JEE Main & Advanced

Laws of Motion

Laws of Motion for JEE Main & Advanced

1
Module 1

Newton's Laws and Applications

Newton's Three Laws, Momentum and ImpulseTopic 1

Newton's First Law (Law of Inertia): Every body remains at rest or in uniform motion in a straight line unless acted upon by a net external force. This law defines the concept of an inertial frame — a frame in which the law holds. Inertia is the inherent property of a body to resist changes in its state of motion, and is quantitatively measured by its mass.

Newton's Second Law: The rate of change of linear momentum of a body equals the net external force acting on it: $$\vec{F}_{\text{net}} = \frac{d\vec{p}}{dt}$$ where $\vec{p} = m\vec{v}$ is linear momentum. For constant mass: $\vec{F} = m\vec{a}$. The second law is a vector equation, and applies independently along each axis: $F_x = ma_x$, $F_y = ma_y$, $F_z = ma_z$.

Newton's Third Law (Action-Reaction): For every action there is an equal and opposite reaction. If body A exerts force $\vec{F}_{AB}$ on body B, then B exerts $\vec{F}_{BA} = -\vec{F}_{AB}$ on A. Crucially, action and reaction act on different bodies and therefore never cancel.

Linear Momentum is a vector $\vec{p} = m\vec{v}$ with dimensions $[MLT^{-1}]$. For a system of particles, $\vec{P}_{\text{total}} = \sum m_i \vec{v}_i = M_{\text{total}} \vec{v}_{cm}$.

Conservation of Linear Momentum: If the net external force on a system is zero, total momentum is conserved: $$\sum m_i \vec{v}_i \bigg|_{\text{before}} = \sum m_i \vec{v}_i \bigg|_{\text{after}}$$

This is a direct consequence of Newton's second + third laws applied to internal forces (which cancel pairwise). It holds independently along each axis.

Impulse is the integral of force over time: $$\vec{J} = \int_{t_1}^{t_2} \vec{F}\,dt = \Delta \vec{p}$$ This impulse-momentum theorem is especially useful for short-duration forces (collisions, kicks) where the precise force profile is unknown but the change in momentum is measurable.

Variable-mass systems (rockets, conveyor belts): Newton's second law in the form $\vec{F} = d\vec{p}/dt$ remains valid. For a rocket with exhaust velocity $u$ relative to itself, and instantaneous mass $m$: $$m\frac{dv}{dt} = -u\frac{dm}{dt} - mg \text{ (with gravity)}$$ leading to the Tsiolkovsky rocket equation: $\Delta v = u \ln(m_0/m_f) - gt$.

Spring Forces: A spring of stiffness $k$ stretched/compressed by $x$ exerts a restoring force $F = -kx$ (Hooke's law). Springs in series: $1/k_{\text{eq}} = 1/k_1 + 1/k_2$. Springs in parallel: $k_{\text{eq}} = k_1 + k_2$.

Worked Examples
1

A bullet of mass $50$ g moving at $200$ m/s strikes a wooden block of mass $950$ g at rest on a smooth surface and embeds in it. Find the velocity of the combined system.

Show solution

This is a perfectly inelastic collision. Conservation of momentum: $$m_1 v_1 + m_2 v_2 = (m_1 + m_2) v$$ $$0.05 \times 200 + 0.95 \times 0 = 1.0 \times v$$ $$v = 10 \text{ m/s}$$

Final Answer: $v = 10$ m/s.

2

A force varies with time as $F = 5t^2 - 3t + 4$ N. Find the impulse delivered between $t = 0$ and $t = 2$ s.

Show solution

$$J = \int_0^2 (5t^2 - 3t + 4)\,dt = \left[\frac{5t^3}{3} - \frac{3t^2}{2} + 4t\right]_0^2$$ $$= \frac{40}{3} - 6 + 8 = \frac{40}{3} + 2 = \frac{40 + 6}{3} = \frac{46}{3} \approx 15.33 \text{ N·s}$$

Final Answer: $J = 46/3$ N·s.

3

A ball of mass $0.1$ kg strikes a wall at $20$ m/s at $60^\circ$ to the normal and rebounds elastically at the same angle. Find the impulse delivered by the wall.

Show solution

Decompose into perpendicular and parallel components to the wall. Only the perpendicular component (along normal) reverses; the parallel component is unchanged.

Initial perpendicular momentum (into wall) = $mv\cos 60^\circ = 0.1 \times 20 \times 0.5 = 1$ kg·m/s. Final perpendicular momentum (out of wall) = $-1$ kg·m/s.

$|\Delta p| = 1 - (-1) = 2$ kg·m/s (directed away from wall).

Final Answer: Impulse = $2$ N·s, perpendicular to wall.

✎ Self-Check — 5 questions0 / 5
Q1.

A body of mass $m$ collides elastically with a wall at velocity $v$ normally. The change in momentum is:

Q2.

If a body is in equilibrium, the net force on it is:

Q3.

A constant force $F$ acts on a body of mass $m$ for time $t$. The change in momentum is:

Q4.

The slope of momentum vs. time graph gives:

Q5.

A man of mass $60$ kg stands on a weighing machine in a lift accelerating upward at $2 \text{ m/s}^2$. Reading on the machine ($g = 10 \text{ m/s}^2$):

Free-Body Diagrams, Tensions, Pulleys and ConstraintsTopic 2

A free-body diagram (FBD) isolates a single body and shows all external forces acting on it. The systematic approach to mechanics problems:

  1. Identify each body and draw its FBD separately.
  2. List all forces (gravity, normal, tension, friction, applied) with correct directions.
  3. Choose convenient axes (often along surfaces or applied forces).
  4. Write $F = ma$ along each axis.
  5. Apply constraint equations relating positions/velocities/accelerations of bodies.
  6. Solve the system algebraically.

Tension in an Ideal (Massless, Inextensible) String: The tension is the same throughout the string. If the string passes over a frictionless pulley, the magnitude of tension is preserved across it. For a string with mass per unit length $\lambda$ hanging under gravity, tension varies linearly with height.

Ideal Pulley is massless and frictionless: it redirects tension without changing magnitude. A movable pulley introduces a $2:1$ velocity ratio: if the support of a single movable pulley moves down by $x$, the load goes up by $x/2$ (and vice versa).

Constraint Equations: For inextensible strings connecting bodies, the total length is constant. Differentiating gives:

  • Velocity constraint: $\sum (\text{velocities along string}) = 0$
  • Acceleration constraint: $\sum (\text{accelerations along string}) = 0$

Examples of Constraints:

(1) Atwood machine (two masses $m_1$, $m_2$ over a fixed pulley):

  • $a_1 = -a_2$ (one goes up while the other goes down)
  • $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2}$
  • $T = \dfrac{2 m_1 m_2 g}{m_1 + m_2}$

(2) Two blocks connected on a smooth table, pulled by force $F$:

  • Equal acceleration $a = F/(m_1 + m_2)$
  • Tension in connecting string: $T = m_2 a = m_2 F/(m_1 + m_2)$

(3) Block on a smooth wedge: The block slides along the inclined surface. Its acceleration along the incline equals $g \sin\theta$ (smooth case), with normal $N = mg\cos\theta$.

(4) Movable pulley with two strings: Constraint $a_1 + a_2 = 2 a_p$ where $a_p$ is pulley's acceleration.

Pseudo-Forces in Non-Inertial Frames: In a frame accelerating with $\vec{a}_0$, an apparent force $\vec{F}_{\text{pseudo}} = -m\vec{a}_0$ must be added to each body to apply $F = ma$ correctly. Pseudo-forces have no Newtonian third-law pair.

Worked Examples
1

Two blocks $m_1 = 3$ kg and $m_2 = 5$ kg are connected by a string over a frictionless pulley (Atwood machine). Find acceleration and tension. Take $g = 10 \text{ m/s}^2$.

Show solution

The heavier mass descends: $$a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(5 - 3) \times 10}{8} = 2.5 \text{ m/s}^2$$

$$T = \frac{2 m_1 m_2 g}{m_1 + m_2} = \frac{2 \times 3 \times 5 \times 10}{8} = \frac{300}{8} = 37.5 \text{ N}$$

Final Answer: $a = 2.5 \text{ m/s}^2$; $T = 37.5$ N.

2

A block of mass $5$ kg is placed on a smooth wedge of inclination $30^\circ$. The wedge is at rest. Find the acceleration of the block down the incline. ($g = 10 \text{ m/s}^2$).

Show solution

Component of gravity along the incline: $g \sin 30^\circ = 10 \times 0.5 = 5 \text{ m/s}^2$.

No friction implies $a = g \sin\theta = 5 \text{ m/s}^2$ down the incline.

Normal reaction: $N = mg \cos 30^\circ = 5 \times 10 \times \frac{\sqrt{3}}{2} = 25\sqrt{3}$ N ≈ $43.3$ N.

Final Answer: $a = 5 \text{ m/s}^2$ down the incline; $N = 25\sqrt{3}$ N.

3

A movable pulley system has one end of the string fixed and the other end attached to a $10$ kg block being pulled down by force $F$. A load $L$ is hung from the movable pulley. Find $F$ required to keep $L = 60$ kg in equilibrium. (Frictionless, massless string and pulleys; $g = 10 \text{ m/s}^2$.)

Show solution

In a single movable pulley system with $L$ supported by two string segments (one fixed, one going up to the fixed pulley and down through the applied force): $$2T = Lg \implies T = Lg/2 = 60 \times 10 / 2 = 300 \text{ N}$$

For the block being pulled down with force $F$, $F = T = 300$ N.

Final Answer: $F = 300$ N.

✎ Self-Check — 5 questions0 / 5
Q1.

In Atwood machine with $m_1 = 2$ kg and $m_2 = 6$ kg, the tension in the string is ($g = 10 \text{ m/s}^2$):

Q2.

A block of mass $2$ kg is pulled along a smooth horizontal surface by a force of $10$ N at $60^\circ$ above horizontal. Acceleration of the block:

Q3.

A man in a lift feels weightless. The lift is:

Q4.

Two blocks $m_1 = 4$ kg and $m_2 = 2$ kg on a smooth table connected by a string. Force $F = 12$ N pulls $m_1$. Tension in string:

Q5.

A block of $10$ kg is suspended by two strings making angles $30^\circ$ and $60^\circ$ with the vertical from a horizontal ceiling. The tensions are ($g = 10 \text{ m/s}^2$):

2
Module 2

Friction and Circular Motion Dynamics

Static, Kinetic Friction and Inclined PlanesTopic 1

Friction is a contact force that opposes relative motion (or tendency of motion) between two surfaces. It originates from microscopic interlocking and adhesion of surface asperities.

Static Friction ($f_s$): Acts when surfaces are in contact but not sliding. It is a self-adjusting force that takes whatever value is needed (up to its limit) to prevent sliding: $$0 \leq f_s \leq \mu_s N$$ where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. The maximum is the limiting friction: $f_{s,\max} = \mu_s N$.

Kinetic Friction ($f_k$): Acts when there is relative sliding. It is approximately constant in magnitude: $$f_k = \mu_k N$$ with $\mu_k$ the coefficient of kinetic friction. Generally, $\mu_k < \mu_s$ for the same pair of surfaces.

Both coefficients depend on the nature of the two surfaces (not on area of contact or normal force, to good approximation), and friction always acts parallel to the surface, opposing relative sliding.

Angle of Friction ($\alpha$): The angle between the resultant of friction and normal forces and the normal: $$\tan \alpha = \mu_s$$

Angle of Repose ($\theta_r$): The maximum inclination angle of a surface at which a body remains in equilibrium without sliding: $$\tan \theta_r = \mu_s$$ So the angle of repose equals the angle of friction.

Block on a Rough Inclined Plane:

  • If $\theta \leq \theta_r$: block remains in equilibrium, friction adjusts to balance $mg \sin\theta$.
  • If $\theta > \theta_r$ and block slides down: $a = g(\sin\theta - \mu_k \cos\theta)$.
  • If block is pushed up at $\theta$ along the slope and released, deceleration = $g(\sin\theta + \mu_k \cos\theta)$.

Pulling vs. Pushing: For a block being dragged on a horizontal surface by a force $F$ at angle $\theta$ above horizontal:

  • Pulling (force at angle $\theta$ upward): $N = mg - F\sin\theta$, so friction $= \mu(mg - F\sin\theta)$.
  • Pushing (force at angle $\theta$ downward into surface): $N = mg + F\sin\theta$, so friction is larger.

Therefore pulling requires less force than pushing at the same angle — a useful practical insight.

Rolling Friction: Acts on a body that rolls without slipping. Much smaller than kinetic friction (coefficient $\mu_r$ typically $10^{-2}$ to $10^{-4}$). It originates from deformation of surfaces.

Worked Examples
1

A block of mass $5$ kg rests on a horizontal surface with $\mu_s = 0.4$. The minimum horizontal force needed to make it slide is ($g = 10 \text{ m/s}^2$):

Show solution

$$F_{\min} = \mu_s N = \mu_s m g = 0.4 \times 5 \times 10 = 20 \text{ N}$$

Final Answer: $20$ N.

2

A block of mass $2$ kg is placed on an incline of $30^\circ$. $\mu_s = 0.3$. Will the block slide? If yes, find the acceleration. ($g = 10$, $\mu_k = 0.2$.)

Show solution

Force trying to slide block down: $mg\sin\theta = 2 \times 10 \times 0.5 = 10$ N. Max static friction (uphill): $\mu_s N = \mu_s mg\cos\theta = 0.3 \times 2 \times 10 \times (\sqrt{3}/2) = 3\sqrt{3} \approx 5.2$ N.

Since $10 > 5.2$, block slides down.

Acceleration: $a = g(\sin\theta - \mu_k\cos\theta) = 10(0.5 - 0.2 \times \sqrt{3}/2) = 10(0.5 - 0.173) = 3.27 \text{ m/s}^2$.

Final Answer: Yes, block slides; $a \approx 3.27 \text{ m/s}^2$ down the incline.

3

A block of mass $10$ kg is being pulled by a force $F$ at $30^\circ$ above horizontal on a rough surface with $\mu = 0.2$. Find $F$ for the block to just slide. ($g = 10 \text{ m/s}^2$.)

Show solution

Just about to slide: $F\cos\theta = \mu N$ where $N = mg - F\sin\theta$. $$F\cos 30^\circ = \mu(mg - F\sin 30^\circ)$$ $$F \cdot \frac{\sqrt{3}}{2} = 0.2(100 - F \cdot 0.5) = 20 - 0.1F$$ $$0.866 F + 0.1F = 20 \implies 0.966 F = 20 \implies F \approx 20.7 \text{ N}$$

Final Answer: $F \approx 20.7$ N.

✎ Self-Check — 5 questions0 / 5
Q1.

The relation between angle of friction $\alpha$ and coefficient of friction $\mu$ is:

Q2.

A body just slides down an incline of $\tan^{-1}(0.5)$. The coefficient of friction is:

Q3.

A block of $2$ kg is pulled along a horizontal surface ($\mu = 0.4$) by horizontal force $10$ N. Acceleration:

Q4.

The coefficient of kinetic friction $\mu_k$ is generally:

Q5.

A block kept on a horizontal surface requires a horizontal force of $20$ N to just slide. If we add a vertical downward force of $30$ N on the block, the minimum horizontal force required is (block of mass $5$ kg, $g = 10 \text{ m/s}^2$):

Centripetal Force, Banking, and Non-Inertial FramesTopic 2

A body moving in a circle of radius $r$ with speed $v$ has centripetal acceleration $a_c = v^2/r$ directed toward the centre. The corresponding force (the centripetal force) is provided by physical agencies like tension, gravity, friction, normal reaction, or their combinations. The centripetal force is not a new fundamental force but a label for the radial component of the net force: $$F_c = \frac{mv^2}{r} = m\omega^2 r$$

Vertical Circular Motion: A bob of mass $m$ on a string of length $\ell$ swung in a vertical circle has variable speed due to gravity. At the topmost point, gravity and tension both act downward, providing centripetal force: $$T_{\text{top}} + mg = \frac{mv_{\text{top}}^2}{\ell} \implies T_{\text{top}} = \frac{mv_{\text{top}}^2}{\ell} - mg$$

For the string to remain taut, $T_{\text{top}} \geq 0$, requiring $v_{\text{top}} \geq \sqrt{g\ell}$. By energy conservation between top and bottom: $$\frac{1}{2}mv_{\text{bot}}^2 = \frac{1}{2}mv_{\text{top}}^2 + 2mg\ell$$

So minimum speed at bottom for complete circle: $v_{\text{bot}} = \sqrt{5g\ell}$.

At the bottom: $T_{\text{bot}} - mg = mv_{\text{bot}}^2/\ell \implies T_{\text{bot}} = mv_{\text{bot}}^2/\ell + mg$. For minimum case: $T_{\text{bot}} = 6mg$.

Conical Pendulum: A bob revolves in a horizontal circle while suspended by a string at angle $\theta$ to vertical. Tension provides both vertical (against gravity) and horizontal (centripetal) components: $$T\cos\theta = mg, \quad T\sin\theta = \frac{mv^2}{r}, \quad r = \ell\sin\theta$$ Period: $T_{\text{period}} = 2\pi\sqrt{\ell\cos\theta/g}$.

Banking of Roads: On a banked turn of angle $\theta$, the normal force has a horizontal component providing centripetal force. For a frictionless banked road: $$\tan\theta = \frac{v^2}{rg} \implies v_{\text{optimal}} = \sqrt{rg\tan\theta}$$

With friction $\mu$, the safe speed range is: $$\sqrt{rg \cdot \frac{\sin\theta - \mu\cos\theta}{\cos\theta + \mu\sin\theta}} \leq v \leq \sqrt{rg \cdot \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}}$$

For an unbanked circular road ($\theta = 0$), friction alone provides centripetal force: $v_{\max} = \sqrt{\mu r g}$.

Pseudo (Inertial) Forces in accelerated frames: In a frame accelerating with $\vec{A}$ relative to an inertial frame, every body of mass $m$ experiences a pseudo-force $\vec{F}_{\text{pseudo}} = -m\vec{A}$ (opposite to frame's acceleration). Newton's laws then formally hold in this rotating/accelerated frame.

In a uniformly rotating frame with angular velocity $\vec{\omega}$:

  • Centrifugal force: $\vec{F}_{\text{cf}} = m\omega^2 \vec{r}$ (outward, radial)
  • Coriolis force: $\vec{F}_{\text{Cor}} = -2m(\vec{\omega} \times \vec{v}_{\text{rel}})$ (perpendicular to velocity in rotating frame)
Worked Examples
1

A car of mass $1000$ kg moves on a horizontal circular track of radius $50$ m. The coefficient of friction is $0.5$. Find the maximum safe speed. ($g = 10 \text{ m/s}^2$.)

Show solution

On an unbanked road, friction provides centripetal force: $$\mu m g \geq \frac{mv^2}{r} \implies v_{\max} = \sqrt{\mu r g} = \sqrt{0.5 \times 50 \times 10} = \sqrt{250} = 5\sqrt{10} \approx 15.8 \text{ m/s}$$

Final Answer: $v_{\max} \approx 15.8$ m/s (or $\sqrt{250}$ m/s).

2

A stone of mass $0.5$ kg attached to a string of length $1$ m is swung in a vertical circle. Find the minimum velocity at the lowest point for the stone to complete the circle. ($g = 10 \text{ m/s}^2$.)

Show solution

$$v_{\text{bot,min}} = \sqrt{5g\ell} = \sqrt{5 \times 10 \times 1} = \sqrt{50} = 5\sqrt{2} \approx 7.07 \text{ m/s}$$

Final Answer: $v_{\text{bot,min}} = 5\sqrt{2}$ m/s.

3

A road is banked at $30^\circ$ for a curve of radius $200$ m. The maximum speed for safe driving with $\mu = 0.5$ ($g = 10 \text{ m/s}^2$):

Show solution

$$v_{\max} = \sqrt{rg \cdot \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}}$$ $$= \sqrt{200 \times 10 \cdot \frac{0.5 + 0.5 \times (\sqrt{3}/2)}{(\sqrt{3}/2) - 0.5 \times 0.5}}$$ $$= \sqrt{2000 \cdot \frac{0.5 + 0.433}{0.866 - 0.25}} = \sqrt{2000 \cdot \frac{0.933}{0.616}}$$ $$= \sqrt{2000 \times 1.514} = \sqrt{3028} \approx 55 \text{ m/s}$$

Final Answer: $v_{\max} \approx 55$ m/s.

✎ Self-Check — 5 questions0 / 5
Q1.

A car can take a turn safely on an unbanked road of radius $r$ with $\mu = 0.4$ at maximum speed:

Q2.

A stone of mass $m$ is whirled in a horizontal circle by a string of length $L$ at angular velocity $\omega$. The tension in the string is:

Q3.

In vertical circular motion, the speed at the highest point is minimum equal to:

Q4.

The banking angle for a curved road of radius $50$ m to allow a vehicle to negotiate it at $10$ m/s ($g = 10$):

Q5.

A conical pendulum of length $1$ m makes an angle $60^\circ$ with the vertical. Its period ($g = 10$):

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