What is the relationship between force and motion according to Newton's laws?

Newton's laws of motion describe the relationship between the forces acting on an object and its motion. The first law states that an object will remain at rest or in uniform motion unless acted upon by a force. The second law quantifies this relationship, stating that force equals mass times acceleration (F=ma). The third law asserts that for every action, there is an equal and opposite reaction. These principles are fundamental in understanding how forces influence movement in Physics, particularly in the Edexcel IGCSE curriculum.

How does momentum relate to force and motion?

Momentum is a measure of the motion of an object and is calculated as the product of its mass and velocity (p=mv). In the context of forces and motion, momentum is crucial because it is conserved in isolated systems, meaning the total momentum before an event is equal to the total momentum after, provided no external forces act on it. This principle is essential in solving problems involving collisions and explosions in Edexcel IGCSE Physics.

How do forces affect the shape of an object?

Forces can cause an object to change shape through compression, tension, bending, or twisting. When a force is applied, it can deform the object, and this deformation can be temporary or permanent depending on the material's properties. Understanding how materials respond to forces is important in Physics, especially when studying elasticity and plasticity in the Edexcel IGCSE curriculum.

What is the difference between speed and velocity in the context of movement?

Speed is a scalar quantity that refers to how fast an object is moving, regardless of its direction. In contrast, velocity is a vector quantity that includes both the speed and direction of an object's movement. This distinction is crucial in Physics, particularly in the Edexcel IGCSE curriculum, as it affects how we calculate and understand motion and forces.

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

The resultant force is the single force that represents the combined effect of all the individual forces acting on an object. To calculate it, you sum up all the forces, taking into account their directions. If the forces are in the same direction, you add them; if they are in opposite directions, you subtract them. Understanding how to calculate the resultant force is essential for analyzing motion and predicting how objects will move under various forces, a key concept in Edexcel IGCSE Physics.

Why is "Treating mass and weight as the same quantity" a common mistake in Forces, Movement, Shape and Momentum?

Why it happens: In everyday speech we 'weigh' things in kg. How to avoid it: Mass is a SCALAR in kg, invariant. Weight is a VECTOR force in N, equal to mg. Same object on Earth and Moon: SAME mass, DIFFERENT weight. Source: 4PH1 Examiner Reports 2022-2024

Why is "Identifying the wrong pair for Newton's third law (Paper 2)" a common mistake in Forces, Movement, Shape and Momentum?

Why it happens: Confusing third-law pairs with the two forces on a single object (e.g. weight + normal reaction on a book on a table). How to avoid it: Third-law pairs ALWAYS act on DIFFERENT bodies, are the SAME TYPE of force, are EQUAL in magnitude and OPPOSITE in direction. The book pushes down on the table; the table pushes up on the book — that pair is Newton's third law. The weight of the book and the table's normal reaction on it are NOT a third-law pair. Source: 4PH1/2P Examiner Reports 2022-2024 (recurring item)

Why is "Forgetting that momentum is a vector" a common mistake in Forces, Movement, Shape and Momentum?

Why it happens: Calculations look like scalar arithmetic. How to avoid it: Define a positive direction at the start of any momentum problem. Take velocities in the opposite direction as negative. State direction in your final answer. Source: 4PH1/2P Examiner Reports 2022-2024

Why is "Applying Hooke's law beyond the limit of proportionality" a common mistake in Forces, Movement, Shape and Momentum?

Why it happens: Treating any force-extension data as obeying F = kx. How to avoid it: Hooke's law applies only to the INITIAL LINEAR REGION (spec 1.23). Beyond the limit of proportionality, F/x is no longer constant and F = kx cannot be used.

Forces and MotionEdexcel IGCSE Physics (4PH1)

Forces, Movement, Shape and Momentum

Q: Why is "Using the wrong distance in the moment formula" a common mistake in Forces, Movement, Shape and Momentum?

Why it happens: Using the slant distance from pivot to the force application point. How to avoid it: d is the PERPENDICULAR distance from the pivot to the line of action of the force. If the force is angled, use the component perpendicular to the lever arm.

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Forces, Movement, Shape and Momentum — Pearson Edexcel International GCSE Physics 4PH1 Study Notes (2026 onwards, Spec Issue 4)

How forces change motion and shape. Newton's laws, $F = ma$ , $W = mg$ , Hooke's law, terminal velocity, stopping distance. Plus Paper-2-only depth: momentum, conservation of momentum, $F = (mv - mu)/t$ , Newton's third law and moments.

What you’ll learn

Mapped to the Cambridge IGCSE 4PH1 syllabus (2026 onwards).

1.11 — Describe the effects of forces between bodies (speed, shape, direction).
1.12 — Identify different types of force (gravitational, electrostatic).
1.13 — Understand how vector quantities differ from scalar quantities.
1.14 — Understand that force is a vector quantity.
1.15 — Calculate the resultant force of forces along a line.
1.16 — Know that friction is a force that opposes motion.
1.17 — Know and use $F = m \times a$ .
1.18 — Know and use $W = m \times g$ .
1.19 — Know stopping distance = thinking distance + braking distance.
1.20 — Describe factors affecting stopping distance (speed, mass, road, reaction time).
1.21 — Describe forces on falling objects and explain terminal velocity.
1.22 — Required practical: investigate extension vs applied force for springs/wires/rubber.
1.23 — Know Hooke's law applies in the initial linear region.
1.24 — Describe elastic behaviour.
1.25P — Use $p = m \times v$ for momentum (Paper 2).
1.26P — Use momentum to explain safety features (Paper 2).
1.27P — Use conservation of momentum (Paper 2).
1.28P — Use $F = (mv - mu)/t$ (Paper 2).
1.29P — Demonstrate Newton's third law (Paper 2).
1.30P — Use $M = F \times d$ for moments (Paper 2).
1.31P — Know weight acts at the centre of gravity (Paper 2).
1.32P — Use the principle of moments for parallel forces in one plane (Paper 2).
1.33P — Understand how upward forces on a light beam vary with the position of a heavy object (Paper 2).

Forces, vectors and resultants (spec 1.11 - 1.16)

What forces do; vector vs scalar.

Effects of forces (spec 1.11). A resultant force can change:

the speed of an object (accelerating or decelerating it);
the direction of motion (turning it);
the shape of an object (stretching, compressing, bending).

Types of force (spec 1.12). Gravitational (e.g. weight), electrostatic (between charges), magnetic, contact (push/pull/normal/tension), frictional.

Scalars vs vectors (spec 1.13, 1.14). A SCALAR has magnitude only (mass, speed, energy, temperature). A VECTOR has magnitude AND direction (velocity, acceleration, force, momentum, displacement).

Resultant force on a line (spec 1.15). For forces acting along the same line, add forces in one direction and subtract opposing forces:

Two forces of 12 N forwards and 5 N backwards → resultant = $12 - 5 = 7$ N forwards.
Two forces of 8 N to the right and 8 N to the left → resultant = 0 N → no acceleration (balanced).

Friction (spec 1.16). A force that OPPOSES motion (or attempted motion) between two surfaces in contact. Air resistance is a frictional force on objects moving through air. Friction transfers kinetic energy to thermal energy (heat) in the surfaces.

Forces change speed, direction OR shape.
Vectors need direction; scalars don't.
Resultant force = sum along the line, signs included.

Newton's second law: $F = ma$ (spec 1.17)

Resultant force = mass × acceleration.

Spec statement 1.17: force = mass × acceleration, applied to the RESULTANT (unbalanced) force.

$F = m \times a$

$F$ in newtons (N)
$m$ in kilograms (kg)
$a$ in metres per second squared (m/s²)

Direction. Resultant force and acceleration ALWAYS point in the same direction. If the question gives one direction, the other follows.

Two interpretations of F = ma.

Scenario	Resultant force	Behaviour
F resultant = 0	balanced	constant velocity (Newton's 1st law)
F resultant ≠ 0	unbalanced	accelerating (positive a) or decelerating (negative a)

Worked walkthrough. A 1200 kg car accelerates at 2.5 m/s². Resultant force $F = 1200 \times 2.5 = 3000$ N forwards. If the engine produces 4000 N tractive force, the friction/drag must be $4000 - 3000 = 1000$ N.

Edexcel tip. Always state the direction of the resultant force in your answer — not just the magnitude.

F = ma applies to the RESULTANT force.
Same direction as acceleration.
If F resultant = 0, velocity is constant.

See the full worked example for forces, movement, shape and momentum →

Weight, mass and $W = mg$ (spec 1.18)

Mass is invariant; weight is a force.

Mass is a measure of how much MATTER an object contains. It is a SCALAR, measured in kg, and is the same everywhere in the universe.

Weight is the FORCE on a mass due to a gravitational field. It is a VECTOR, measured in newtons, pointing toward the centre of the gravitating body.

$W = m \times g$

where $g$ is the gravitational field strength in N/kg.

Planet/Moon	$g$ (N/kg)
Earth	≈ 10 (use 9.81 if data sheet says so)
Moon	1.6
Mars	3.7
Jupiter	25

Classic Edexcel question. "An astronaut has a mass of 80 kg on Earth. State the mass and weight of the astronaut on the Moon."

Mass on Moon = 80 kg (unchanged — mass is invariant).
Weight on Moon = $80 \times 1.6 = 128$ N.
(Weight on Earth = $80 \times 10 = 800$ N.)

Failing to recognise mass as invariant is the most common error in this section.

Mass: scalar, kg, invariant.
Weight: vector force, N, depends on $g$ .
$W = m \times g$ .

See the full worked example for forces, movement, shape and momentum →

Stopping distance and terminal velocity (spec 1.19 - 1.21)

Two-stage stopping; balanced forces give constant fall speed.

Stopping distance (spec 1.19) is split into two stages:

$\text{stopping distance} = \text{thinking distance} + \text{braking distance}$

Thinking distance — the distance travelled at constant speed during the driver's REACTION TIME (typically 0.5-1.0 s).
Braking distance — the distance travelled while the brakes decelerate the car to rest.

Factors affecting each (spec 1.20).

Affects...	Factor
Thinking distance	reaction time → alcohol, drugs, tiredness, distraction, age, illness
Braking distance	speed (large effect — proportional to $v^2$ ), mass of vehicle, road condition (wet, icy, loose), tyre condition, brake condition
Both	initial speed

Edexcel pitfall. Alcohol does NOT change braking distance — it increases REACTION TIME, hence thinking distance. Match each factor to the correct stage.

Terminal velocity (spec 1.21). A falling object initially accelerates downward under its weight. As speed increases, air resistance (drag) increases. Eventually drag = weight → resultant force = 0 → acceleration = 0 → constant velocity. This constant velocity is the terminal velocity.

A typical velocity-time graph for a skydiver:

Free fall: weight only → $a = g$ → straight rising line.
Increasing drag: $a$ decreases → curve flattening.
Terminal velocity: drag = weight → horizontal line.
Parachute opens: drag suddenly larger than weight → deceleration → smaller terminal velocity reached.

Stopping = thinking + braking.
Reaction-time factors → thinking; everything else → braking.
Terminal velocity: drag = weight.

See the full worked example for forces, movement, shape and momentum →

Hooke's law, springs and the required practical (spec 1.22 - 1.24)

$F = kx$ in the linear region.

Hooke's law (spec 1.23). Within the INITIAL LINEAR REGION of a force-extension graph:

$F = k \times x$

$F$ = applied force (N)
$x$ = extension (m)
$k$ = spring constant (N/m) — the gradient of the F-x graph

Beyond the limit of proportionality the graph curves — Hooke's law no longer applies.

Required practical (spec 1.22). Investigate how extension varies with applied force for helical springs, metal wires and rubber bands.

Apparatus: clamp stand, spring/wire/rubber, ruler, slotted masses, pointer (to read against the ruler).

Method.

Suspend the spring from the clamp stand; record the initial length.
Add masses in equal steps (e.g. 100 g = 1 N at $g = 10$ N/kg).
After each mass, record the new length and calculate extension = new length − initial length.
Plot a graph of force (y-axis) vs extension (x-axis).

Analysis.

Straight line through origin → Hooke's law obeyed; gradient $= k$ .
Curve flattening → past the limit of proportionality.
Rubber bands DO NOT obey Hooke's law over their entire range — the F-x graph is curved and the unloading curve differs from the loading curve (hysteresis).

Elastic behaviour (spec 1.24) = the ability of a material to RECOVER its original shape after the deforming force is removed. Beyond the elastic limit the material is permanently deformed (plastic behaviour).

$F = kx$ in linear region only.
Required practical: 5 measurements, plot F vs x.
Rubber bands don't obey Hooke's law.

See the full worked example for forces, movement, shape and momentum →

Momentum and conservation (Paper 2 only, spec 1.25-1.27P)

$p = mv$ ; conserved in collisions.

Momentum (Paper 2 only — bold "P" content):

$p = m \times v$

$p$ in kg m/s
$m$ in kg
$v$ in m/s

Momentum is a VECTOR. Direction must be stated.

Principle of conservation of momentum (spec 1.27P). In any collision or explosion, provided no external force acts, the TOTAL momentum of the system BEFORE = TOTAL momentum AFTER.

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

Three exam-favourite cases.

Stick-together collision. Two objects collide and move off as one. Use $m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$ .
Explosion / separation. Two objects initially at rest fly apart (e.g. a gun and bullet). $0 = m_1 v_1 + m_2 v_2$ → $v_1$ and $v_2$ point in OPPOSITE directions.
Bouncing collision. Both objects continue separately. Use the full equation with sign care.

Safety features (spec 1.26P). Airbags, seat belts, crumple zones, gymnasium mats, cushioned trainer soles all work by INCREASING the time over which momentum changes. By rearranging spec 1.28P, $F = \Delta p / t$ : same $\Delta p$ spread over longer $t$ → smaller $F$ → less damage to the passenger.

$p = mv$ (vector).
Total momentum conserved in any closed collision.
Safety features extend $t$ to reduce $F$ .

See the full worked example for forces, movement, shape and momentum →

Force as rate of change of momentum (Paper 2 only, spec 1.28P)

$F = (mv - mu)/t$ .

Spec 1.28P: force = change in momentum / time taken.

$F = \dfrac{mv - mu}{t}$

This is Newton's second law expressed in its most general form. ( $F = ma$ is a special case for constant mass.)

Sign. If $v < u$ (object slowing down), the force is NEGATIVE — meaning it acts opposite to the direction of motion. Quote magnitude and direction in answers.

Application. Crumple zones in a car increase the time $t$ over which the passenger's momentum decreases. Same change in momentum, longer time → smaller force → less injury.

Worked example. A 1500 kg car slows from 24 m/s to rest in 4.0 s. Average force = $\dfrac{1500(0 - 24)}{4.0} = -9000$ N. The minus sign indicates the force opposes motion (braking force). With a SHORTER stopping time (say, 0.2 s in a crash without crumple zones) the force would be $1500 \times 24 / 0.2 = 180\,000$ N — twenty times worse.

$F = \Delta p / t = (mv-mu)/t$ .
Longer $t$ → smaller $F$ .
Always state direction.

See the full worked example for forces, movement, shape and momentum →

Newton's third law (Paper 2 only, spec 1.29P)

Equal and opposite, on DIFFERENT bodies.

Newton's third law. When body A exerts a force on body B, body B exerts an equal-and-opposite force on body A.

Critical property. The two forces in a third-law pair always:

act on DIFFERENT objects,
are the SAME TYPE of force (both contact, both gravitational, …),
are equal in MAGNITUDE,
are opposite in DIRECTION.

Worked example. A rocket pushes hot gases downward; the gases push the rocket upward with an equal and opposite force. That upward force is the THRUST that lifts the rocket.

Classic Edexcel trap. A book on a table experiences (i) its weight pulling it down and (ii) the normal contact force from the table pushing it up. These are EQUAL and OPPOSITE — but they are NOT a Newton's-third-law pair, because they act on the SAME OBJECT (the book). The third-law pair to (i) is the gravitational pull the BOOK exerts on the EARTH. The third-law pair to (ii) is the force the BOOK exerts on the TABLE (pressing down on it).

Equal and opposite, on DIFFERENT bodies.
Same TYPE of force.
Book-on-table: pairs cross between book/Earth and book/table.

Moments and the principle of moments (Paper 2 only, spec 1.30-1.33P)

$M = F \times d_\perp$ ; balance = clockwise = anticlockwise.

Moment of a force (spec 1.30P) = turning effect.

$M = F \times d$

where $d$ is the PERPENDICULAR distance from the pivot to the line of action of the force. Units: N m.

Centre of gravity (spec 1.31P). The single point through which the entire weight of a body acts. For a uniform regular object (rod, beam, disc) the centre of gravity lies at the geometric centre.

Principle of moments (spec 1.32P). For a body in rotational equilibrium about a pivot:

$\sum M_{\text{clockwise}} = \sum M_{\text{anticlockwise}}$

Worked example. A uniform 40 N beam pivoted at its centre. Hang 15 N at 0.80 m to the left. Where does a 10 N weight balance it?

Beam weight: acts AT the pivot → zero moment.
Anticlockwise moment (15 N left): $15 \times 0.80 = 12$ N m.
Clockwise moment (10 N right at distance $d$ ): $10 \times d$ .
Balance: $10d = 12 \Rightarrow d = 1.2$ m to the right.

Light beam on two supports (spec 1.33P). A light beam supported at its two ends has a heavy object placed on it at distance $x$ from the left support. Use moments about each support in turn to find the upward force at each end.

If the supports are at the ends of a beam of length $L$ and the load $W$ sits at distance $x$ from the left:

Taking moments about the LEFT support: $R_R \times L = W \times x \Rightarrow R_R = Wx/L$ .
Taking moments about the RIGHT support: $R_L \times L = W \times (L - x) \Rightarrow R_L = W(L-x)/L$ .

As the load slides toward one end, that end's support takes a larger share of the weight — and the other end's reaction approaches zero.

$M = F \times d_\perp$ in N m.
Balance: clockwise = anticlockwise.
Pivot at the centre of a uniform beam → beam weight contributes nothing.

See the full worked example for forces, movement, shape and momentum →

How it’s examined

Topic 1 (c) is THE highest-yielding section of Topic 1. Typical Paper 1 marks: 8-12 across $F=ma$ , $W=mg$ and stopping distance/terminal velocity. Typical Paper 2 marks: 10-15 across momentum, conservation, $F = \Delta p/t$ , and moments. Required practical (spec 1.22, Hooke's law) appears as a 5-7 mark extended-response question on roughly half of recent papers.

Sources: Pearson Edexcel International GCSE Physics 4PH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4PH1 Examiner Reports 2022-2024; 4PH1/1P May/Jun 2024 question paper and mark scheme; 4PH1/2P May/Jun 2024 question paper and mark scheme; 4PH1/1P January 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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Step-by-step worked examples — Forces, Movement, Shape and Momentum

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

1Applying $F = ma$ (4 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• F=ma, spec-1.17
Question
A 1200 kg car accelerates at 2.5 m/s². Calculate the resultant force on the car. (4 marks)
Step-by-step solution
1. Step 1
  State spec 1.17 relationship.
  $F = m \times a$
2. Step 2
  Substitute.
  $F = 1200 \times 2.5$
3. Step 3
  Evaluate with unit.
  $F = 3000\ \text{N}$
4. Step 4
  State direction (1 mark). The resultant force acts in the direction of the acceleration (forwards).
Answer
$F = 3000$ N, acting in the direction of motion.
Examiner tip
On 4-mark $F=ma$ items, Edexcel usually splits marks as {formula} / {substitution} / {answer with unit} / {direction or context comment}. Don't stop at the number — add the direction or describe the force (e.g. 'forwards', 'tractive', 'on the car') to bank the final mark.
2Weight and mass — same object on Earth and Mars (3 marks, Paper 1)
Core• Adapted from 4PH1/1P January 2024• weight, W=mg, spec-1.18
Question
A robot has a mass of 250 kg. Earth's gravitational field strength is 10 N/kg; Mars is 3.7 N/kg. (a) Calculate the weight of the robot on Earth. (b) State its mass on Mars. (3 marks)
Step-by-step solution
1. Step 1
  (a) Spec 1.18 relationship.
  $W = m \times g = 250 \times 10 = 2500\ \text{N}$
2. Step 2
  (b) Mass is INVARIANT — it does not depend on the gravitational field. So the mass on Mars is still 250 kg. Only the WEIGHT changes ( $W = 250 \times 3.7 = 925$ N on Mars).
Answer
(a) Weight on Earth = 2500 N. (b) Mass on Mars = 250 kg (unchanged).
Examiner tip
Edexcel examiner reports consistently flag this as a 'sleeper' misconception — students change BOTH mass and weight when only weight changes. State explicitly that mass is constant for full marks.
3Hooke's law and spring constant (5 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• Hooke's law, spec-1.22, spec-1.23
Question
A spring extends by 4.0 cm when a 2.0 N weight is hung from it, and by 8.0 cm when 4.0 N is added. (a) Show that the spring obeys Hooke's law. (b) Calculate the spring constant in N/m. (5 marks)
Step-by-step solution
1. Step 1
  (a) Hooke's law (spec 1.23) states extension is directly proportional to applied force, in the INITIAL LINEAR REGION. Check the ratio $F/x$ is constant:
  $\dfrac{2.0}{0.040} = 50\ \text{N/m} \;\;\text{and}\;\; \dfrac{4.0}{0.080} = 50\ \text{N/m}$
2. Step 2
  (a) conclusion. Both ratios are 50 N/m → $F \propto x$ , so the spring obeys Hooke's law over this range.
3. Step 3
  (b) Spring constant $k = F/x = 50$ N/m.
Answer
(a) $F/x$ is constant (= 50 N/m) → Hooke's law obeyed. (b) $k = 50$ N/m.
Examiner tip
Always convert cm to m BEFORE computing the spring constant. Edexcel mark schemes accept $k$ in N/cm (= 0.50 N/cm here) ONLY if the units are stated; missing units = lost mark.
4Stopping distance components (4 marks, Paper 1)
Core• Adapted from 4PH1/1P January 2024• stopping distance, spec-1.19, spec-1.20
Question
A car driver reacts in 0.7 s while travelling at 20 m/s. After braking, the car decelerates uniformly to rest in 30 m. Calculate the total stopping distance. State one factor that increases the braking distance. (4 marks)
Step-by-step solution
1. Step 1
  Thinking distance = speed × reaction time (the car continues at constant speed during this).
  $d_{\text{think}} = 20 \times 0.7 = 14\ \text{m}$
2. Step 2
  Braking distance = 30 m (given).
3. Step 3
  Total stopping distance (spec 1.19) = thinking + braking.
  $d_{\text{total}} = 14 + 30 = 44\ \text{m}$
4. Step 4
  Factor (spec 1.20) that increases the braking distance: wet/icy road, worn tyres or brakes, increased mass (e.g. fully loaded vehicle), higher initial speed.
Answer
Total stopping distance = 44 m. Example factor: wet road increases braking distance.
Examiner tip
Mark schemes credit ANY one valid factor for the final mark. Reaction-time factors (alcohol, drugs, tiredness, distraction) affect THINKING distance — not braking distance. Make sure the factor you give matches what the question asks about.
5Conservation of momentum in a collision (6 marks, Paper 2 — P content)
Extended• Adapted from 4PH1/2P May/Jun 2024• momentum, spec-1.25P, spec-1.27P, Paper 2
Question
A 1200 kg car travelling east at 15 m/s collides with a stationary 800 kg car. The cars stick together after the collision. Calculate the velocity of the combined wreck immediately after the collision. State the direction. (6 marks)
Step-by-step solution
1. Step 1
  Define momentum (spec 1.25P). $p = m \times v$ .
2. Step 2
  Total momentum BEFORE. Only the moving car has momentum.
  $p_{\text{before}} = 1200 \times 15 + 800 \times 0 = 18\,000\ \text{kg m/s east}$
3. Step 3
  Conservation of momentum (spec 1.27P). Total momentum is conserved in a collision with no external forces.
  $p_{\text{after}} = p_{\text{before}} = 18\,000\ \text{kg m/s}$
4. Step 4
  Combined mass after collision = 1200 + 800 = 2000 kg.
5. Step 5
  Solve for $v$ .
  $v = \dfrac{p_{\text{after}}}{m_{\text{total}}} = \dfrac{18\,000}{2000} = 9.0\ \text{m/s}$
6. Step 6
  Direction. Same direction as the initial moving car — i.e. EAST.
Answer
Combined velocity = 9.0 m/s east.
Examiner tip
Edexcel 4PH1/2P momentum questions reward FOUR explicit steps: state the law, calculate $p_{\text{before}}$ , apply conservation, solve for final velocity. State direction explicitly — momentum is a vector and the direction mark is non-negotiable.
6 $F = (mv - mu)/t$ for a stopping car (5 marks, Paper 2 — P content)
Extended• impulse, spec-1.28P, Paper 2
Question
A 1500 kg car travelling at 24 m/s is brought to rest in 4.0 s by braking. Calculate the average braking force. (5 marks)
Step-by-step solution
1. Step 1
  Spec 1.28P: force = change in momentum / time taken.
  $F = \dfrac{mv - mu}{t}$
2. Step 2
  Substitute. $v = 0$ (the car comes to rest); $u = 24$ .
  $F = \dfrac{1500(0) - 1500(24)}{4.0}$
3. Step 3
  Evaluate.
  $F = \dfrac{-36\,000}{4.0} = -9000\ \text{N}$
4. Step 4
  Interpret the sign. The minus indicates the force acts OPPOSITE to the direction of motion. Magnitude = 9000 N.
Answer
Average braking force = 9000 N, opposing motion.
Examiner tip
Edexcel 4PH1/2P uses this equation to ask about safety features (airbags, crumple zones, seat belts). Same change of momentum spread over a LONGER time gives a SMALLER force — that's the safety principle, and it gets a marked sentence in extended responses (spec 1.26P).
7Principle of moments — a balanced beam (6 marks, Paper 2 — P content)
Extended• Adapted from 4PH1/2P May/Jun 2024• moments, spec-1.30P, spec-1.32P, Paper 2
Question
A uniform beam of weight 40 N rests on a pivot at its centre. A 15 N weight is hung 0.80 m to the left of the pivot. Where must a 10 N weight be hung to balance the beam? (6 marks)
Step-by-step solution
1. Step 1
  State the principle of moments (spec 1.32P). When balanced: sum of clockwise moments = sum of anticlockwise moments about any pivot.
2. Step 2
  Beam weight acts at the centre. Because the pivot is AT the centre and the beam is uniform, the beam's own weight produces ZERO moment about the pivot. Ignore it.
3. Step 3
  Set up the equation. Let $d$ be the distance to the right of the pivot at which the 10 N weight hangs.
  $15 \times 0.80 = 10 \times d$
4. Step 4
  Evaluate.
  $12 = 10d \;\;\Rightarrow\;\; d = 1.2\ \text{m}$
Answer
10 N weight must hang 1.2 m to the right of the pivot.
Examiner tip
Edexcel mark scheme requires the principle of moments to be STATED before any numbers go in (1 mark). Always identify which moments are clockwise and which are anticlockwise. If the beam is NOT uniform or NOT pivoted at its centre, include the beam's weight × distance term (spec 1.31P).

Key Formulae — Forces, Movement, Shape and Momentum

The formulae you need to memorise for forces, movement, shape and momentum on the Pearson Edexcel IGCSE 4PH1 paper, with every variable defined in plain English and a note on when to use it.

Newton's second law (spec 1.17)
$F = m \times a$
When to use
Resultant (unbalanced) force = mass × acceleration. Force in N, mass in kg, acceleration in m/s². Direction of $F$ matches direction of $a$ .
Weight (spec 1.18)
$W = m \times g$
When to use
$W$ in newtons, $m$ in kg, $g$ in N/kg (≈ 10 N/kg on Earth, 3.7 N/kg on Mars, 1.6 N/kg on the Moon). Weight is the FORCE due to a gravitational field; mass is NOT affected by $g$ .
Spring constant (spec 1.23)
$F = k \times x$
When to use
In the initial linear region of a force-extension graph (Hooke's law). $k$ has units N/m. Use ONLY below the limit of proportionality.
Momentum (Paper 2, spec 1.25P)
$p = m \times v$
When to use
Vector quantity. Units kg m/s. Direction must be stated. Used with conservation of momentum (spec 1.27P) in collisions and explosions.
Force as rate of change of momentum (Paper 2, spec 1.28P)
$F = \dfrac{mv - mu}{t}$
When to use
Same change of momentum over a LONGER time gives a SMALLER force — the principle behind seat belts, airbags and crumple zones (spec 1.26P).
Moment of a force (Paper 2, spec 1.30P)
$M = F \times d$
When to use
$d$ is the PERPENDICULAR distance from the pivot to the line of action of the force. Units: N m. Clockwise and anticlockwise moments are equal in magnitude when an object balances (spec 1.32P).

Key Definitions and Keywords — Forces, Movement, Shape and Momentum

Definitions to memorise and the exact keywords mark schemes credit for forces, movement, shape and momentum answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4PH1 sitting.

Scalar vs vector (spec 1.13, 1.14)
Examiner keyword
A SCALAR has magnitude only (e.g. mass, speed, energy). A VECTOR has magnitude AND direction (e.g. velocity, acceleration, force, momentum). Force is a vector — both its size AND its direction must be stated.
Resultant force (spec 1.15)
Examiner keyword
The single force that has the same effect as all the individual forces acting on an object combined. For forces along the same line, add forces in the same direction and subtract opposing ones. A non-zero resultant force causes acceleration ( $F = ma$ ).
Friction (spec 1.16)
A force that opposes motion (or attempted motion) between two surfaces in contact. Air resistance is a type of frictional force on objects moving through air.
Weight
Examiner keyword
The force on a mass due to a gravitational field, $W = mg$ . A VECTOR pointing toward the centre of the gravitating body (downward on Earth). Distinct from mass (a scalar, measured in kg).
Stopping distance (spec 1.19)
Examiner keyword
The sum of the THINKING distance (the distance travelled during the driver's reaction time, at constant speed) and the BRAKING distance (the distance travelled while the brakes decelerate the car to rest).
Terminal velocity (spec 1.21)
Examiner keyword
The constant velocity reached by a falling object when the drag force (air resistance) acting upwards equals the weight acting downwards, so the resultant force is zero and the acceleration is zero.
Hooke's law (spec 1.23)
Examiner keyword
Within the initial LINEAR region of a force-extension graph, the extension of a spring is directly proportional to the applied force: $F = kx$ , where $k$ is the spring constant.
Elastic behaviour (spec 1.24)
Examiner keyword
The ability of a material to RECOVER its original shape after the forces causing the deformation have been removed.
Momentum (Paper 2, spec 1.25P)
Examiner keyword
The product of mass and velocity, $p = mv$ . A VECTOR quantity. Units: kg m/s. Conserved in any closed system where no external force acts.
Newton's third law (Paper 2, spec 1.29P)
Examiner keyword
When body A exerts a force on body B, body B exerts an equal and opposite force on body A. The two forces act on DIFFERENT objects (so they do not cancel each other).
Moment of a force (Paper 2, spec 1.30P)
Examiner keyword
The turning effect of a force about a pivot, $M = F \times d$ , where $d$ is the perpendicular distance from the pivot to the line of action of the force. Units: N m.
Centre of gravity (Paper 2, spec 1.31P)
Examiner keyword
The single point through which the entire weight of a body can be considered to act. For a uniform regular object it lies at the geometric centre.

Common Mistakes and Misconceptions — Forces, Movement, Shape and Momentum

The traps other students keep falling into on forces, movement, shape and momentum questions — taken from recent Pearson Edexcel IGCSE 4PH1 examiner reports and mark schemes — and how to avoid them.

✕Treating mass and weight as the same quantity
4PH1 Examiner Reports 2022-2024
Why it happens
In everyday speech we 'weigh' things in kg.
How to avoid it
Mass is a SCALAR in kg, invariant. Weight is a VECTOR force in N, equal to $mg$ . Same object on Earth and Moon: SAME mass, DIFFERENT weight.
✕Identifying the wrong pair for Newton's third law (Paper 2)
4PH1/2P Examiner Reports 2022-2024 (recurring item)
Why it happens
Confusing third-law pairs with the two forces on a single object (e.g. weight + normal reaction on a book on a table).
How to avoid it
Third-law pairs ALWAYS act on DIFFERENT bodies, are the SAME TYPE of force, are EQUAL in magnitude and OPPOSITE in direction. The book pushes down on the table; the table pushes up on the book — that pair is Newton's third law. The weight of the book and the table's normal reaction on it are NOT a third-law pair.
✕Using the wrong distance in the moment formula
Why it happens
Using the slant distance from pivot to the force application point.
How to avoid it
$d$ is the PERPENDICULAR distance from the pivot to the line of action of the force. If the force is angled, use the component perpendicular to the lever arm.
✕Forgetting that momentum is a vector
4PH1/2P Examiner Reports 2022-2024
Why it happens
Calculations look like scalar arithmetic.
How to avoid it
Define a positive direction at the start of any momentum problem. Take velocities in the opposite direction as negative. State direction in your final answer.
✕Applying Hooke's law beyond the limit of proportionality
Why it happens
Treating any force-extension data as obeying $F = kx$ .
How to avoid it
Hooke's law applies only to the INITIAL LINEAR REGION (spec 1.23). Beyond the limit of proportionality, $F/x$ is no longer constant and $F = kx$ cannot be used.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Forces, Movement, Shape and Momentum — frequently asked questions

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

Forces, Movement, Shape and Momentum

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Applying F=maF = maF=ma (4 marks, Paper 1)

2Weight and mass — same object on Earth and Mars (3 marks, Paper 1)

3Hooke's law and spring constant (5 marks, Paper 1)

4Stopping distance components (4 marks, Paper 1)

5Conservation of momentum in a collision (6 marks, Paper 2 — P content)

6F=(mv−mu)/tF = (mv - mu)/tF=(mv−mu)/t for a stopping car (5 marks, Paper 2 — P content)

7Principle of moments — a balanced beam (6 marks, Paper 2 — P content)

Newton's second law (spec 1.17)

Weight (spec 1.18)

Spring constant (spec 1.23)

Momentum (Paper 2, spec 1.25P)

Force as rate of change of momentum (Paper 2, spec 1.28P)

Moment of a force (Paper 2, spec 1.30P)

Scalar vs vector (spec 1.13, 1.14)

Resultant force (spec 1.15)

Friction (spec 1.16)

Weight

Stopping distance (spec 1.19)

Terminal velocity (spec 1.21)

Hooke's law (spec 1.23)

Elastic behaviour (spec 1.24)

Momentum (Paper 2, spec 1.25P)

Newton's third law (Paper 2, spec 1.29P)

Moment of a force (Paper 2, spec 1.30P)

Centre of gravity (Paper 2, spec 1.31P)

✕Treating mass and weight as the same quantity

✕Identifying the wrong pair for Newton's third law (Paper 2)

✕Using the wrong distance in the moment formula

✕Forgetting that momentum is a vector

✕Applying Hooke's law beyond the limit of proportionality

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Forces, Movement, Shape and Momentum — frequently asked questions

1Applying $F = ma$ (4 marks, Paper 1)

6 $F = (mv - mu)/t$ for a stopping car (5 marks, Paper 2 — P content)