What is the basic concept of dynamics in the context of a particle moving in a straight line?

In the context of Edexcel International A Levels Mathematics, specifically Mechanics 1, dynamics refers to the study of forces and their effects on the motion of a particle. When a particle moves in a straight line, its dynamics are governed by Newton's laws of motion, which describe how forces influence the velocity and acceleration of the particle.

How do you calculate the acceleration of a particle moving in a straight line?

To calculate the acceleration of a particle moving in a straight line, you can use the formula a = (v - u) / t, where 'a' is the acceleration, 'v' is the final velocity, 'u' is the initial velocity, and 't' is the time taken for the change in velocity. This formula is derived from the basic principles of kinematics, which are a key part of the Mechanics 1 syllabus in Edexcel International A Levels.

What role does friction play in the dynamics of a particle moving in a plane?

Friction is a force that opposes the motion of a particle moving in a plane. It plays a crucial role in dynamics by affecting the net force acting on the particle. In the context of Edexcel International A Levels Mechanics 1, understanding friction is essential for solving problems related to motion, as it influences both the acceleration and the overall motion of the particle.

How can you determine the resultant force acting on a particle in a plane?

To determine the resultant force acting on a particle in a plane, you need to consider all the individual forces acting on the particle and use vector addition to find their sum. This involves breaking down forces into their components and using trigonometry to calculate the resultant vector. This concept is fundamental in the study of dynamics within the Edexcel International A Levels Mechanics 1 curriculum.

What is the significance of Newton's second law in the dynamics of a particle moving in a straight line?

Newton's second law is pivotal in understanding the dynamics of a particle moving in a straight line. It states that the force acting on a particle is equal to the mass of the particle multiplied by its acceleration (F = ma). This law provides the foundation for analyzing how forces affect the motion of particles, making it a critical component of the Mechanics 1 module in Edexcel International A Levels Mathematics.

Why is "Drawing the tension on a hanging particle as 'downward'" a common mistake in Dynamics of a particle moving in a straight line or plane?

Why it happens: Confusion about what 'tension' means — students think of it as a weight-like pull. How to avoid it: Tension always acts **along the string, away from the particle, toward the pulley/anchor**. For a particle hanging from a string above, tension is UPWARD on the particle.

Why is "Using $T_1$ and $T_2$ for the two sides of a smooth pulley" a common mistake in Dynamics of a particle moving in a straight line or plane?

Why it happens: Mechanical intuition that 'heavier side pulls harder'. How to avoid it: For a **light inextensible** string over a **smooth** pulley, tension is the SAME throughout — use one symbol T. Source: WME01/01 examiner reports — annual

Why is "Writing $T - 5g = 5a$ for the heavier (downward-moving) particle" a common mistake in Dynamics of a particle moving in a straight line or plane?

Why it happens: Reflex to put T first regardless of direction. How to avoid it: Pick a positive direction for each particle BEFORE writing equations. For a particle moving DOWN, take downward positive: weight - tension = ma. For a particle moving UP, upward positive: tension - weight = ma.

Why is "Including weight as a horizontal force when resolving" a common mistake in Dynamics of a particle moving in a straight line or plane?

Why it happens: Mechanical reflex. How to avoid it: Weight always acts VERTICALLY. When resolving horizontally, the weight does NOT appear unless the surface is inclined.

Why is "Including friction in equations for a 'smooth' surface" a common mistake in Dynamics of a particle moving in a straight line or plane?

Why it happens: Force-diagram habit. How to avoid it: **Smooth** means **no friction**. The only contact force is the normal reaction perpendicular to the surface.

Mechanics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Dynamics of a particle moving in a straight line or plane

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Dynamics of a Particle Study Notes — Edexcel IAL Mathematics M1 (WME01/01, 2018 spec — 2026 onwards)

Newton's three laws and their application to single particles, connected particles over smooth pulleys, lifts (apparent weight), and forces applied at angles. Force diagrams, resolving into perpendicular components and $\mathbf{F} = m\mathbf{a}$ are the universal workflow.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

M1 4.1 — State and apply Newton's three laws of motion.
M1 4.2 — Apply $\mathbf{F} = m\mathbf{a}$ to a particle moving in a straight line under one or more forces.
M1 4.3 — Draw force diagrams for single particles and connected-particle systems.
M1 4.4 — Solve connected-particle problems with a light inextensible string over a smooth fixed pulley.
M1 4.5 — Solve lift problems and find the normal reaction (apparent weight).

Newton's three laws

Inertia, $\mathbf{F} = m\mathbf{a}$ , action-reaction.

Newton's first law (inertia). A particle continues in its state of rest or uniform motion in a straight line unless acted upon by a resultant external force.

If $\Sigma \mathbf{F} = \mathbf{0}$ , then $\mathbf{a} = \mathbf{0}$ .
Equivalent statements: 'in equilibrium' and 'moving with constant velocity' have the same force-balance.

Newton's second law. The resultant force on a particle equals its mass times its acceleration:

$\mathbf{F}_{\text{net}} = m\mathbf{a}.$

Vector equation; apply COMPONENT-WISE if forces are not all collinear.
$\mathbf{a}$ is in the direction of $\mathbf{F}_{\text{net}}$ .
Units: $\mathbf{F}$ in newtons (N), $m$ in kg, $\mathbf{a}$ in m s $^{-2}$ . $1\,\text{N} = 1\,\text{kg m s}^{-2}$ .

Newton's third law. If body $A$ exerts a force on body $B$ , then $B$ exerts an equal and opposite force on $A$ .

Used in normal reactions: the ground pushes the box up with force $R$ ; the box pushes the ground down with force $R$ .
Used in connected-particle systems: the string pulls $A$ toward $B$ with tension $T$ ; equally, the string pulls $B$ toward $A$ with the same $T$ .

Universal workflow for an M1 dynamics problem:

Draw the force diagram with every force labelled.
Choose axes (along/perpendicular to motion, or horizontal/vertical).
Resolve each force into those axes.
Apply $\mathbf{F} = m\mathbf{a}$ along each axis.
Solve the resulting equation(s).

Law 1: no net force = no acceleration.
Law 2: $\mathbf{F}_{\text{net}} = m\mathbf{a}$ (vector).
Law 3: equal and opposite reactions.
Workflow: diagram, axes, resolve, $F = ma$ , solve.

Single particle on a horizontal surface

Vertical equilibrium gives $R$ ; horizontal $F = ma$ gives the acceleration.

The simplest scenario. A particle on a horizontal surface, possibly with an applied force.

Forces on the particle:

Weight $W = mg$ vertically down.
Normal reaction $R$ vertically up.
Applied force $P$ in some direction.
Friction $F$ (if the surface is rough) along the surface, opposing motion.

Vertical equilibrium (the particle stays on the surface): sum of vertical forces = 0.

$R + P_y - mg = 0,$

where $P_y$ is the vertical component of $P$ (zero if $P$ is purely horizontal).

Horizontal $F = ma$ (the particle accelerates along the surface):

$P_x - F = m a.$

Worked example. Box of mass $10$ kg pulled by a rope at $30^{\circ}$ above horizontal, tension $50$ N, on a smooth floor (no friction).

$P_x = 50\cos 30^{\circ} \approx 43.30$ N.
$P_y = 50\sin 30^{\circ} = 25$ N (upward).
Vertically: $R + 25 - 98 = 0 \Rightarrow R = 73$ N. (Lifting the rope REDUCES the floor reaction.)
Horizontally: $43.30 = 10 a \Rightarrow a \approx 4.33\,\text{m s}^{-2}$ .

Why does the vertical tension reduce $R$ ? Because the rope SUPPORTS some of the weight. If the rope were vertical (lifting), $R$ would be zero.

Vertical: $R = mg - P_y$ (or $R = mg$ if $P$ is horizontal).
Horizontal: $F_{\text{net,horizontal}} = ma$ .
Applied force at an angle: resolve into components.
Lifting force REDUCES $R$ .

Connected particles over a smooth pulley

Same $T$ throughout; common $|a|$ . Two equations, two unknowns ( $T$ , $a$ ).

Two hanging particles (Atwood machine).

Setup: particles $m_1$ and $m_2$ ( $m_2 > m_1$ ) on opposite sides of a light inextensible string over a smooth fixed pulley.

Force diagrams:

$m_1$ : tension $T$ up, weight $m_1 g$ down. Moves UP with acceleration $a$ .
$m_2$ : tension $T$ up, weight $m_2 g$ down. Moves DOWN with acceleration $a$ .

Equations (positive = direction of motion for each):

$T - m_1 g = m_1 a$ (1)
$m_2 g - T = m_2 a$ (2)

Add (1) + (2):

$(m_2 - m_1) g = (m_1 + m_2) a \Rightarrow a = \dfrac{(m_2 - m_1) g}{m_1 + m_2}.$

Tension (substitute into (1)):

$T = m_1(a + g) = \dfrac{2 m_1 m_2 g}{m_1 + m_2}.$

Force on pulley = $2T$ downward (both sides of the string pull down on the pulley).

Particle on table + hanging particle:

Setup: $m_A$ on smooth horizontal table connected over a pulley at the table's edge to $m_B$ hanging vertically.

$T = m_A a$ (1, no friction)
$m_B g - T = m_B a$ (2)

Add: $m_B g = (m_A + m_B) a \Rightarrow a = m_B g/(m_A + m_B)$ .

Worked example. $m_A = 2$ , $m_B = 3$ , $g = 9.8$ :

$a = 3 \cdot 9.8 / 5 = 5.88\,\text{m s}^{-2}$ .
$T = 2 \cdot 5.88 = 11.76$ N.

Same $T$ on both sides of a smooth pulley (light inextensible string).
Common acceleration magnitude $|a|$ for connected particles.
Equation for each particle; ADD to eliminate $T$ .
Force on pulley = $2T$ (when both sides pull in same direction).

Lifts and apparent weight

$R - mg = ma$ . Up-accelerating lift feels heavier.

The model. A passenger of mass $m$ stands inside a lift. Forces on the passenger:

Weight $W = mg$ down.
Normal reaction $R$ from the lift floor, up.

The passenger has the same acceleration as the lift, $a$ (positive upward).

Newton's second law on the passenger:

$R - mg = m a \Rightarrow R = m(g + a).$

Lift state	$a$	$R$
Stationary or constant velocity	$0$	$mg$ (normal weight)
Accelerating upward	$+$	$> mg$ (heavier feel)
Accelerating downward	$-$	$< mg$ (lighter feel)
Free fall	$-g$	$0$ (weightless)

Apparent weight = $R$ . It's what a bathroom scale in the lift would read.

Worked example. $70$ kg passenger; lift accelerates up at $1.4$ m/s².

$R = 70(9.8 + 1.4) = 70 \cdot 11.2 = 784$ N.
Compared to normal $686$ N — passenger feels heavier by $98$ N.

Sign convention matters: 'lift decelerates while going up' means velocity is up but acceleration is DOWN. Same equation, $a < 0$ , $R < mg$ .

$R = m(g + a)$ with $a > 0$ for upward acceleration.
Up-accelerating lift: $R > mg$ (heavier).
Down-accelerating lift: $R < mg$ (lighter).
Free fall: $R = 0$ (weightless).

Resolving forces at an angle

Component-wise application of Newton's second law.

When to resolve. Any time a force is NOT parallel to one of your chosen axes.

The basic decomposition: a force $F$ at angle $\theta$ above the horizontal has components

$F_x = F\cos\theta, \qquad F_y = F\sin\theta.$

Then apply $F = ma$ in each direction.

Standard problem types:

Particle pulled by a rope at an angle on horizontal floor — resolve tension; vertical balance gives $R$ ; horizontal $F = ma$ gives $a$ .
Particle on an inclined plane — choose axes along/perpendicular to the slope:
- Along: weight contributes $mg\sin\theta$ ; perpendicular: weight contributes $mg\cos\theta$ .
- Perpendicular balance gives $R = mg\cos\theta$ .
- Along: $mg\sin\theta \pm \text{(other forces)} = ma$ .
Particle on an inclined plane with an applied force — also resolve the applied force along/perpendicular.
String at an angle to the ceiling (statics) — vertical balance gives one equation; horizontal balance gives another.

Picking axes: for inclined-plane problems, ALWAYS use axes along/perpendicular to the slope. The 'no perpendicular motion' condition makes one equation simple.

Force at angle $\theta$ : $F_x = F\cos\theta$ , $F_y = F\sin\theta$ .
Apply $F = ma$ in each direction.
Inclined plane: use axes along/perpendicular to slope.
Perpendicular equation usually gives $R$ .

How it’s examined

Dynamics is the largest topic block in WME01/01 — typically $25$ - $35\%$ of the paper. Expect at least one connected-particle question ( $10$ - $14$ marks) and one lift / single-particle problem ( $6$ - $8$ marks). Mark schemes credit separate force diagrams for each particle (B1 each), explicit $F = ma$ equations with sign convention (M1 each) and clean simultaneous solving (A-marks). The most common error: writing $T - 5g = 5a$ for the heavier downward-moving particle (wrong sign on weight). Force on the pulley appears in $\sim 30\%$ of pulley questions and is often missed.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WME01/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Dynamics of a particle moving in a straight line or plane

Step-by-step solutions to past-paper-style questions on dynamics of a particle moving in a straight line or plane, written exactly the way a tutor would explain them at the board.

1Single particle on a horizontal surface (5 marks, M1)
Core• Adapted from WME01/01 January 2024 Q4• dynamics, Newton's second law, horizontal
Question
A particle of mass $4\,\text{kg}$ rests on a smooth horizontal surface. A horizontal force of $12\,\text{N}$ is applied to it. Find the acceleration and the normal reaction.
Step-by-step solution
1. Step 1
  Draw a force diagram. Particle on a horizontal surface: weight $W = 4g = 39.2\,\text{N}$ down, normal reaction $R$ up, applied force $12\,\text{N}$ horizontal.
2. Step 2
  Resolve vertically. The particle has no vertical acceleration (stays on the surface), so vertical forces balance.
  $R - W = 0 \Rightarrow R = 39.2\,\text{N}$
3. Step 3
  Apply Newton's second law horizontally. $F = ma$ .
  $12 = 4 a \Rightarrow a = 3\,\text{m s}^{-2}$
4. Step 4
  Direction: $a = 3$ m s $^{-2}$ horizontally, in the direction of the applied force.
Answer
Acceleration $a = 3\,\text{m s}^{-2}$ ; normal reaction $R = 39.2\,\text{N}$ .
Examiner tip
B1 for force diagram. M1 for resolving vertically. A1 for $R = 39.2$ . M1 for $F = ma$ horizontally. A1 for $a = 3$ . Mark schemes accept $R = 4g$ as an alternative form. Always state direction of acceleration.
2Connected particles over a smooth pulley (10 marks, M1)
Extended• Adapted from WME01/01 June 2024 Q6• dynamics, pulley, connected particles
Question
Particles $A$ (mass $3\,\text{kg}$ ) and $B$ (mass $5\,\text{kg}$ ) are connected by a light inextensible string passing over a smooth fixed pulley. Both particles hang vertically. The system is released from rest. Taking $g = 9.8\,\text{m s}^{-2}$ , find (a) the acceleration of each particle, (b) the tension in the string, (c) the force exerted on the pulley by the string.
Step-by-step solution
1. Step 1
  Draw force diagrams for $A$ and $B$ separately. On $A$ : tension $T$ upward, weight $3g$ downward. On $B$ : tension $T$ upward, weight $5g$ downward. The heavier $B$ accelerates downward; $A$ accelerates upward with the same magnitude $a$ .
2. Step 2
  Apply $F = ma$ to $A$ (upward positive for $A$ ):
  $T - 3g = 3a \quad \text{(1)}$
3. Step 3
  Apply $F = ma$ to $B$ (downward positive for $B$ ):
  $5g - T = 5a \quad \text{(2)}$
4. Step 4
  Add (1) and (2) to eliminate $T$ :
  $5g - 3g = 5a + 3a \Rightarrow 2g = 8a \Rightarrow a = g/4 = 2.45\,\text{m s}^{-2}$
5. Step 5
  Substitute back into (1):
  $T = 3a + 3g = 3 \cdot 2.45 + 3 \cdot 9.8 = 7.35 + 29.4 = 36.75\,\text{N}$
6. Step 6
  Force on pulley: both sides of the string pull downward on the pulley with tension $T$ . So force on pulley = $2T = 73.5\,\text{N}$ vertically downward.
Answer
(a) $a = 2.45\,\text{m s}^{-2}$ ( $A$ up, $B$ down). (b) $T = 36.75\,\text{N}$ . (c) Force on pulley $= 73.5\,\text{N}$ downward.
Examiner tip
B1 for force diagrams labelled. M1 for equation on $A$ . A1. M1 for equation on $B$ . A1. M1 for solving simultaneously. A1 for $a$ . A1 for $T$ . M1 for force on pulley $= 2T$ . A1. Common error: writing $T - 5g = 5a$ for $B$ (wrong sign) — heavier particle's weight EXCEEDS tension, so weight - tension = $ma$ .
3Particle on table connected to hanging particle (10 marks, M1)
Extended• Adapted from WME01/01 January 2023 Q5• dynamics, pulley, smooth table
Question
Particle $A$ (mass $2\,\text{kg}$ ) rests on a smooth horizontal table. It is connected by a light inextensible string passing over a smooth pulley at the edge of the table to particle $B$ (mass $3\,\text{kg}$ ) hanging vertically. The system is released from rest. Find (a) the acceleration of the system, (b) the tension in the string.
Step-by-step solution
1. Step 1
  Draw force diagrams. $A$ on the table: tension $T$ horizontally toward the pulley, weight $2g$ down, normal reaction $R$ up. Smooth, so no friction. $B$ hanging: tension $T$ up, weight $3g$ down.
2. Step 2
  Both particles share acceleration magnitude $a$ : $A$ horizontally toward pulley, $B$ vertically downward.
3. Step 3
  Equation for $A$ (horizontal, toward pulley positive):
  $T = 2 a \quad \text{(1)}$
4. Step 4
  Equation for $B$ (downward positive):
  $3g - T = 3a \quad \text{(2)}$
5. Step 5
  Add (1) and (2):
  $3g = 5a \Rightarrow a = \dfrac{3g}{5} = \dfrac{3 \cdot 9.8}{5} = 5.88\,\text{m s}^{-2}$
6. Step 6
  Tension:
  $T = 2a = 2 \cdot 5.88 = 11.76\,\text{N}$
Answer
(a) Acceleration $a = 5.88\,\text{m s}^{-2}$ . (b) Tension $T = 11.76\,\text{N}$ .
Examiner tip
B1 for diagrams. M1A1 for equation on $A$ (with no friction term). M1A1 for equation on $B$ . M1A1 for solving. A1 for $a$ and A1 for $T$ . Watch the geometry: the tension on $A$ is horizontal (acting along the table toward the pulley); it's NOT $T\cos\theta$ for any angle.
4Lift acceleration and apparent weight (8 marks, M1)
Extended• Adapted from WME01/01 June 2023 Q3• dynamics, lift, apparent weight
Question
A passenger of mass $70\,\text{kg}$ stands inside a lift. The lift accelerates upward at $1.4\,\text{m s}^{-2}$ . Find the magnitude of the normal reaction exerted on the passenger by the lift floor. State whether the passenger feels 'heavier' or 'lighter' than usual and explain why.
Step-by-step solution
1. Step 1
  Draw the force diagram for the passenger. Two forces: normal reaction $R$ upward (from floor), weight $W = 70g$ downward.
2. Step 2
  Apply Newton's second law to the passenger, taking upward positive. The passenger accelerates upward at $1.4\,\text{m s}^{-2}$ .
  $R - 70g = 70 \cdot 1.4$
3. Step 3
  Solve for $R$ :
  $R = 70 \cdot 9.8 + 70 \cdot 1.4 = 686 + 98 = 784\,\text{N}$
4. Step 4
  Compare with the usual weight $686\,\text{N}$ : the floor pushes UP harder than usual, so the passenger feels heavier — the 'apparent weight' is $784\,\text{N}$ .
5. Step 5
  Physical interpretation: the passenger's body must be accelerated upward at $1.4\,\text{m s}^{-2}$ , so the floor must exert an extra force $ma = 98\,\text{N}$ on top of supporting the weight.
Answer
$R = 784\,\text{N}$ . The passenger feels heavier because the floor exerts a greater upward force than the weight (the extra $98\,\text{N}$ produces the upward acceleration).
Examiner tip
B1 for force diagram. M1 for $R - mg = ma$ . A1 for $R = 784$ . B1 for 'heavier'. B1 for the explanation. If the lift accelerated DOWNWARD at $1.4$ , the answer would be $R = 70g - 70 \cdot 1.4 = 588\,\text{N}$ (lighter). If the lift were in free fall ( $a = -g$ ), $R = 0$ (weightless).
5Particle dragged at an angle (8 marks, M1)
Extended• Adapted from WME01/01 January 2024 Q7• dynamics, resolving, applied force
Question
A box of mass $10\,\text{kg}$ is dragged across a smooth horizontal floor by a rope inclined at $30^{\circ}$ above the horizontal. The tension in the rope is $50\,\text{N}$ . Find (a) the acceleration of the box, (b) the normal reaction between the box and the floor.
Step-by-step solution
1. Step 1
  Draw a force diagram. On the box: weight $W = 10g = 98\,\text{N}$ downward, normal reaction $R$ upward, tension $T = 50\,\text{N}$ along the rope (at $30^{\circ}$ above horizontal).
2. Step 2
  Resolve the tension into horizontal and vertical components.
  $T_x = 50\cos 30^{\circ} = 25\sqrt{3} \approx 43.30\,\text{N}, \quad T_y = 50\sin 30^{\circ} = 25\,\text{N}$
3. Step 3
  Resolve vertically (no vertical acceleration — box stays on floor):
  $R + T_y - W = 0 \Rightarrow R = 98 - 25 = 73\,\text{N}$
4. Step 4
  Apply $F = ma$ horizontally (smooth floor, no friction):
  $T_x = 10 a \Rightarrow 43.30 = 10 a \Rightarrow a \approx 4.33\,\text{m s}^{-2}$
5. Step 5
  Note that lifting the rope reduces $R$ (the vertical tension component helps support the box) — useful for problems with friction, since reduced $R$ means reduced friction.
Answer
(a) Acceleration $a \approx 4.33\,\text{m s}^{-2}$ horizontally. (b) Normal reaction $R = 73\,\text{N}$ .
Examiner tip
B1 for diagram. M1 for resolving tension into components. A1 for $T_x$ , A1 for $T_y$ . M1 for resolving vertically. A1 for $R = 73$ . M1 for $F = ma$ horizontally. A1 for $a$ . A common error: thinking the vertical component of tension adds to the weight, giving $R = 98 + 25 = 123$ . Draw arrows carefully — the vertical tension is UP, so it REDUCES the floor reaction needed.

Key Formulae — Dynamics of a particle moving in a straight line or plane

The formulae you need to memorise for dynamics of a particle moving in a straight line or plane on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Newton's second law (vector form)
$\mathbf{F}_{\text{net}} = m\mathbf{a}$
$\mathbf{F}_{\text{net}}$
resultant (net) force (N)
$m$
mass (kg)
$\mathbf{a}$
acceleration (m s $^{-2}$ )
When to use
Apply once you have all the forces on a particle. Sum forces (with signs/components) and equate to $m\mathbf{a}$ . NOT in the IAL formula booklet — assumed knowledge.
Example
Box of mass $10$ kg pulled by $50$ N at $30^{\circ}$ horizontally on smooth floor: $50\cos 30^{\circ} = 10 a \Rightarrow a = 4.33\,\text{m s}^{-2}$ .
Resolving forces into components
$F_x = F\cos\theta, \quad F_y = F\sin\theta$
$F$
magnitude of force
$\theta$
angle from the chosen axis (typically horizontal)
When to use
Whenever a force is applied at an angle — split into perpendicular components, then apply Newton's second law in each direction. NOT in the formula booklet.
Example
$50$ N at $30^{\circ}$ above horizontal: $F_x = 50\cos 30^{\circ} \approx 43.3$ , $F_y = 50\sin 30^{\circ} = 25$ .
Weight
$W = mg$
$W$
weight (N)
$m$
mass (kg)
$g$
$9.8\,\text{m s}^{-2}$ (Edexcel convention)
When to use
Every force diagram begins with the weight, acting vertically downward at the centre of mass. Distinct from mass (which is in kg).
Example
$70$ kg passenger: $W = 70 \times 9.8 = 686\,\text{N}$ .
Connected particles equations
$T - m_1 g = m_1 a \text{ (lighter, moving up)}, \quad m_2 g - T = m_2 a \text{ (heavier, moving down)}$
$T$
common tension in the light inextensible string
$m_1, m_2$
masses on either side
$a$
common acceleration magnitude
When to use
Two particles connected by a light inextensible string over a smooth fixed pulley. Add the two equations to eliminate $T$ : $a = (m_2 - m_1)g/(m_1 + m_2)$ . Substitute back for $T$ .
Example
$m_1 = 3$ , $m_2 = 5$ : $a = 2g/8 = g/4 \approx 2.45$ ; $T = 3(a + g) \approx 36.75$ N.
Force on a smooth pulley
$F_{\text{pulley}} = 2T$
$T$
tension in the string
$F_{\text{pulley}}$
magnitude of the force exerted on the pulley by the string
When to use
When both sides of the string pull in the same direction on the pulley (e.g. vertically when the pulley is at the top and both particles hang). For a string going over a corner the two tensions may not be collinear — then use vector addition with the angle between them.
Example
$T = 36.75\,\text{N}$ on a pulley at top of system: force on pulley $= 2 \cdot 36.75 = 73.5\,\text{N}$ downward.

Key Definitions and Keywords — Dynamics of a particle moving in a straight line or plane

Definitions to memorise and the exact keywords mark schemes credit for dynamics of a particle moving in a straight line or plane answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Newton's first law
Examiner keyword
A particle continues in a state of rest or of uniform motion in a straight line unless acted upon by a resultant external force. Equivalent: zero net force $\Rightarrow$ zero acceleration.
Newton's second law
Examiner keyword
The resultant force on a particle equals mass times acceleration: $\mathbf{F} = m\mathbf{a}$ . The acceleration is in the direction of the resultant force.
Newton's third law
Examiner keyword
If body $A$ exerts a force on body $B$ , then $B$ exerts an equal and opposite force on $A$ . Used implicitly when modelling normal reactions (the ground pushes the box up; the box pushes the ground down with equal force).
Resultant force
Examiner keyword
The vector sum of all forces acting on the particle. Used in Newton's second law: $\mathbf{F}_{\text{net}} = m\mathbf{a}$ .
Tension
Examiner keyword
The internal force in a stretched string or rod, pulling along its length. In a light inextensible string over a smooth pulley, tension is the same throughout.
Normal reaction
Examiner keyword
The contact force from a surface, acting perpendicular ('normal') to the surface, pushing the particle away. Magnitude adjusts so the particle does not penetrate the surface.
Apparent weight
The magnitude of the normal reaction exerted on a person by the floor (or seat) of an accelerating vehicle/lift. Equal to actual weight at rest or constant velocity; greater when accelerating upward; less when accelerating downward; zero in free fall.

Common Mistakes and Misconceptions — Dynamics of a particle moving in a straight line or plane

The traps other students keep falling into on dynamics of a particle moving in a straight line or plane questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Drawing the tension on a hanging particle as 'downward'
Why it happens
Confusion about what 'tension' means — students think of it as a weight-like pull.
How to avoid it
Tension always acts along the string, away from the particle, toward the pulley/anchor. For a particle hanging from a string above, tension is UPWARD on the particle.
✕Using $T_1$ and $T_2$ for the two sides of a smooth pulley
WME01/01 examiner reports — annual
Why it happens
Mechanical intuition that 'heavier side pulls harder'.
How to avoid it
For a light inextensible string over a smooth pulley, tension is the SAME throughout — use one symbol $T$ .
✕Writing $T - 5g = 5a$ for the heavier (downward-moving) particle
Why it happens
Reflex to put $T$ first regardless of direction.
How to avoid it
Pick a positive direction for each particle BEFORE writing equations. For a particle moving DOWN, take downward positive: weight $-$ tension $= ma$ . For a particle moving UP, upward positive: tension $-$ weight $= ma$ .
✕Including weight as a horizontal force when resolving
Why it happens
Mechanical reflex.
How to avoid it
Weight always acts VERTICALLY. When resolving horizontally, the weight does NOT appear unless the surface is inclined.
✕Including friction in equations for a 'smooth' surface
Why it happens
Force-diagram habit.
How to avoid it
Smooth means no friction. The only contact force is the normal reaction perpendicular to the surface.

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.