelastic = momentum conserved, KE not.
Terminal velocity: drag = weight → resultant force zero → constant velocity.

How it’s examined

Forces and momentum is core Unit 1 (WPH11) content and appears on the Unit 1 paper (Mechanics and Materials). Expect: state/define questions on Newton's laws and momentum (1–2 marks, AO1); identify-the-third-law-pair reasoning (2–3 marks — watch the weight/normal-reaction trap); $\Sigma F = ma$ and $W = mg$ calculations (2–3 marks, full M/A working with units and 3 s.f.); impulse and crumple-zone questions ( $F\Delta t = \Delta p$ , often with a velocity reversal); conservation-of-momentum collisions and explosions where you must keep vector signs and may have to test elasticity by comparing kinetic energy; and a levels-based 6-marker such as 'explain conservation of momentum using Newton's laws' or 'explain how the motion changes until terminal velocity is reached', marked with continuous linked prose. Show the equation, substitution with units, rearrangement and a final answer to 3 s.f. — no working risks zero.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 1 examiner reports; Edexcel IAL Physics Unit 1 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

Step-by-step worked examples — Forces and Momentum

Step-by-step solutions to past-paper-style questions on forces and momentum, written exactly the way a tutor would explain them at the board.

Question type:

1Calculating weight from mass (2 marks)
Getting startedDirect calculation• weight, AO2, Unit 1
Question
A laptop has a mass of $1.80\ \text{kg}$ . Calculate its weight on Earth, where $g = 9.81\ \text{m s}^{-2}$ . (2 marks)
Step-by-step solution
1. Step 1
  Weight is the gravitational force on a mass: use $W = mg$ with $m$ in kilograms and $g$ in $\text{m s}^{-2}$ .
  $W = mg$
2. Step 2
  Substitute $m = 1.80\ \text{kg}$ and $g = 9.81\ \text{m s}^{-2}$ and evaluate to 3 significant figures.
  $W = 1.80 \times 9.81 = 17.7\ \text{N}$
Answer
$W = 17.7\ \text{N}$ (downwards, towards the centre of the Earth).
Examiner tip
Mark scheme: (1) correct substitution into $W = mg$ ; (2) answer $17.7\ \text{N}$ to 3 s.f. with the newton. Writing $1.80\ \text{kg}$ as the 'weight' loses the mark — weight is a force in newtons, not a mass in kilograms.
2Momentum of a moving car (2 marks)
Getting startedDirect calculation• momentum, AO2, Unit 1
Question
A car of mass $1200\ \text{kg}$ travels at $25.0\ \text{m s}^{-1}$ . Calculate its momentum. (2 marks)
Step-by-step solution
1. Step 1
  Momentum is the product of mass and velocity. It is a vector, so it acts in the direction of motion.
  $p = mv$
2. Step 2
  Substitute $m = 1200\ \text{kg}$ and $v = 25.0\ \text{m s}^{-1}$ .
  $p = 1200 \times 25.0 = 3.00 \times 10^{4}\ \text{kg m s}^{-1}$
Answer
$p = 3.00 \times 10^{4}\ \text{kg m s}^{-1}$ in the direction of travel.
Examiner tip
Mark scheme: (1) $p = mv$ with substitution; (2) $3.00 \times 10^{4}\ \text{kg m s}^{-1}$ (accept $30\,000$ ). The unit is $\text{kg m s}^{-1}$ (equivalently $\text{N s}$ ) — a missing or wrong unit loses the accuracy mark.
3Spotting a Newton's third-law pair (3 marks)
Getting startedIdentify & classify• Newton's third law, AO1, Unit 1
Question
A book rests on a table. A student says the weight of the book and the normal contact force from the table on the book are a Newton's third-law pair. Explain whether the student is correct. (3 marks)
Step-by-step solution
1. Step 1
  A Newton's third-law pair must act on two different objects, be the same type of force, and be equal in magnitude but opposite in direction.
2. Step 2
  Both the weight and the normal contact force act on the same object (the book), and they are different types of force (gravitational vs contact). So they are NOT a third-law pair — they merely balance because the book is in equilibrium (Newton's first law).
3. Step 3
  The true third-law partner of the book's weight is the gravitational pull of the book on the Earth; the partner of the table's normal force on the book is the normal force of the book pushing down on the table.
Answer
The student is wrong: those two forces act on the SAME object and are different types, so they cannot be a third-law pair — they balance by Newton's first law. The correct pairs act on different objects (book–Earth gravity; book–table contact).
Examiner tip
Mark scheme: (1) state the criteria (same type, opposite direction, different objects); (2) note both forces act on the book / are different types → not a pair; (3) identify a correct partner force. This is the single most common third-law misconception in Edexcel reports.
4Resultant force and acceleration (3 marks)
Building confidenceMulti-step problem• Newton's second law, AO2, Unit 1
Question
A $1500\ \text{kg}$ car experiences a forward driving force of $4200\ \text{N}$ and a total resistive force of $900\ \text{N}$ . Calculate the acceleration of the car. (3 marks)
Step-by-step solution
1. Step 1
  Find the resultant (net) force by subtracting the resistive force from the driving force, since they act in opposite directions.
  $\Sigma F = 4200 - 900 = 3300\ \text{N}$
2. Step 2
  Apply Newton's second law in the form $\Sigma F = ma$ and rearrange for acceleration.
  $a = \frac{\Sigma F}{m}$
3. Step 3
  Substitute the resultant force and mass.
  $a = \frac{3300}{1500} = 2.20\ \text{m s}^{-2}$
Answer
$a = 2.20\ \text{m s}^{-2}$ in the direction of motion.
Examiner tip
Mark scheme: (1) resultant force $= 3300\ \text{N}$ ; (2) use of $\Sigma F = ma$ ; (3) $a = 2.20\ \text{m s}^{-2}$ . The classic error is using the $4200\ \text{N}$ driving force alone — Newton's second law uses the RESULTANT force, not a single applied force.
5Impulse and the force on a struck ball (4 marks)
Building confidenceMulti-step problem• impulse, momentum, AO2, Unit 1
Question
A $0.45\ \text{kg}$ football moving at $12.0\ \text{m s}^{-1}$ is kicked straight back the way it came and leaves the boot at $15.0\ \text{m s}^{-1}$ . The boot is in contact with the ball for $0.018\ \text{s}$ . Calculate the average force exerted by the boot on the ball. (4 marks)
Step-by-step solution
1. Step 1
  Take the final direction of travel as positive. The initial velocity is then $-12.0\ \text{m s}^{-1}$ and the final velocity $+15.0\ \text{m s}^{-1}$ (the ball reverses direction).
2. Step 2
  Find the change in momentum, $\Delta p = m v - m u$ , keeping the signs.
  $\Delta p = 0.45 \times (15.0 - (-12.0)) = 0.45 \times 27.0 = 12.15\ \text{kg m s}^{-1}$
3. Step 3
  Impulse equals change in momentum: $F\Delta t = \Delta p$ . Rearrange for the average force.
  $F = \frac{\Delta p}{\Delta t}$
4. Step 4
  Substitute and evaluate to 3 significant figures.
  $F = \frac{12.15}{0.018} = 675\ \text{N}$
Answer
$F = 675\ \text{N}$ , acting in the direction the ball leaves the boot.
Examiner tip
Mark scheme: (1) take a direction as positive / reverse the sign of one velocity; (2) $\Delta p = 12.15\ \text{kg m s}^{-1}$ using $15 - (-12) = 27$ ; (3) use of $F = \Delta p / \Delta t$ ; (4) $675\ \text{N}$ . The fatal slip is using $15 - 12 = 3$ — because the ball reverses, the speeds ADD in the momentum change.
6A perfectly inelastic collision (4 marks)
Building confidenceMulti-step problem• conservation of momentum, inelastic, AO2, Unit 1
Question
A railway truck of mass $3.0 \times 10^{3}\ \text{kg}$ moving at $4.0\ \text{m s}^{-1}$ collides with a stationary truck of mass $2.0 \times 10^{3}\ \text{kg}$ . The trucks couple together. Calculate their common velocity after the collision. (4 marks)
Step-by-step solution
1. Step 1
  Linear momentum is conserved (no external resultant force). Write total momentum before = total momentum after.
  $m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
2. Step 2
  Total momentum before: only the first truck is moving.
  $p_{before} = (3.0 \times 10^{3})(4.0) + (2.0 \times 10^{3})(0) = 1.2 \times 10^{4}\ \text{kg m s}^{-1}$
3. Step 3
  After coupling, the combined mass is $5.0 \times 10^{3}\ \text{kg}$ , moving at the common velocity $v$ .
  $1.2 \times 10^{4} = (5.0 \times 10^{3}) v$
4. Step 4
  Rearrange for $v$ .
  $v = \frac{1.2 \times 10^{4}}{5.0 \times 10^{3}} = 2.4\ \text{m s}^{-1}$
Answer
$v = 2.4\ \text{m s}^{-1}$ in the original direction of the first truck.
Examiner tip
Mark scheme: (1) state momentum conserved / before = after; (2) total momentum before $= 1.2 \times 10^{4}\ \text{kg m s}^{-1}$ ; (3) use combined mass $5.0 \times 10^{3}\ \text{kg}$ ; (4) $v = 2.4\ \text{m s}^{-1}$ . Forgetting to add the masses after coupling is the usual error.
7Conservation of momentum with a rebound (5 marks)
Building confidenceMulti-step problem• conservation of momentum, vectors, AO2, Unit 1
Question
Trolley A ( $0.80\ \text{kg}$ ) moves right at $3.0\ \text{m s}^{-1}$ and collides with trolley B ( $1.2\ \text{kg}$ ) moving left at $1.5\ \text{m s}^{-1}$ . After the collision A is brought to rest. Calculate the velocity of B after the collision. (5 marks)
Step-by-step solution
1. Step 1
  Take 'to the right' as positive. So $u_A = +3.0\ \text{m s}^{-1}$ and $u_B = -1.5\ \text{m s}^{-1}$ .
2. Step 2
  Total momentum before (signs kept):
  $p_{before} = (0.80)(3.0) + (1.2)(-1.5) = 2.4 - 1.8 = 0.60\ \text{kg m s}^{-1}$
3. Step 3
  After the collision A is at rest ( $v_A = 0$ ), so the total momentum is carried entirely by B.
  $p_{after} = (0.80)(0) + (1.2) v_B$
4. Step 4
  Apply conservation of momentum, $p_{before} = p_{after}$ .
  $0.60 = 1.2 \, v_B$
5. Step 5
  Solve for $v_B$ . The positive sign means B now moves to the right.
  $v_B = \frac{0.60}{1.2} = 0.50\ \text{m s}^{-1}$
Answer
$v_B = +0.50\ \text{m s}^{-1}$ — B moves to the right at $0.50\ \text{m s}^{-1}$ .
Examiner tip
Mark scheme: (1) consistent sign convention; (2) momentum before $= 0.60\ \text{kg m s}^{-1}$ using the negative for B; (3)/(4) set before = after; (5) $v_B = +0.50\ \text{m s}^{-1}$ with direction stated. Treating $1.5\ \text{m s}^{-1}$ as positive gives the wrong total — vectors in opposite directions must carry opposite signs.
8Show that terminal velocity occurs when forces balance (4 marks)
StretchShow that / prove• terminal velocity, drag, AO2, Unit 1
Question
A $0.30\ \text{kg}$ ball falls through oil. The drag force is given by $F_{drag} = kv$ where $k = 0.49\ \text{N s m}^{-1}$ . Show that the ball reaches a terminal velocity of about $6\ \text{m s}^{-1}$ . Use $g = 9.81\ \text{m s}^{-2}$ . (4 marks)
Step-by-step solution
1. Step 1
  At terminal velocity the ball moves at constant velocity, so the resultant force is zero: the upward drag equals the downward weight (ignoring upthrust).
  $F_{drag} = W$
2. Step 2
  Write the weight using $W = mg$ .
  $W = 0.30 \times 9.81 = 2.943\ \text{N}$
3. Step 3
  Set $kv = W$ and rearrange for the terminal velocity $v$ .
  $v = \frac{W}{k} = \frac{2.943}{0.49}$
4. Step 4
  Evaluate, carrying one extra significant figure as this is a 'show that'.
  $v = 6.01\ \text{m s}^{-1} \approx 6\ \text{m s}^{-1}$
Answer
$v_{terminal} = 6.01\ \text{m s}^{-1} \approx 6\ \text{m s}^{-1}$ , when drag = weight and the resultant force is zero.
Examiner tip
Mark scheme: (1) state resultant force zero / drag = weight at terminal velocity; (2) $W = 2.94\ \text{N}$ ; (3) $v = W/k$ ; (4) $6.0\ \text{m s}^{-1}$ (quoted to an extra s.f. to confirm the 'show that'). Never just quote $6\ \text{m s}^{-1}$ — you must reach it from the physics.
9Testing whether a collision is elastic (6 marks)
StretchMulti-step problem• elastic, kinetic energy, AO2, Unit 1
Question
A $2.0\ \text{kg}$ ball moving at $6.0\ \text{m s}^{-1}$ strikes a stationary $4.0\ \text{kg}$ ball. After the collision the $2.0\ \text{kg}$ ball moves at $-2.0\ \text{m s}^{-1}$ and the $4.0\ \text{kg}$ ball moves at $4.0\ \text{m s}^{-1}$ . (a) Show that momentum is conserved. (b) Determine, with calculation, whether the collision is elastic. (6 marks)
Step-by-step solution
1. Step 1
  (a) Momentum before: only the $2.0\ \text{kg}$ ball moves.
  $p_{before} = (2.0)(6.0) + (4.0)(0) = 12\ \text{kg m s}^{-1}$
2. Step 2
  Momentum after (keep signs):
  $p_{after} = (2.0)(-2.0) + (4.0)(4.0) = -4.0 + 16 = 12\ \text{kg m s}^{-1}$
3. Step 3
  Since $p_{before} = p_{after} = 12\ \text{kg m s}^{-1}$ , momentum is conserved.
4. Step 4
  (b) A collision is elastic only if total kinetic energy is also conserved. KE before:
  $E_{k,before} = \tfrac{1}{2}(2.0)(6.0)^2 = 36\ \text{J}$
5. Step 5
  KE after:
  $E_{k,after} = \tfrac{1}{2}(2.0)(2.0)^2 + \tfrac{1}{2}(4.0)(4.0)^2 = 4.0 + 32 = 36\ \text{J}$
6. Step 6
  Since the total kinetic energy is unchanged ( $36\ \text{J} = 36\ \text{J}$ ) as well as the momentum, the collision is elastic.
Answer
(a) $p_{before} = p_{after} = 12\ \text{kg m s}^{-1}$ → momentum conserved. (b) $E_k$ before = after = $36\ \text{J}$ → kinetic energy conserved, so the collision is elastic.
Examiner tip
Mark scheme: (a) 2 marks — momentum before $= 12$ , after $= 12$ (signs correct). (b) 4 marks — KE before $= 36\ \text{J}$ ; KE after $= 36\ \text{J}$ (note $\tfrac{1}{2}mv^2$ uses $v^2$ , so the sign of $-2.0$ disappears); conclusion 'elastic because KE conserved'. Squaring the velocity removes the negative — a frequent slip is keeping the minus sign inside the square.
10Explain conservation of momentum from Newton's laws (6 marks)
StretchShow that / prove• Adapted from Edexcel IAL Physics Unit 1• conservation of momentum, Newton's laws, AO1, Unit 1
Question
Two objects A and B collide. Using Newton's second and third laws, explain why the total momentum of the system is conserved during the collision. (6 marks)
Step-by-step solution
1. Step 1
  During the collision A pushes on B and B pushes on A. By Newton's third law these two forces are equal in magnitude and opposite in direction: $F_{A \to B} = -F_{B \to A}$ .
2. Step 2
  The two objects are in contact for the same time $\Delta t$ , so the impulse each receives is equal and opposite: $F_{A \to B}\,\Delta t = -F_{B \to A}\,\Delta t$ .
3. Step 3
  By Newton's second law in the form $F\Delta t = \Delta p$ , impulse equals change in momentum. So B gains exactly the momentum that A loses: $\Delta p_B = -\Delta p_A$ .
4. Step 4
  Therefore the changes cancel: $\Delta p_A + \Delta p_B = 0$ , meaning the total momentum of the system does not change during the collision.
5. Step 5
  This holds provided no external resultant force acts on the system (the forces between A and B are internal). Momentum is then conserved.
Answer
Newton's 3rd law → equal and opposite forces; same contact time → equal and opposite impulses; Newton's 2nd law ( $F\Delta t = \Delta p$ ) → B gains the momentum A loses; so total momentum is unchanged provided no external resultant force acts.
Examiner tip
Top-band levels answer: (1) Newton's 3rd law gives equal/opposite forces; (2) acting for the same time → equal/opposite impulse; (3) link impulse to $\Delta p$ via Newton's 2nd law; (4) momentum gained by one = lost by the other → total constant; (5) condition: no external resultant force. Continuous linked reasoning — not a bullet list — is what reaches Level 3.

Model Answers — Forces and Momentum

High-scoring sample answers for forces and momentum on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). State Newton's first law of motion.
Model answer
An object will remain at rest, or continue to move at a constant velocity (constant speed in a straight line), unless acted on by a resultant (net) external force.

In other words, if the resultant force on a body is zero, its velocity does not change — it neither speeds up, slows down, nor changes direction.
Why this scores
Why this scores 2/2. (1) stays at rest OR moves at constant velocity; (2) unless acted on by a resultant force. Saying 'an object keeps moving until a force stops it' is too loose — the word resultant (or net) and 'constant velocity' (not just 'speed') are required.
Question 2
3 marks
Q (3 marks). Define linear momentum and state what is meant by the impulse of a force. Give the equation linking impulse and momentum.
Model answer
Linear momentum is the product of an object's mass and its velocity, $p = mv$ . It is a vector quantity with the unit $\text{kg m s}^{-1}$ .

The impulse of a force is the product of the force and the time for which it acts, $F\Delta t$ . It is also a vector, measured in $\text{N s}$ .

The two are linked by impulse = change in momentum: $F\Delta t = \Delta p = mv - mu$
Why this scores
Why this scores 3/3. (1) momentum = mass × velocity ( $p = mv$ ), a vector; (2) impulse = force × time ( $F\Delta t$ ); (3) the linking equation $F\Delta t = \Delta p$ . Stating that both are vectors is good practice and underpins the sign work in collision questions.
Question 3
3 marks
Q (3 marks). A swimmer pushes backwards against the water and moves forwards. Use Newton's third law to explain this, and state the two conditions a third-law pair of forces must satisfy.
Model answer
By Newton's third law, when the swimmer's hands and feet push backwards on the water, the water pushes forwards on the swimmer with an equal and opposite force. This forward force on the swimmer (the reaction) is what propels them forwards.

A Newton's third-law pair of forces must be:

equal in magnitude and opposite in direction, and the same type of force (here, both are contact/push forces);

acting on two different objects (one on the water, one on the swimmer).
Why this scores
Why this scores 3/3. (1) swimmer pushes water back → water pushes swimmer forward (equal and opposite); (2) forces equal/opposite, same type; (3) act on two different objects. The marker looks for the two forces acting on different bodies — pairs acting on the same object are the giveaway error.
Question 4
4 marks
Q (4 marks). A box of weight $50\ \text{N}$ rests on a rough slope inclined at $30°$ to the horizontal and does not move. Draw and label a free-body diagram, and name the three forces acting on the box.
Model answer
The three forces acting on the box are:

its weight, $50\ \text{N}$ , acting vertically downwards from the centre of mass;

the normal contact force $N$ , acting perpendicular to the slope surface (away from the surface);

the friction force $f$ , acting up the slope, parallel to the surface, opposing the tendency to slide down.

Because the box is stationary the forces are in equilibrium: the component of weight down the slope, $50\sin 30° = 25\ \text{N}$ , is balanced by friction, and the component into the slope, $50\cos 30° = 43.3\ \text{N}$ , is balanced by the normal contact force.
Why this scores
Why this scores 4/4. (1) weight vertically down labelled $50\ \text{N}$ ; (2) normal force perpendicular to the surface; (3) friction up the slope; (4) recognition that the forces balance / correct resolution ( $50\sin 30° = 25\ \text{N}$ down the slope). Drawing the normal force vertically (instead of perpendicular to the slope) is the classic free-body error.
Question 5
5 marks
Q (5 marks). A car of mass $900\ \text{kg}$ travelling at $20\ \text{m s}^{-1}$ is brought to rest in a collision lasting $0.12\ \text{s}$ . (a) Calculate the average force on the car. (b) Explain, in terms of impulse, why a crumple zone that increases the collision time reduces the force on the occupants.
Model answer
(a) Change in momentum: $\Delta p = mv - mu = 900 \times (0 - 20) = -1.8 \times 10^{4}\ \text{kg m s}^{-1}$ Using $F = \Delta p / \Delta t$ : $F = \frac{-1.8 \times 10^{4}}{0.12} = -1.5 \times 10^{5}\ \text{N}$ The magnitude of the average force is $1.5 \times 10^{5}\ \text{N}$ , directed opposite to the original motion (a decelerating force).

(b) The change in momentum, $\Delta p$ , is fixed by the mass and the speed change — the occupants must lose the same momentum however they stop. Since $F = \Delta p / \Delta t$ , increasing the time $\Delta t$ over which the momentum changes decreases the average force $F$ . A crumple zone extends $\Delta t$ , so the force on the occupants — and hence the risk of injury — is reduced.
Why this scores
Why this scores 5/5. (a) 3 marks — $\Delta p = -1.8 \times 10^{4}\ \text{kg m s}^{-1}$ ; use of $F = \Delta p/\Delta t$ ; magnitude $1.5 \times 10^{5}\ \text{N}$ . (b) 2 marks — $\Delta p$ fixed; larger $\Delta t$ → smaller $F$ . The high-value point is recognising $\Delta p$ is unchanged so the benefit comes purely from extending the time.
Question 6
5 marks
Q (5 marks). A stationary $0.80\ \text{kg}$ trolley contains a spring-loaded plunger. When released, the plunger pushes a $0.20\ \text{kg}$ block away at $6.0\ \text{m s}^{-1}$ . Calculate the recoil velocity of the trolley, and explain why the trolley and block move in opposite directions.
Model answer
Before release the total momentum is zero (everything is at rest). Momentum is conserved because the spring forces are internal and there is no external resultant force, so the total momentum after release must also be zero.

Taking the block's direction as positive: $0 = m_{block}v_{block} + m_{trolley}v_{trolley}$ $0 = (0.20)(6.0) + (0.80)\,v_{trolley}$ $0 = 1.2 + 0.80\,v_{trolley}$ $v_{trolley} = \frac{-1.2}{0.80} = -1.5\ \text{m s}^{-1}$

The trolley recoils at $1.5\ \text{m s}^{-1}$ in the opposite direction to the block. By Newton's third law the plunger pushes the block one way and the block pushes the plunger (and trolley) the other way with an equal and opposite force, so their momenta are equal and opposite and sum to zero.
Why this scores
Why this scores 5/5. (1) total momentum zero before and after; (2) correct conservation equation with signs; (3) $v_{trolley} = -1.5\ \text{m s}^{-1}$ ; (4) state opposite direction; (5) Newton's third law explanation of the equal-and-opposite momenta. The minus sign is the physics — it tells you the recoil direction, so it must be interpreted, not dropped.
Question 7
6 marks
Q (6 marks). A skydiver jumps from a stationary balloon and falls. Explain, in terms of the forces acting, how the skydiver's motion changes from the moment of jumping until they reach terminal velocity. (This question is marked using levels.)
Model answer
At the instant of jumping the only significant downward force is the skydiver's weight, $W = mg$ . Air resistance (drag) is zero because the speed is zero, so there is a large downward resultant force. By Newton's second law, $\Sigma F = ma$ , the skydiver accelerates downwards at close to $g$ .

As the speed increases, the drag force increases (drag rises with speed). The drag acts upwards, opposing the motion, so the resultant downward force decreases. The skydiver still accelerates, but the acceleration becomes progressively smaller, so the speed rises ever more gently.

Eventually the drag grows until it is equal in magnitude to the weight. Now the upward and downward forces balance, the resultant force is zero, and by Newton's first law the skydiver stops accelerating and falls at a constant maximum speed — the terminal velocity. At this point $W = F_{drag}$ , so $mg =$ drag.

Throughout, the weight stays constant; it is the growing drag that brings the resultant force — and therefore the acceleration — down to zero.
Why this scores
Why this reaches the top band (5–6/6). Three linked stages: (i) at start, only weight acts → large resultant → acceleration $\approx g$ ; (ii) as speed rises, drag increases → resultant decreases → acceleration decreases; (iii) drag = weight → resultant zero → constant terminal velocity (Newton's first law). Continuous, correctly sequenced reasoning that names 'resultant force' and links it to acceleration at each stage. Saying drag 'pushes the diver back up' (implying upward motion) is a level-capping error — drag only reduces the downward acceleration.
Question 8
6 marks
Q (6 marks). A $0.50\ \text{kg}$ trolley moving at $4.0\ \text{m s}^{-1}$ collides head-on with a stationary $1.5\ \text{kg}$ trolley and the two move off together. (a) Calculate their common velocity. (b) Show, by calculation, that the collision is inelastic. (c) State what happens to the 'lost' kinetic energy.
Model answer
(a) Conservation of momentum (before = after): $(0.50)(4.0) + (1.5)(0) = (0.50 + 1.5)\,v$ $2.0 = 2.0\,v \implies v = 1.0\ \text{m s}^{-1}$ The combined trolleys move off at $1.0\ \text{m s}^{-1}$ .

(b) Kinetic energy before: $E_{k,before} = \tfrac{1}{2}(0.50)(4.0)^2 = 4.0\ \text{J}$ Kinetic energy after: $E_{k,after} = \tfrac{1}{2}(2.0)(1.0)^2 = 1.0\ \text{J}$ Since $E_{k,after} = 1.0\ \text{J} < E_{k,before} = 4.0\ \text{J}$ , total kinetic energy is not conserved, so the collision is inelastic.

(c) The 'lost' $3.0\ \text{J}$ of kinetic energy is transferred to other forms — mainly heat (internal/thermal energy) and sound as the trolleys deform and couple. Energy is still conserved overall; only kinetic energy is not.
Why this scores
Why this scores 6/6. (a) 2 marks — conservation set-up and $v = 1.0\ \text{m s}^{-1}$ . (b) 3 marks — KE before $= 4.0\ \text{J}$ , KE after $= 1.0\ \text{J}$ , conclusion 'inelastic because KE not conserved'. (c) 1 mark — KE → heat/sound/deformation, total energy still conserved. The discriminator is stating that momentum is conserved but kinetic energy is not in an inelastic collision.
Question 9
6 marks
Q (6 marks). A ball is dropped from rest into a tall column of liquid. (a) Sketch the velocity–time graph for the ball from release until it reaches terminal velocity, and describe its shape. (b) Explain, using the graph and Newton's laws, what the gradient of the graph represents and why it changes as it does.
Model answer
(a) The velocity–time graph starts at the origin (released from rest). It rises steeply at first with a gradient close to $g$ , then curves and flattens as the gradient decreases, finally becoming a horizontal line at the constant terminal velocity. The curve is concave-down, approaching the horizontal asymptote.

(b) The gradient of a velocity–time graph is the acceleration.

At the start, only weight acts (drag is zero), so the resultant force is large and, by $\Sigma F = ma$ , the acceleration — and hence the gradient — is at its maximum (≈ $g$ ).

As the ball speeds up, drag increases, reducing the resultant downward force, so the acceleration falls and the gradient decreases — the line bends over.

When drag has risen to equal the weight, the resultant force is zero, so the acceleration is zero and the gradient becomes flat: the ball moves at constant terminal velocity (Newton's first law).
Why this scores
Why this scores 6/6. (a) 3 marks — starts at origin, steep initial gradient, curve flattening to a horizontal line at terminal velocity. (b) 3 marks — gradient = acceleration; gradient decreases because drag increases → resultant force decreases; gradient zero when drag = weight (resultant zero). A* answers explicitly identify the horizontal asymptote as terminal velocity and never let the velocity decrease — the graph levels off, it does not dip.

Key Formulae — Forces and Momentum

The formulae you need to memorise for forces and momentum on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Newton's second law (resultant force)
$\Sigma F = ma$
$\Sigma F$
resultant (net) force / $\text{N}$
$m$
mass / $\text{kg}$
$a$
acceleration / $\text{m s}^{-2}$
When to use
When a known resultant force acts on a mass, or to find the resultant force from an acceleration. ALWAYS use the resultant of all forces, not a single applied force.
Example
Resultant $3300\ \text{N}$ on a $1500\ \text{kg}$ car gives $a = 3300/1500 = 2.20\ \text{m s}^{-2}$ .
Weight
$W = mg$
$W$
weight (a force) / $\text{N}$
$m$
mass / $\text{kg}$
$g$
gravitational field strength $= 9.81\ \text{m s}^{-2}$ on Earth
When to use
To find the gravitational force on a mass — e.g. the downward force on a free-body diagram, or the weight balanced by drag at terminal velocity.
Example
A $0.30\ \text{kg}$ ball has weight $W = 0.30 \times 9.81 = 2.94\ \text{N}$ .
Linear momentum
$p = mv$
$p$
momentum (a vector) / $\text{kg m s}^{-1}$
$m$
mass / $\text{kg}$
$v$
velocity / $\text{m s}^{-1}$
When to use
To calculate momentum and to set up conservation-of-momentum equations (total before = total after). Momentum is a vector — assign signs by direction.
Example
A $1200\ \text{kg}$ car at $25\ \text{m s}^{-1}$ has $p = 1200 \times 25 = 3.0 \times 10^{4}\ \text{kg m s}^{-1}$ .
Impulse = change in momentum
$F\Delta t = \Delta p = mv - mu$
$F$
average resultant force / $\text{N}$
$\Delta t$
time the force acts / $\text{s}$
$\Delta p$
change in momentum / $\text{kg m s}^{-1}$ (or $\text{N s}$ )
$u, v$
initial and final velocity / $\text{m s}^{-1}$
When to use
When a force acts for a measured time — to find the force, the impulse, or the velocity change. Keep velocity signs; a reversal makes the speeds add.
Example
A boot delivering $\Delta p = 12.15\ \text{kg m s}^{-1}$ in $0.018\ \text{s}$ exerts $F = 12.15/0.018 = 675\ \text{N}$ .

Key Definitions and Keywords — Forces and Momentum

Definitions to memorise and the exact keywords mark schemes credit for forces and momentum answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Newton's first law of motion
Examiner keyword
An object remains at rest or continues to move with constant velocity unless acted on by a resultant (net) force. Zero resultant force means zero change in velocity.
Newton's second law of motion
Examiner keyword
The resultant force on an object equals the rate of change of its momentum; for constant mass this gives $\Sigma F = ma$ . Force and acceleration act in the same direction.
Newton's third law of motion
Examiner keyword
When object A exerts a force on object B, object B exerts an equal and opposite force of the same type on A. The two forces act on different objects.
Resultant force
Examiner keyword
The single force that has the same effect as all the forces acting on an object combined (their vector sum). A zero resultant force means the object is in equilibrium.
Weight
Examiner keyword
The gravitational force on an object, $W = mg$ , measured in newtons. It acts vertically downwards from the centre of mass and depends on the gravitational field strength $g$ .
Mass
A measure of the amount of matter in an object (in kilograms) and of its inertia — its resistance to a change in motion. Mass is a scalar and is the same everywhere.
Free-body diagram
Examiner keyword
A diagram showing a single object with all the external forces acting ON it drawn as labelled arrows (magnitude and direction), and no forces it exerts on other objects.
Linear momentum
Examiner keyword
The product of an object's mass and velocity, $p = mv$ . It is a vector with the unit $\text{kg m s}^{-1}$ and points in the direction of the velocity.
Impulse
Examiner keyword
The product of a force and the time for which it acts, $F\Delta t$ . Impulse equals the change in momentum, $F\Delta t = \Delta p$ . Unit: $\text{N s}$ (equivalently $\text{kg m s}^{-1}$ ).
Principle of conservation of linear momentum
Examiner keyword
In a closed system with no external resultant force, the total linear momentum before an interaction equals the total linear momentum after it.
Elastic collision
Examiner keyword
A collision in which both total momentum and total kinetic energy are conserved. No kinetic energy is transferred to other forms.
Inelastic collision
Examiner keyword
A collision in which total momentum is conserved but total kinetic energy is not — some kinetic energy is transferred to heat, sound or deformation. (A perfectly inelastic collision is one in which the objects move off together.)
Drag (air/fluid resistance)
Examiner keyword
A resistive force on an object moving through a fluid, acting opposite to the motion. Its magnitude increases with the object's speed.
Terminal velocity
Examiner keyword
The constant maximum velocity reached by a falling object when the drag force has grown to equal its weight, giving zero resultant force and hence zero acceleration.
Equilibrium
The state of an object on which the resultant force (and resultant moment) is zero, so it remains at rest or moves at constant velocity.

Common Mistakes and Misconceptions — Forces and Momentum

The traps other students keep falling into on forces and momentum questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using a single applied force in $\Sigma F = ma$ instead of the resultant force
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students see one force in the question (e.g. the driving force) and substitute it directly, ignoring friction or drag.
How to avoid it
Always find the resultant force first by combining all the forces (subtract opposing forces). Newton's second law uses the net force, never one force in isolation.
✕Quoting weight and normal contact force as a Newton's third-law pair
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Both are equal and opposite when an object rests on a surface, so they look like a pair.
How to avoid it
A third-law pair acts on two different objects and is the same type of force. Weight (gravitational) and normal force (contact) both act on the same object, so they only balance by Newton's first law — they are not a pair.
✕Ignoring direction (signs) when calculating change of momentum or applying conservation
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Momentum is treated as a scalar, so velocities in opposite directions are added with the same sign.
How to avoid it
Momentum is a vector. Choose a positive direction and give opposite velocities opposite signs. When a ball rebounds, $\Delta v = v - (-u)$ , so the speeds add — e.g. $15 - (-12) = 27$ , not $3$ .
✕Keeping the negative sign when squaring a velocity in $\tfrac{1}{2}mv^2$
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Carrying the vector sign across from a momentum calculation into a kinetic-energy calculation.
How to avoid it
Kinetic energy uses $v^2$ , so the sign disappears: $(-2.0)^2 = +4.0$ . KE is a scalar — direction is irrelevant once the velocity is squared.
✕Saying a falling object 'decelerates' or 'slows down' as it approaches terminal velocity
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Confusing decreasing acceleration with deceleration once drag grows large.
How to avoid it
The object keeps speeding up — its acceleration decreases towards zero. At terminal velocity the resultant force is zero, so the velocity is constant (a flat line on a v–t graph), not falling.
✕Confusing mass (kg) with weight (N) or omitting units
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
The everyday word 'weight' is used loosely for mass, and $g$ is sometimes forgotten.
How to avoid it
Mass is in kilograms; weight is a force in newtons, $W = mg$ . Always multiply by $g = 9.81\ \text{m s}^{-2}$ to convert a mass to a weight, and label every numerical answer with its correct unit for the A marks.

Forces and Momentum — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Forces and Momentum

Short Notes - Forces and Momentum

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Calculating weight from mass (2 marks)

2Momentum of a moving car (2 marks)

3Spotting a Newton's third-law pair (3 marks)

4Resultant force and acceleration (3 marks)

5Impulse and the force on a struck ball (4 marks)

6A perfectly inelastic collision (4 marks)

7Conservation of momentum with a rebound (5 marks)

8Show that terminal velocity occurs when forces balance (4 marks)

9Testing whether a collision is elastic (6 marks)

10Explain conservation of momentum from Newton's laws (6 marks)

Newton's second law (resultant force)

Weight

Linear momentum

Impulse = change in momentum

Newton's first law of motion

Newton's second law of motion

Newton's third law of motion

Resultant force

Weight

Mass

Free-body diagram

Linear momentum

Impulse

Principle of conservation of linear momentum

Elastic collision

Inelastic collision

Drag (air/fluid resistance)

Terminal velocity

Equilibrium

✕Using a single applied force in ΣF=ma\Sigma F = maΣF=ma instead of the resultant force

✕Quoting weight and normal contact force as a Newton's third-law pair

✕Ignoring direction (signs) when calculating change of momentum or applying conservation

✕Keeping the negative sign when squaring a velocity in 12mv2\tfrac{1}{2}mv^221​mv2

✕Saying a falling object 'decelerates' or 'slows down' as it approaches terminal velocity

✕Confusing mass (kg) with weight (N) or omitting units

Forces and Momentum — frequently asked questions

✕Using a single applied force in $\Sigma F = ma$ instead of the resultant force

✕Keeping the negative sign when squaring a velocity in $\tfrac{1}{2}mv^2$