Newton’s laws of motion: force, mass, acceleration, connected particles

Newton’s Laws of Motion and Their Application to Connected Particles

1. Newton’s First Law – Law of Inertia

• A particle remains at rest or moves with constant velocity unless a net external force acts on it.
• Mathematical form $$\sum\mathbf{F}=0\;\Longrightarrow\;\mathbf{v}= \text{constant}$$
• Key idea: a body resists any change to its state of motion (inertia).

2. Newton’s Second Law – Law of Acceleration

• Rate of change of momentum of a particle equals the net external force, and the force is directed along the change of momentum.
• General form $$\sum\mathbf{F}= \frac{d\mathbf{p}}{dt}$$
• If the mass \(m\) is constant, this reduces to the familiar form $$\sum\mathbf{F}=m\mathbf{a},\qquad \mathbf{a}= \frac{d\mathbf{v}}{dt}$$
• Units
  • Force \(F\) – newton (N) = kg·m·s\(^{-2}\)
  • Mass \(m\) – kilogram (kg)
  • Acceleration \(a\) – metre per second squared (m·s\(^{-2}\))

3. Newton’s Third Law – Action–Reaction

• For every action force there is an equal and opposite reaction force.
• Mathematical expression $$\mathbf{F}_{AB}= -\mathbf{F}_{BA}$$
• The two forces act on different bodies; therefore they never cancel in the free‑body diagram of a single particle.

4. Forces & Equilibrium (Syllabus 4.1)

4.1 Definition of a Force

• A vector quantity that can cause a change in the state of motion of a body.
• Typical forces: weight, normal reaction, tension, friction, applied force, spring force.
• Exam tip: Vector notation (bold or arrow symbols) is not required in the Cambridge papers – plain letters are acceptable as long as the direction is clear.

4.2 Free‑Body Diagrams (FBDs)

• Draw the object as a point (or simple shape) and represent every external force acting on it as an arrow.
• Adopt a consistent convention for direction (e.g. right → positive x, up → positive y).
• Label each force (W, N, T, \(f\), \(F\), etc.).

4.3 Resolving Forces into Components

When a force \(\mathbf{F}\) makes an angle \(\theta\) with the chosen axes:

\[ F_x = F\cos\theta,\qquad F_y = F\sin\theta \]

Use right‑hand trigonometry; remember the sign convention.

4.4 Resultant Force

• Resultant in the x‑direction: \(\displaystyle R_x = \sum F_{x}\)
• Resultant in the y‑direction: \(\displaystyle R_y = \sum F_{y}\)
• Magnitude and direction: \[ R = \sqrt{R_x^{2}+R_y^{2}},\qquad \tan\phi = \frac{R_y}{R_x} \]

4.5 Equilibrium Conditions

A body is in static equilibrium when the vector sum of all external forces is zero:

\[ \sum\mathbf{F}= \mathbf{0}\;\Longrightarrow\; \begin{cases} \sum F_x = 0\\[2mm] \sum F_y = 0 \end{cases} \]

4.6 Friction

• Limiting (maximum) static friction: \(f_{\max}= \mu_s N\)
• Kinetic friction (when sliding): \(f_k = \mu_k N\)
• \(\mu_s\) and \(\mu_k\) are dimensionless coefficients of static and kinetic friction respectively.
• Normal reaction \(N\) is the component of the contact force perpendicular to the surface.

4.7 Smooth (Frictionless) Contact

• If a surface is smooth, the frictional component is zero; only the normal reaction acts perpendicular to the surface.
• This simplification is common in pulley and inclined‑plane questions.

4.8 Lami’s Theorem (Optional)

If three concurrent forces keep a particle in equilibrium, each force is proportional to the sine of the angle between the other two:

\[ \frac{F_1}{\sin\alpha}= \frac{F_2}{\sin\beta}= \frac{F_3}{\sin\gamma} \]

4.9 Example – Block on a Rough Incline

1. Draw the FBD: weight \(W=mg\) (vertical), normal \(N\) (perpendicular to plane), friction \(f\) (up the plane if the block tends to slide down), and any applied force \(P\) parallel to the plane.
2. Resolve weight: \[ W_{\parallel}= mg\sin\theta,\qquad W_{\perp}= mg\cos\theta \]
3. Equilibrium equations (taking up the plane as positive): \[ \sum F_{\parallel}=0:\; P + f - mg\sin\theta =0 \] \[ \sum F_{\perp}=0:\; N - mg\cos\theta =0 \]
4. If the block is just about to move, set \(f = \mu_s N\) and solve for the required \(P\) or for the limiting angle \(\theta\).

5. Kinematics of Motion in a Straight Line (Syllabus 4.2)

5.1 Basic Quantities

Symbol	Quantity	Units
s, x	Displacement (vector) / distance (scalar)	metre (m)
v, u	Velocity / initial velocity	metre per second (m s\(^{-1}\))
a, a_0	Acceleration / initial acceleration	metre per second squared (m s\(^{-2}\))
t	Time	second (s)

5.2 Graphical Interpretation

• Velocity–time graph: gradient = acceleration, area = displacement.
• Displacement–time graph: gradient = velocity.
• For constant acceleration, the velocity–time graph is a straight line; the displacement–time graph is a parabola.

5.3 Constant‑Acceleration Equations (derived from calculus)

\[ \begin{aligned} v &= u + at\\[2mm] s &= ut + \tfrac12 at^{2}\\[2mm] v^{2} &= u^{2} + 2as \end{aligned} \]

5.4 Derivation from Differentiation (Pure‑Mathematics 1 link)

Starting from the definition of displacement \(x(t)\):

\[ v(t)=\frac{dx}{dt},\qquad a(t)=\frac{dv}{dt}= \frac{d^{2}x}{dt^{2}} \]

Integrating under the assumption of constant \(a\) gives the three equations above. This connection is useful when the exam asks for a brief justification.

5.5 Example – Uniformly Accelerated Motion

A car starts from rest, accelerates uniformly at \(2\;\text{m s}^{-2}\) for \(5\) s.

1. Final velocity: \(v = 0 + (2)(5) = 10\;\text{m s}^{-1}\).
2. Distance travelled: \(s = 0\cdot5 + \tfrac12(2)(5)^{2}=25\;\text{m}\).
3. On a velocity–time graph the line rises from \((0,0)\) to \((5\text{ s},10\text{ m s}^{-1})\); the triangular area under the line equals \(25\) m, confirming the calculation.

6. Momentum (Syllabus 4.3)

6.1 Linear Momentum

• Defined for a particle of mass \(m\) moving with velocity \(\mathbf{v}\) as \(\mathbf{p}=m\mathbf{v}\) (kg·m·s\(^{-1}\)).
• Newton’s second law can be written \(\displaystyle\sum\mathbf{F}= \frac{d\mathbf{p}}{dt}\).

6.2 Impulse

• Impulse \(\mathbf{J}\) is the integral of net force over the time interval \(\Delta t\): \[ \mathbf{J}= \int_{t_1}^{t_2}\mathbf{F}\,dt = \Delta\mathbf{p} \]
• For a constant force, \(\displaystyle\mathbf{J}= \mathbf{F}\Delta t\).

6.3 Conservation of Momentum (One‑Dimensional Collisions)

• If the net external force on a system of particles is zero, the total momentum of the system is constant:
\[ \sum\mathbf{p}_{\text{initial}} = \sum\mathbf{p}_{\text{final}} \]
• Applicable to elastic and inelastic collisions in the Cambridge papers.

6.4 Example – Elastic Collision on a Frictionless Track

Two carts of masses \(m_1=2\;\text{kg}\) and \(m_2=3\;\text{kg}\) move along a straight, frictionless track. Cart 1 moves to the right with \(u_1=4\;\text{m s}^{-1}\); cart 2 is initially at rest. After a perfectly elastic collision, cart 1 rebounds with speed \(v_1\) and cart 2 moves forward with speed \(v_2\).

1. Conservation of momentum: \[ m_1u_1 = m_1v_1 + m_2v_2 \]
2. Conservation of kinetic energy (elastic case): \[ \tfrac12 m_1u_1^{2}= \tfrac12 m_1v_1^{2}+ \tfrac12 m_2v_2^{2} \]
3. Solving gives \(v_1 = -\dfrac{m_1-m_2}{m_1+m_2}u_1 = -\dfrac{2-3}{5}\times4 = 0.8\;\text{m s}^{-1}\) (to the left) and \(v_2 = \dfrac{2m_1}{m_1+m_2}u_1 = \dfrac{4}{5}\times4 = 3.2\;\text{m s}^{-1}\) (to the right).

7. Newton’s Laws Applied to Particles & Connected Particles (Syllabus 4.4)

7.1 System of Particles

A system consists of two or more particles that may interact via internal forces and may be acted upon by external forces.

7.2 Centre of Mass (CM)

\[ \mathbf{R}= \frac{\displaystyle\sum_{i} m_i\mathbf{r}_i}{\displaystyle\sum_{i} m_i}= \frac{1}{M}\sum_i m_i\mathbf{r}_i, \qquad M=\sum_i m_i. \]

7.3 Newton’s Second Law for a System

\[ \sum\mathbf{F}_{\text{ext}} = M\mathbf{A}_{\text{CM}},\qquad \mathbf{A}_{\text{CM}} = \frac{d^{2}\mathbf{R}}{dt^{2}}. \]

7.4 Internal Forces

• Internal forces occur in equal‑and‑opposite pairs (Newton’s third law): \(\mathbf{F}_{ij}= -\mathbf{F}_{ji}\).
• Consequently \(\displaystyle\sum\mathbf{F}_{\text{int}} = \mathbf{0}\); they do not affect the motion of the centre of mass.

7.5 When to Use the System (Centre‑of‑Mass) Approach – Checklist

• All particles share the same linear acceleration (e.g. linked by light strings or rigid rods).
• The question asks directly for the acceleration of the whole system or for the motion of the centre of mass.
• Internal forces (tension, normal contact) are difficult or unnecessary to calculate individually.
• External forces are easily identified and can be summed vectorially.

7.6 Example – Two‑Mass System on a Frictionless Surface

Two blocks of masses \(m_1\) and \(m_2\) are connected by a light string. A horizontal force \(F\) is applied to \(m_1\).

1. Particle‑by‑particle method \[ \begin{aligned} &m_1:\; F - T = m_1 a\\ &m_2:\; T = m_2 a \end{aligned} \] Eliminating the tension \(T\) gives \(a = \dfrac{F}{m_1+m_2}\).
2. System (CM) method \[ \sum F_{\text{ext}} = F = (m_1+m_2)a \;\Longrightarrow\; a = \frac{F}{m_1+m_2} \] The same result is obtained, illustrating that internal tension forces cancel.

8. Work, Energy & Power (Syllabus 4.5 – brief overview)

8.1 Work

• Work done by a constant force \(\mathbf{F}\) acting through a displacement \(\mathbf{s}\) at an angle \(\theta\): \[ W = F s\cos\theta = \mathbf{F}\!\cdot\!\mathbf{s} \]
• Unit: joule (J) = N·m = kg·m\(^2\)·s\(^{-2}\).

8.2 Kinetic Energy

\[ E_k = \tfrac12 mv^{2} \]

Work‑Energy Theorem: net work done on a particle equals the change in its kinetic energy, \(\displaystyle W_{\text{net}} = \Delta E_k\).

8.3 Potential Energy (Gravitational)

\[ E_p = mgh \]

For a conservative force, the work done equals the negative change in potential energy, \(W = -\Delta E_p\).

8.4 Power

\[ P = \frac{W}{t} = \mathbf{F}\!\cdot\!\mathbf{v} \]

Unit: watt (W) = J·s\(^{-1}\).

9. Summary Table of Newton’s Laws

Law	Statement	Mathematical Form	Key Implications
First	A body remains at rest or in uniform straight‑line motion unless acted upon by a net external force.	\(\displaystyle\sum\mathbf{F}=0\;\Longrightarrow\;\mathbf{v}= \text{constant}\)	Defines inertia; introduces the concept of net force.
Second	The net external force on a body equals the rate of change of its momentum.	\(\displaystyle\sum\mathbf{F}= \frac{d\mathbf{p}}{dt}=m\mathbf{a}\)	Quantitative link between force, mass and acceleration; basis for system analysis.
Third	For every action there is an equal and opposite reaction.	\(\displaystyle\mathbf{F}_{AB}= -\mathbf{F}_{BA}\)	Internal forces cancel in a system; essential for connected‑particle problems.

10. Key Points to Remember

• All forces are vectors – resolve them into components before applying \(\sum F_x =0\) and \(\sum F_y =0\).
• Mass is invariant in Newtonian mechanics (no relativistic effects in the Cambridge syllabus).
• Newton’s laws are valid only in inertial frames; ensure the chosen reference frame is not accelerating.
• When analysing a system, treat the centre of mass as a single particle acted on by the resultant external force.
• Internal forces obey Newton’s third law and never appear in \(\sum\mathbf{F}_{\text{ext}} = M\mathbf{A}_{\text{CM}}\).
• Vector notation is optional in the exam – clear labeling of direction suffices.
• For momentum problems, remember: \(\displaystyle\Delta\mathbf{p}= \mathbf{F}\Delta t\) and total momentum is conserved when \(\sum\mathbf{F}_{\text{ext}}=0\).
• Work‑energy and power relations provide an alternative route to solving dynamics problems, especially when forces vary.