Dynamics (Extended)

This document provides a rigorous treatment of Newton’s laws, connected particles, pulleys, friction, Inclined planes, momentum, and impulse.

Newton’s First Law (Law of Inertia). A body remains at rest or continues to move with constant Velocity unless acted upon by a resultant external force.

Newton’s Second Law. The resultant force acting on a body is equal to the rate of change of Momentum:

$\mathbf{F} = \frac{d\mathbf{p}}{dt} = \frac{d(m\mathbf{v})}{dt}$

For constant mass $m$:

$\boxed{\mathbf{F} = m\mathbf{a}}$

Newton’s Third Law. If body $A$ exerts a force on body $B$Then body $B$ exerts an equal and Opposite force on body $A$:

$\mathbf{F}_{AB} = -\mathbf{F}_{BA}$

The weight of a body of mass $m$ near the Earth’s surface is:

$W = mg$

Directed vertically downward, where $g \approx 9.8\;\mathrm{m\,s^{-2}}$.

The normal reaction $R$ is the force exerted by a surface on a body, always perpendicular to the Surface.

Problem. A woman of mass $70\;\mathrm{kg}$ stands in a lift. Find the normal reaction from the Floor of the lift when: (a) the lift accelerates upward at $2\;\mathrm{m\,s^{-2}}$; (b) the lift Moves at constant velocity; (c) the lift decelerates at $3\;\mathrm{m\,s^{-2}}$ while moving upward.

Let $R$ be the upward normal reaction. Taking upward as positive:

$R - mg = ma$

(a) $R = 70(9.8 + 2) = 70 \times 11.8 = 826\;\mathrm{N}$

(b) $R = 70(9.8 + 0) = 686\;\mathrm{N}$

(c) Decelerating upward means $a = -3$: $R = 70(9.8 - 3) = 70 \times 6.8 = 476\;\mathrm{N}$

For systems of connected particles:

1. Draw a separate free-body diagram for each particle.
2. Identify the tension $T$ in any connecting string (same magnitude on each particle for an inextensible string).
3. Apply Newton’s second law to each particle separately.
4. If the string is inextensible, both particles have the same magnitude of acceleration.
5. Solve the resulting simultaneous equations.

Problem. Two particles of masses $3\;\mathrm{kg}$ and $5\;\mathrm{kg}$ are connected by a light Inextensible string. The $3\;\mathrm{kg}$ particle lies on a smooth horizontal table and the $5\;\mathrm{kg}$ particle hangs freely over the edge. Find the acceleration of the system and the Tension in the string.

For the $5\;\mathrm{kg}$ particle (taking downward as positive):

5g - T = 5a \tag{1}

For the $3\;\mathrm{kg}$ particle (taking the direction of motion as positive):

T = 3a \tag{2}

Adding (1) and (2):

$5g = 8a \implies a = \frac{5g}{8} = \frac{49}{8} = 6.125\;\mathrm{m\,s^{-2}}$

$T = 3 \times 6.125 = 18.375\;\mathrm{N}$

Problem. A car of mass $1200\;\mathrm{kg}$ tows a trailer of mass $400\;\mathrm{kg}$ along a Horizontal road. The engine provides a driving force of $3200\;\mathrm{N}$. Resistance forces of $200\;\mathrm{N}$ and $100\;\mathrm{N}$ act on the car and trailer respectively. Find the Acceleration and the tension in the tow bar.

System as a whole:

$(1200 + 400)a = 3200 - 200 - 100 = 2900$

$a = \frac{2900}{1600} = 1.8125\;\mathrm{m\,s^{-2}}$

Trailer alone:

$T - 100 = 400a = 400 \times 1.8125 = 725$

$T = 825\;\mathrm{N}$

A smooth, light pulley changes only the direction of the tension — the tension is the same on Both sides of the string.

Problem. Two masses $4\;\mathrm{kg}$ and $6\;\mathrm{kg}$ are connected by a light inextensible String over a smooth, light pulley. Find the acceleration and the tension.

The heavier mass accelerates downward. Let $a$ be the acceleration magnitude.

For the $6\;\mathrm{kg}$ mass (downward positive):

6g - T = 6a \tag{1}

For the $4\;\mathrm{kg}$ mass (upward positive):

T - 4g = 4a \tag{2}

Adding (1) and (2):

$2g = 10a \implies a = \frac{g}{5} = 1.96\;\mathrm{m\,s^{-2}}$

From (2): $T = 4(g + a) = 4 \times 11.76 = 47.04\;\mathrm{N}$

Problem. A particle of mass $8\;\mathrm{kg}$ on a smooth plane inclined at $30^\circ$ is Connected by a light inextensible string over a smooth pulley at the top of the plane to a particle Of mass $5\;\mathrm{kg}$ hanging freely. Find the acceleration and tension.

Assuming the $8\;\mathrm{kg}$ mass moves up the plane (we will check this assumption):

For the $8\;\mathrm{kg}$ mass (up the plane positive):

T - 8g\sin 30^\circ = 8a \implies T - 4g = 8a \tag{1}

For the $5\;\mathrm{kg}$ mass (downward positive):

5g - T = 5a \tag{2}

Adding: $5g - 4g = 13a \implies a = \dfrac{g}{13} \approx 0.754\;\mathrm{m\,s^{-2}}$.

Since $a \gt 0$Our assumption was correct.

$T = 5(g - a) = 5(9.8 - 0.754) = 45.23\;\mathrm{N}$

Static friction $F_s$ prevents a body from starting to move. It satisfies:

$F_s \leq \mu_s R$

Where $\mu_s$ is the coefficient of static friction and $R$ is the normal reaction.

Dynamic (kinetic) friction $F_d$ acts when a body is sliding:

$F_d = \mu_d R$

Where $\mu_d$ is the coefficient of dynamic friction. In practice, $\mu_d \lt \mu_s$.

At A Level, a single coefficient $\mu$ is used, and we write:

$F \leq \mu R \quad (\mathrm{limiting\ equilibrium})$

$F = \mu R \quad (\mathrm{when\ sliding})$

The angle of friction $\lambda$ is defined by $\tan\lambda = \mu$.

When a body is on the point of sliding on an inclined plane, the angle of the plane equals the Angle of friction:

$\tan\alpha = \mu$

Problem. A block of mass $10\;\mathrm{kg}$ rests on a rough plane inclined at $25^\circ$ to the Horizontal. The coefficient of friction is $\mu = 0.3$. A horizontal force of $P\;\mathrm{N}$ is Applied to the block. Find the range of values of $P$ for which the block remains in equilibrium.

Resolving perpendicular to the plane (upward from plane positive):

$R - 10g\cos 25^\circ - P\sin 25^\circ = 0$

$R = 10g\cos 25^\circ + P\sin 25^\circ$

Resolving up the plane:

$P\cos 25^\circ - 10g\sin 25^\circ - F = 0$

Case 1: on the point of sliding up the plane ($F = \mu R$Acting down the plane):

$P\cos 25^\circ - 10g\sin 25^\circ = \mu(10g\cos 25^\circ + P\sin 25^\circ)$

$P\cos 25^\circ - \mu P\sin 25^\circ = 10g\sin 25^\circ + 10\mu g\cos 25^\circ$

$P(\cos 25^\circ - 0.3\sin 25^\circ) = 10g(\sin 25^\circ + 0.3\cos 25^\circ)$

$P = \frac{10(9.8)(0.4226 + 0.3 \times 0.9063)}{0.9063 - 0.3 \times 0.4226} = \frac{98 \times 0.6945}{0.7795} \approx 87.3\;\mathrm{N}$

Case 2: on the point of sliding down the plane ($F = \mu R$Acting up the plane):

$P\cos 25^\circ - 10g\sin 25^\circ + \mu R = 0$

$P\cos 25^\circ + \mu(10g\cos 25^\circ + P\sin 25^\circ) = 10g\sin 25^\circ$

$P(\cos 25^\circ + 0.3\sin 25^\circ) = 10g(\sin 25^\circ - 0.3\cos 25^\circ)$

$P = \frac{98(0.4226 - 0.2719)}{0.9063 + 0.1268} = \frac{98 \times 0.1507}{1.0331} \approx 14.3\;\mathrm{N}$

Therefore, for equilibrium: $14.3 \leq P \leq 87.3\;\mathrm{N}$.

The linear momentum of a body of mass $m$ moving with velocity $\mathbf{v}$ is:

$\mathbf{p} = m\mathbf{v}$

Momentum is a vector quantity measured in $\mathrm{kg\,m\,s^{-1}}$ (or $\mathrm{Ns}$).

Theorem. If no external resultant force acts on a system of particles, the total momentum of the System is conserved.

$m_1\mathbf{u}_1 + m_2\mathbf{u}_2 = m_1\mathbf{v}_1 + m_2\mathbf{v}_2$

Proof. Newton’s second law for the system: $\mathbf{F}_{\mathrm{ext}} = \dfrac{d\mathbf{p}_{\mathrm{total}}}{dt}$.

If $\mathbf{F}_{\mathrm{ext}} = \mathbf{0}$Then $\mathbf{p}_{\mathrm{total}}$ is constant. $\blacksquare$

The impulse of a constant force $\mathbf{F}$ acting over a time interval $\Delta t$ is:

$\mathbf{I} = \mathbf{F}\,\Delta t$

By Newton’s second law, impulse equals change in momentum:

$\boxed{\mathbf{I} = \mathbf{F}\,\Delta t = m\mathbf{v} - m\mathbf{u} = \Delta\mathbf{p}}$

For a variable force:

$\mathbf{I} = \int_{t_1}^{t_2} \mathbf{F}\,dt$

Problem. A particle of mass $2\;\mathrm{kg}$ moving at $5\;\mathrm{m\,s^{-1}}$ collides with a Stationary particle of mass $3\;\mathrm{kg}$. After the collision, the $2\;\mathrm{kg}$ particle Moves at $1\;\mathrm{m\,s^{-1}}$ in its original direction. Find the velocity of the $3\;\mathrm{kg}$ Particle and the magnitude of the impulse exerted on it.

Conservation of momentum (one dimension):

$2(5) + 3(0) = 2(1) + 3v$

$10 = 2 + 3v \implies v = \frac{8}{3} \approx 2.67\;\mathrm{m\,s^{-1}}$

Impulse on the $3\;\mathrm{kg}$ particle $= \Delta p = 3v - 0 = 3 \times \dfrac{8}{3} = 8\;\mathrm{Ns}$.

The coefficient of restitution $e$ for a collision is defined as:

$e = \frac{\mathrm{relative\ speed\ of\ separation}}{\mathrm{relative\ speed\ of\ approach}}$

For a direct impact between two particles:

$e = \frac{v_2 - v_1}{u_1 - u_2}$

Where $u_1, u_2$ are velocities before and $v_1, v_2$ are velocities after the collision. The value $e = 1$ corresponds to a perfectly elastic collision; $e = 0$ to a perfectly inelastic collision.

Problem. Two particles of masses $2\;\mathrm{kg}$ and $3\;\mathrm{kg}$ move towards each other With speeds $6\;\mathrm{m\,s^{-1}}$ and $4\;\mathrm{m\,s^{-1}}$ respectively. They collide directly With coefficient of restitution $e = 0.5$. Find their velocities after the collision.

Taking the direction of the $2\;\mathrm{kg}$ particle as positive, $u_1 = 6$, $u_2 = -4$.

Conservation of momentum:

2(6) + 3(-4) = 2v_1 + 3v_2 \implies 12 - 12 = 2v_1 + 3v_2 \implies 2v_1 + 3v_2 = 0 \tag{1}

Restitution equation:

v_2 - v_1 = 0.5(6 - (-4)) = 0.5(10) = 5 \tag{2}

From (2): $v_2 = v_1 + 5$. Substituting into (1):

$2v_1 + 3(v_1 + 5) = 0 \implies 5v_1 = -15 \implies v_1 = -3\;\mathrm{m\,s^{-1}}$

$v_2 = -3 + 5 = 2\;\mathrm{m\,s^{-1}}$

Newton’s third law pairs must satisfy:

1. They act on different bodies.
2. They are equal in magnitude.
3. They are opposite in direction.
4. They are of the same type (both gravitational, both contact, etc.).

Common error: The weight of a book on a table and the normal reaction are not a Newton’s Third law pair. The weight is the gravitational pull of the Earth on the book; the normal reaction Is the contact force of the table on the book. The correct pair for the book’s weight is the Gravitational pull of the book on the Earth.

Problem. A book of mass $2\;\mathrm{kg}$ rests on a table of mass $10\;\mathrm{kg}$ inside a Lift accelerating upward at $3\;\mathrm{m\,s^{-2}}$. Find: (a) the force exerted by the book on The table; (b) the force exerted by the table on the book; (c) the tension in the lift cable.

(a) and (b) are a Newton’s third law pair, so they are equal in magnitude.

For the book: $R_{\mathrm{table\ on\ book}} - 2g = 2(3) = 6$

$R = 2(9.8) + 6 = 25.6\;\mathrm{N}$

The book exerts $25.6\;\mathrm{N}$ downward on the table.

(c) For the entire system (book + table + lift platform):

Let the total mass being accelerated be $M$ (including the lift structure). If we consider just The book and table: total mass $= 12\;\mathrm{kg}$Acceleration $= 3\;\mathrm{m\,s^{-2}}$.

$T - 12g = 12(3) \implies T = 12(12.8) = 153.6\;\mathrm{N}$

(If the lift structure itself has mass, this would need to be included.)

A block of mass $5\;\mathrm{kg}$ is pushed along a rough horizontal surface by a horizontal force Of $30\;\mathrm{N}$. The coefficient of friction is $0.4$. Find the acceleration of the block.

Two particles of masses $3\;\mathrm{kg}$ and $7\;\mathrm{kg}$ are connected by a light Inextensible string passing over a smooth pulley. The system is released from rest. Find the speed Of the particles after they have moved $2\;\mathrm{m}$ and the tension in the string.

A particle of mass $4\;\mathrm{kg}$ is on a rough plane inclined at $30^\circ$ to the horizontal, Connected by a string over a pulley to a particle of mass $6\;\mathrm{kg}$ hanging vertically. The Coefficient of friction between the $4\;\mathrm{kg}$ mass and the plane is $0.2$. Find the Acceleration of the system.

A ball of mass $0.15\;\mathrm{kg}$ is struck by a bat. Immediately before the impact, the ball is Moving at $20\;\mathrm{m\,s^{-1}}$ towards the bat. Immediately after, it moves at $30\;\mathrm{m\,s^{-1}}$ Away from the bat. The impact lasts $0.01\;\mathrm{s}$. Find the magnitude of the force exerted by The bat on the ball.

A $5\;\mathrm{kg}$ particle moving at $8\;\mathrm{m\,s^{-1}}$ collides directly with a $3\;\mathrm{kg}$ Particle moving at $2\;\mathrm{m\,s^{-1}}$ in the same direction. The coefficient of restitution is $0.6$. Find the velocities after collision and the loss of kinetic energy.

Dynamics describes how forces shape motion. Newton’s laws are the rules of the game: objects resist changes to their state, forces cause acceleration proportional to mass, and every interaction produces equal and opposite reactions. Friction is like invisible grit between surfaces that opposes sliding, proportional to how hard surfaces press together. Momentum is the inertia of motion, conserved when no external force acts. Impulse is a momentary push that changes momentum, and the coefficient of restitution measures how bouncy a collision is.

1. Confusing scalar and vector quantities. Always check whether direction matters for the quantity in question.

2. Using the wrong equation from the data sheet. Take time to read the full equation, including conditions and variable definitions.

3. Neglecting air resistance or assuming ideal conditions when the question specifies a real-world scenario.

4. Misidentifying the system boundary when applying conservation laws. Define what is included before writing equations.

Problem. Two masses $m_1 = 5\ \mathrm{kg}$ and $m_2 = 3\ \mathrm{kg}$ are connected by a light inextensible string over a smooth pulley. The $5\ \mathrm{kg}$ mass rests on a rough horizontal surface ($\mu = 0.2$) and the $3\ \mathrm{kg}$ mass hangs vertically over the edge. Find the acceleration and tension.

Solution. For the $3\ \mathrm{kg}$ mass (downward positive): $3g - T = 3a$.

For the $5\ \mathrm{kg}$ mass: $R = 5g$, $F = \mu R = 0.2 \times 5g = 9.8\ \mathrm{N}$. So $T - 9.8 = 5a$.

Adding: $3g - 9.8 = 8a$, giving $a = \frac{29.4 - 9.8}{8} = \frac{19.6}{8} = 2.45\ \mathrm{m\,s^{-2}}$.

$T = 3(g - a) = 3(9.8 - 2.45) = 3 \times 7.35 = 22.05\ \mathrm{N}$.

$\blacksquare$

Problem. A ball of mass $2\ \mathrm{kg}$ moving at $10\ \mathrm{m\,s^{-1}}$ collides with a stationary ball of mass $3\ \mathrm{kg}$. The coefficient of restitution is $e = 0.8$. Find the velocities after collision and the kinetic energy lost.

Solution. Conservation of momentum: $2(10) + 3(0) = 2v_1 + 3v_2$, so $20 = 2v_1 + 3v_2$.

Restitution: $v_2 - v_1 = 0.8(10 - 0) = 8$, so $v_2 = v_1 + 8$.

Substituting: $20 = 2v_1 + 3(v_1 + 8) = 5v_1 + 24$, giving $v_1 = -0.8\ \mathrm{m\,s^{-1}}$.

$v_2 = -0.8 + 8 = 7.2\ \mathrm{m\,s^{-1}}$.

Initial KE $= \frac{1}{2}(2)(100) = 100\ \mathrm{J}$. Final KE $= \frac{1}{2}(2)(0.64) + \frac{1}{2}(3)(51.84) = 0.64 + 77.76 = 78.4\ \mathrm{J}$.

Energy lost $= 100 - 78.4 = 21.6\ \mathrm{J}$.

$\blacksquare$

• Newton’s second law: $\mathbf{F} = m\mathbf{a}$; always draw a free-body diagram before writing equations.
• Connected particles share the same acceleration magnitude for inextensible strings; solve simultaneous equations for each body.
• Friction: $F \leq \mu R$ (static); $F = \mu R$ (sliding). Always check the direction of friction.
• Conservation of momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ applies when no external force acts.
• Impulse: $I = \Delta p = m(v - u)$; coefficient of restitution $e = \frac{v_2 - v_1}{u_1 - u_2}$.

• Forces and Newton’s Laws — This extended treatment builds on the core forces syllabus with deeper analysis of friction, pulleys, and inclined planes.
• Kinematics — Dynamics applies forces to explain the accelerations and motions described kinematically.
• Moments — The equilibrium conditions for rigid bodies extend Newton’s laws to rotational dynamics.
• Energy and Work — The work-energy principle provides an alternative to F=ma for solving dynamics problems involving displacement.