⚛ Physics · Mechanics 🔥 Day 1

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Mechanics — Physics

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Units & Measurement

The language of physics — units, dimensions, and honest error reporting

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1. Physical Quantities: Fundamental vs Derived

A physical quantity is anything that can be measured and expressed as a number times a unit (e.g. 5 kg). Fundamental (base) quantities — mass, length, time — cannot be expressed in terms of any other quantity. Derived quantities are built by combining them (speed = length ÷ time).

Analogy: fundamental quantities are like primary colours; derived quantities are every colour you mix from them.

2. Systems of Units & the Seven SI Base Units

Quantity	Unit	Symbol
Length	metre	m
Mass	kilogram	kg
Time	second	s
Electric current	ampere	A
Temperature	kelvin	K
Amount of substance	mole	mol
Luminous intensity	candela	cd

The two supplementary units — radian (plane angle) and steradian (solid angle) — are dimensionless $[M^0L^0T^0]$ despite being proper SI units. Older systems (CGS, MKS, FPS) still surface in problems via terms like dyne, erg, calorie.

3. Significant Figures & Rounding

All non-zero digits are significant (247 → 3 sig figs).
Zeros between non-zero digits count (1005 → 4).
Leading zeros never count (0.0025 → 2).
Trailing zeros count only with a decimal point (100 → 1; 100.0 → 4).

Rounding (NCERT convention): drop digit $< 5$ → unchanged; $> 5$ → raise by 1; exactly 5 → round to the nearest even digit (banker's rounding).

4. Accuracy vs Precision

Accuracy is how close a measurement is to the true value; precision is how close repeated measurements are to each other. On a dartboard: tight cluster off the bullseye = precise but not accurate; a systematic error can make readings very precise yet consistently inaccurate.

5. Least Count & Errors

The least count is the smallest value an instrument can reliably read. Errors are systematic (consistent bias, e.g. zero error), random (unpredictable fluctuations), or gross (human carelessness). For a sum/difference the absolute errors add; for a product/quotient the relative errors add; for a power the relative error multiplies by the exponent.

6. Dimensions & Dimensional Analysis

The dimension of a quantity shows how it is built from the base quantities, e.g. speed $= [M^0L^1T^{-1}]$. Dimensional analysis lets you (i) check whether an equation is dimensionally consistent, (ii) convert units between systems via $n_1u_1 = n_2u_2$, and (iii) derive relationships — but it cannot find dimensionless constants (the $\tfrac12$ in $KE=\tfrac12mv^2$), handle trig/exponential/log terms, or distinguish quantities that share a dimension (torque and energy are both $[ML^2T^{-2}]$).

💡 Exam Tip: Dimensional analysis only rules out impossible equations — it never proves one fully correct. Use it as a fast sanity check on every numerical, but never as a substitute for derivation.

⚠ Common Mistake: Applying the percentage-error rule to a sum/difference. For $Z = A \pm B$ it is the absolute errors that add ($\Delta Z = \Delta A + \Delta B$), not the percentage errors.

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Units & Measurement — Formulas

Error propagation, least count and unit conversion

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Error Analysis

Mean Absolute Error

$$ \overline{\Delta a} = \frac{1}{n}\sum_{i=1}^{n} |a_{mean} - a_i| $$

Average of the absolute deviations

same as $a$

Percentage Error

$$ \frac{\overline{\Delta a}}{a_{mean}} \times 100\% $$

Relative error expressed as a percent

Error in Sum / Difference

$$ \Delta Z = \Delta A + \Delta B $$

Absolute errors add (even for $A-B$)

same as $Z$

Error in Product / Quotient

$$ \frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B} $$

Relative errors add

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Error in Power

$$ \frac{\Delta Z}{Z} = n\,\frac{\Delta A}{A} \quad (Z = A^n) $$

Relative error multiplies by the exponent

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Instruments & Unit Conversion

Least Count (Vernier)

$$ \text{LC} = 1\,\text{MSD} - 1\,\text{VSD} $$

Main scale div − vernier scale div

cm / mm

Least Count (Screw Gauge)

$$ \text{LC} = \frac{\text{Pitch}}{\text{No. of circular divisions}} $$

Pitch = distance per full rotation

Unit Conversion

$$ n_2 = n_1 \left(\frac{M_1}{M_2}\right)^{a}\left(\frac{L_1}{L_2}\right)^{b}\left(\frac{T_1}{T_2}\right)^{c} $$

$a,b,c$ = dimensions of the quantity

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Units & Measurement — Solved Examples

JEE Main · NEET level

How many significant figures are in 0.00506, and what is the order of magnitude of $5.8 \times 10^{-4}$?

Approach hint

Leading zeros never count. For order of magnitude, write as $a \times 10^b$ with $1 \le a < 10$; round the power up if $a \ge 5$.

0.00506: the digits 5, 0, 6 are significant → 3 significant figures.
$5.8 \times 10^{-4}$: since $a = 5.8 \ge 5$, round the power up → order of magnitude $= 10^{-3}$.

✓ 3 significant figures | order of magnitude $= 10^{-3}$

If $Z = \dfrac{A^2 B}{C^3}$ and the errors in A, B, C are 2%, 3%, 1% respectively, find the percentage error in Z.

Approach hint

Relative errors add; each power multiplies its variable's relative error.

$\dfrac{\Delta Z}{Z} = 2\left(\dfrac{\Delta A}{A}\right) + \dfrac{\Delta B}{B} + 3\left(\dfrac{\Delta C}{C}\right)$
$= 2(2\%) + 3\% + 3(1\%) = 4\% + 3\% + 3\% = 10\%$

✓ Percentage error in Z $= 10\%$

Convert 1 newton into dyne using dimensional analysis.

Approach hint

Force has dimension $[MLT^{-2}]$. Use $n_2 = n_1 (\dfrac{M_1}{M_2})(L_1/L_2)(T_1/T_2)^{-2}$ with SI → CGS.

SI: $M_1=1\,\text{kg}, L_1=1\,\text{m}, T_1=1\,\text{s}$; CGS: $M_2=1\,\text{g}, L_2=1\,\text{cm}, T_2=1\,\text{s}$.
$n_2 = 1 \times \left(\dfrac{1000}{1}\right)\left(\dfrac{100}{1}\right)(1)^{-2} = 1000 \times 100 = 10^5$

✓ $1\,\text{N} = 10^5\,\text{dyne}$

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Units & Measurement — Quick Quiz

3 questions · tap an option to reveal answer

Question 1

Which of the following is a derived unit?

Newton $= \text{kg}\cdot\text{m}\cdot\text{s}^{-2}$, built from base units. kg, mol and ampere are all fundamental SI units — a common trap is to confuse "commonly used" with "fundamental".

Question 2

The pair of physical quantities having the same dimensional formula is:

Impulse and momentum both have dimension $[MLT^{-1}]$. Torque $[ML^2T^{-2}]$ ≠ Force $[MLT^{-2}]$; Energy and Power differ by $T^{-1}$; Pressure and Force differ by $L^{-2}$.

Question 3

A screw gauge has pitch 0.5 mm and 50 divisions on its circular scale. Its least count is:

$\text{LC} = \dfrac{\text{Pitch}}{\text{No. of divisions}} = \dfrac{0.5\,\text{mm}}{50} = 0.01\,\text{mm}$.

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Kinematics

Motion without cause — describes how objects move

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1. Scalars vs Vectors

Quantity	Type	Symbol	SI Unit
Distance	Scalar	s	m
Displacement	Vector	s	m
Speed	Scalar	v	m/s
Velocity	Vector	v	m/s
Acceleration	Vector	a	m/s²

2. Equations of Motion (constant acceleration only)

These four SUVAT equations apply only when acceleration is constant. Never use them when acceleration changes with time or position.

$v = u + at$ — velocity after time $t$
$s = ut + \tfrac{1}{2}at^{2}$ — displacement in time $t$
$v^{2} = u^{2} + 2as$ — velocity after displacement $s$
$s = (u + v)\dfrac{t}{2}$ — displacement using average velocity
$s_n = u + \dfrac{a(2n - 1)}{2}$ — displacement in the nth second

3. Motion Graphs

Graph	Slope gives	Area gives
Displacement–time (s-t)	Velocity	—
Velocity–time (v-t)	Acceleration	Displacement
Acceleration–time (a-t)	—	Change in velocity

4. Projectile Motion

A projectile thrown at angle $\theta$ with speed $u$ splits into independent horizontal and vertical components. Horizontal: constant velocity. Vertical: free fall under gravity.

Horizontal component: $u_x = u\cos\theta$ (constant throughout)
Vertical component: $u_y = u\sin\theta$ (decreases at rate $g$)
At maximum height, vertical velocity $= 0$; horizontal velocity $= u\cos\theta$
Maximum range occurs at $\theta = 45^{\circ}$
Ranges are equal for complementary angles: $\theta$ and $(90° - \theta)$ give same $R$

5. Relative Velocity

The velocity of A relative to B is $\vec v_{AB} = \vec v_A - \vec v_B$.

Same direction: relative speed $= |v_1 - v_2|$
Opposite direction: relative speed $= v_1 + v_2$
Rain-man problems: hold umbrella at angle $\theta = \tan^{-1}(\dfrac{v_{rain}}{v_{man}})$ toward direction of motion

💡 JEE Tip: In projectile problems, always split into x and y components first. The most common error is applying SUVAT to the full velocity instead of the vertical component only. Also remember: horizontal velocity is constant — acceleration only acts vertically.

⚠ Common Mistake: The sₙ formula gives displacement in the nth second, NOT the distance from origin. It is derived by subtracting s(n−1) from s(n). It can be negative for deceleration.

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Kinematics — Formulas

All key relations at a glance

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1D Motion (constant acceleration)

Velocity–Time

$$ v = u + at $$

$u$ = initial vel, $a$ = constant acceleration

m/s

Displacement–Time

$$ s = ut + \tfrac{1}{2}at^{2} $$

s measured from initial position

Velocity–Displacement

$$ v^{2} = u^{2} + 2as $$

Time-independent relation

m²/s²

nth Second

$$ s_n = u + \frac{a(2n - 1)}{2} $$

Displacement during nth second only

Projectile Motion

Time of Flight

$$ T = 2u \sin\theta / g $$

Total time in air; doubles if $\theta$ doubles (approx)

Maximum Height

$$ H = u^{2} \sin^{2}\theta / 2g $$

Max when $\theta = 90°$; zero when $\theta = 0°$

Horizontal Range

$$ R = u^{2} \sin2\theta / g $$

Max at $\theta = 45°$; $R(\theta) = R(90° - \theta)$

R at Maximum

$$ R_{max} = u^{2} / g $$

When $\theta = 45°$

Velocity at any t

$$ v_{x} = u\cos\theta,\quad v_y = u\sin\theta - gt $$

$v_x$ constant; $v_y$ changes with gravity

m/s

Equation of Path

$$ y = x \tan \theta - gx^2 / (2u^2\cos ^2\theta ) $$

Parabolic trajectory equation

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Relative Motion

Relative Velocity

$$ v_{AB} = v_{A} - v_{B} $$

Velocity of A as seen from B

m/s

River–Boat Angle

$$ \sin \theta = v_{river} / v_{boat} $$

To cross river in minimum time; angle upstream

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Kinematics — Solved Examples

JEE Main · JEE Advanced level

A stone is thrown vertically upward with speed 20 m/s from the top of a building 25 m high. Find: (a) maximum height above ground, (b) time to reach the ground. Take $g = 10$ m/s².

Approach hint

Take upward as positive. At max height, $v = 0$. Use $v^2 = u^2 - 2g(\Delta h)$ to find height above the throw point, then add building height. For time to ground, use $s = ut - \tfrac{1}{2}g t^2$ where $s = -25$ m (displacement is downward).

Max height above throw point: $v^2 = u^2 - 2gH \to 0 = 400 - 20H \to H = 20$ m
Max height above ground $= 25 + 20 =$ 45 m
Time to ground: take ground as origin, initial position $+25$ m, $u = +20$ m/s $-25 = 20t - \tfrac{1}{2}(10)t^2 \to 5t^2 - 20t - 25 = 0 \to t^2 - 4t - 5 = 0$
$(t - 5)(t + 1) = 0 \to t = 5$ s (reject $t = -1$)

✓ Max height $= 45$ m above ground | Time to reach ground $= 5$ s

A projectile is fired at 60° to the horizontal with speed 40 m/s. Find: (a) time of flight, (b) horizontal range, (c) maximum height. [$g = 10$ m/s²]

Approach hint

Resolve: $u\sin\theta = 40\sin 60° = 20\sqrt 3$ m/s (vertical), $u\cos\theta = 40\cos 60° = 20$ m/s (horizontal). Apply projectile formulas directly.

$u\sin 60° = 40 \times (\sqrt 3/2) = 20\sqrt 3$ m/s | $u\cos 60° = 40 \times 0.5 = 20$ m/s
Time of flight: $T = 2u\sin\theta / g = 2 \times 20\sqrt 3 / 10 = 4\sqrt 3 \approx$ 6.93 s
Range: $R = u^2\sin 2\theta / g = 1600 \times \sin 120° / 10 = 1600 \times (\sqrt 3/2) / 10 = 80\sqrt 3 \approx$ 138.6 m
Max height: $H = u^2\sin^2\theta / (2g) = 1600 \times 0.75 / 20 =$ 60 m

✓ $T = 4\sqrt 3$ s | $R = 80\sqrt 3$ m | $H = 60$ m

Ball A is dropped from rest from the top of a 80 m tower. At the same instant, Ball B is thrown vertically upward from the ground with speed 30 m/s. At what height above the ground do they meet? [$g = 10$ m/s²]

Approach hint

Let them meet at time t. Ball A falls from 80 m (starts at rest, downward). Ball B rises from 0 m. Both have the same acceleration g downward. Set height equations equal to find t, then height.

Height of A above ground: $h_{A} = 80 - \tfrac{1}{2}(10)t^{2} = 80 - 5t^{2}$
Height of B above ground: $h_{B} = 30t - \tfrac{1}{2}(10)t^{2} = 30t - 5t^{2}$
At meeting: $h_{A} = h_{B} \to 80 - 5t^{2} = 30t - 5t^{2}$
$80 = 30t \to t = \dfrac{8}{3}$ s $\approx 2.67$ s
Height $= 30 \times (\dfrac{8}{3}) - 5 \times (\dfrac{8}{3})^2 = 80 - 5 \times \dfrac{64}{9} = 80 - 35.56 \approx$ 44.4 m

✓ They meet at height $\approx 44.4$ m above ground (at $t = \dfrac{8}{3}$ s)

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Kinematics — Diagrams

Click any diagram to zoom

Projectile motion: parabolic trajectory with maximum height H and horizontal range R

v-t graph: slope = acceleration, shaded area = displacement (valid for any shape)

Horizontal projectile: horizontal velocity stays constant, vertical velocity increases due to gravity

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Kinematics — Quick Quiz

3 questions · tap an option to reveal answer

Question 1

A body travels 10 m in the first 2 seconds and 20 m in the next 4 seconds under constant acceleration. What is the acceleration?

Average velocity in first 2 s $= \dfrac{10}{2} = 5$ m/s. Average velocity in next 4 s $= \dfrac{20}{4} = 5$ m/s. Same average velocity means uniform motion (if avg vel is same for different intervals under constant $a$, then $a = 0$). Verified: $u = 5$ m/s, $a = 0 \to$ $s$ in 6 s $= 30$ m ✓

Question 2

A projectile is thrown at 30° with the horizontal. The ratio of maximum height to horizontal range is:

$H = u^2\sin^2 30°/(2g) = u^2 \times \dfrac{0.25}{2g}$. $R = u^2\sin 60°/g = u^2\dfrac{\times(\sqrt 3/2)}{g}$. $\dfrac{H}{R} = \dfrac{\dfrac{u^2}{8g}}{\dfrac{u^2\sqrt 3}{2g}} = \dfrac{\dfrac{1}{8}}{\sqrt 3/2} = \dfrac{1}{4\sqrt 3}$. Ratio $H:R = 1:4\sqrt 3$ ✓

Question 3

A ball is projected horizontally from a height of 45 m. It lands 90 m horizontally from the base. What is the initial horizontal speed? [$g = 10$ m/s²]

Time to fall: $h = \tfrac{1}{2}g t^2 \to 45 = 5t^2 \to t = 3$ s. Horizontal: $x = u\times t \to 90 = u\times 3 \to u = 30$ m/s ✓

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Laws of Motion

Newton's laws, friction, pseudo forces

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Newton's Three Laws

Law	Statement	Mathematical Form
1st (Inertia)	A body stays at rest or uniform motion unless acted on by a net external force	If $F_{net} = 0$, then $a = 0$
2nd (F=ma)	Net force equals rate of change of momentum	$F = ma = \dfrac{dp}{dt}$
3rd (Action-Reaction)	Every action has an equal and opposite reaction — on different bodies	$\vec F_{AB} = -\vec F_{BA}$

Free Body Diagram (FBD) — The Key Skill

An FBD isolates one body and shows all external forces acting on it. Steps:

Isolate the body (draw it alone)
Draw all contact forces: Normal (⊥ to surface), Friction (‖ to surface, opposing relative motion), Tension (along string, away from body)
Draw body forces: Weight mg (downward)
Apply F = ma along chosen axes

Friction

Type	Formula	When?	Note
Static	$f_s \le \mu_s N$	Body at rest; value adjusts to prevent motion	Can be less than $\mu_s N$
Kinetic	$f_k = \mu_k N$	Body in motion relative to surface	Always $\mu_k < \mu_s$
Rolling	$f_r = \mu_r N$	Rolling contact	$\mu_r \ll \mu_k$

Connected Bodies and Pseudo Force

String constraint: If two bodies are connected by an inextensible string over a pulley, they have the same magnitude of acceleration and |T| is same throughout a massless string.

Pseudo force: In a non-inertial (accelerating) reference frame, add a fictitious force $\vec F_{pseudo} = -m\vec A$ on every object, where $\vec A$ is the frame's acceleration. This lets you apply Newton's laws in that frame.

Lift accelerating up: apparent weight $= m(g + a)$
Lift accelerating down: apparent weight $= m(g - a)$
Free fall ($a = g$ downward): apparent weight $= 0$ (weightlessness)

Dynamics of Circular Motion — Banking of Roads

A body in uniform circular motion needs a centripetal force $F_c = \dfrac{mv^2}{r}$ directed toward the centre. For a vehicle on a curve, this force must come from somewhere:

Level (unbanked) road: friction alone supplies it, so the maximum safe speed is $v_{max} = \sqrt{\mu_s\, r\, g}$.
Banked road (frictionless): tilting the road by angle $\theta$ lets the horizontal component of the normal force provide the centripetal force, giving $\tan\theta = \dfrac{v^2}{rg}$, i.e. the ideal speed $v = \sqrt{rg\tan\theta}$.
Banked road with friction: the safe-speed range widens to $v_{max} = \sqrt{rg\,\dfrac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}}$.

Banking lets vehicles take curves faster and reduces reliance on (and wear from) friction — which is why highway and railway curves are banked.

💡 JEE Tip: Action-reaction pairs act on different bodies — they never cancel each other. The most tested trap: "block on table — weight of block and normal from table are NOT action-reaction pairs." The reaction to the block's weight is the block pulling Earth upward.

⚠ Common Mistake: When finding friction on an incline, $N = mg\cos\theta$ (NOT $mg$). Friction $= \mu \times N = \mu mg\cos\theta$. Students often write $\mu mg$, which is wrong on inclines.

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Laws of Motion — Formulas

Newton's Laws & Momentum

Newton's 2nd Law

$$ F_{net} = ma = \frac{dp}{dt} $$

Net force $=$ mass $\times$ acceleration; also rate of change of momentum

Impulse–Momentum

$$ J = F\Delta t = \Delta p = m(v - u) $$

Impulse $=$ change in momentum; useful when $F$ acts for short time

N·s

Conservation of Momentum

$$ m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2} $$

When $F_{net,ext} = 0$ on the system

kg·m/s

Pseudo Force

$$ F_{p} = -mA_{frame} $$

Added to every body when working in non-inertial frame

Friction

Max Static Friction

$$ f_{s,max} = \mu_{s}N $$

Threshold to start motion; $f_s$ can be anything from 0 to this

Kinetic Friction

$$ f_{k} = \mu_{k}N $$

Constant once motion starts; $\mu_k < \mu_s$

Incline — Normal Force

$$ N = mg \cos \theta $$

Component of weight perpendicular to incline surface

Angle of Friction

$$ \tan\lambda = \mu_{s} $$

$\lambda$ = angle of friction; body slides if incline angle $> \lambda$

Circular Motion

Centripetal Force

$$ F_{c} = \frac{mv^{2}}{r} = m\omega^{2}r $$

Net force toward center; NOT a new force — provided by existing forces

Banking Angle

$$ \tan\theta = \frac{v^{2}}{rg} $$

Angle for no friction required; $v = \sqrt{rg\tan\theta}$

Vertical Circle — Min. Speed at Top

$$ v_{min} = \sqrt{gR} $$

Minimum speed at top so string stays taut ($T \ge 0$)

m/s

Conical Pendulum Period

$$ T = 2\pi\sqrt{L \cos\theta / g} $$

$L$ = string length, $\theta$ = half-angle with vertical

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Laws of Motion — Solved Examples

A 10 kg block rests on a horizontal surface ($\mu_s = 0.4, \mu_k = 0.3$). A horizontal force of 30 N is applied. Find: (a) does the block move? (b) acceleration if it moves. [$g = 10$ m/s²]

Approach hint

Find max static friction first. Compare with applied force. If $F > f_{s,max}$, motion occurs and use kinetic friction for acceleration.

Normal force: $N = mg = 10 \times 10 = 100$ N
Max static friction: $f_{s,max} = \mu_s N = 0.4 \times 100 = 40$ N
Applied force $= 30$ N $< 40$ N → block does NOT move
Static friction adjusts to exactly 30 N to balance the applied force

✓ Block does not move. Static friction $= 30$ N (not 40 N). Acceleration $= 0$.

Two blocks A (3 kg) and B (5 kg) are connected by a string over a frictionless pulley (Atwood machine). Find: (a) acceleration of the system, (b) tension in the string. [$g = 10$ m/s²]

Approach hint

Heavier mass (B) descends. Let $a$ = acceleration. For each mass, write $F = ma$ separately, then solve simultaneously for $a$ and $T$.

For mass A (going up): $T - m_A g = m_A a \to T - 30 = 3a$ ... (i)
For mass B (going down): $m_B g - T = m_B a \to 50 - T = 5a$ ... (ii)
Adding (i) and (ii): $50 - 30 = 8a \to a = \dfrac{20}{8} =$ 2.5 m/s²
From (i): $T = 30 + 3 \times 2.5 = 30 + 7.5 =$ 37.5 N

✓ $a = 2.5$ m/s² | $T = 37.5$ N

A block of mass $m$ rests on a wedge (mass $M$) on a frictionless floor. The wedge angle is $\theta$. Find the horizontal acceleration $A$ of the wedge and the normal force $N$ between the block and wedge. (All surfaces frictionless.)

Approach hint

The block slides on the frictionless wedge face. Constraint: horizontal displacement of block $=$ horizontal displacement of wedge $+$ component of block's motion along wedge. Use FBD for each body and Newton's 2nd law in $x$ and $y$ directions separately.

FBD of block: Normal $N$ from wedge at angle $\theta$ to vertical. No friction. $x$: $-N\sin\theta = m\times a_{block,x}$ ... (but $a_{block,x}$ is not simple)
Use constraint: if wedge moves right by $A$, block moves right by $A$ and also slides down the wedge face. In wedge frame: block acceleration $= g\sin\theta$ along the face (frictionless).
In ground frame, applying Newton's law on the system horizontally: No net horizontal external force $\to (M + m)A_{cm,x} = 0$ cannot be used (wedge is on frictionless floor).
Result after solving the constraint equations: $A = mg\sin\theta\cos\theta/(M + m\sin^2\theta)$ and $N = Mmg\cos\theta/(M + m\sin^2\theta)$

✓ $A = mg\sin\theta\cos\theta / (M + m\sin^2\theta)$ | $N = Mmg\cos\theta / (M + m\sin^2\theta)$

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Laws of Motion — Diagrams

Click to zoom

FBD on incline: $N = mg\cos\theta$ (perpendicular to surface), $mg\sin\theta$ acts along the slope

Atwood machine: heavier mass descends; both have same acceleration magnitude and tension T

Friction vs applied force: friction increases up to $\mu_s N$ (static limit), then drops to constant $\mu_k N$ (kinetic)

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Laws of Motion — Quick Quiz

Question 1

A 5 kg block is on a surface with $\mu_s = 0.5$ and $\mu_k = 0.4$. A 20 N horizontal force is applied. What is the friction force? [$g = 10$ m/s²]

$N = 50$ N. $f_{s,max} = 0.5\times 50 = 25$ N. Applied force $= 20$ N $< 25$ N → block does NOT move. Static friction adjusts to exactly balance applied force $= 20$ N. The block is stationary.

Question 2

A person is in a lift accelerating upward at 2 m/s². Their true mass is 60 kg. What does a spring balance in the lift read? [$g = 10$ m/s²]

Apparent weight $= m(g + a) = 60(10 + 2) = 720$ N. Spring balance reads $\dfrac{720}{10} = 72$ kg. The balance reads apparent weight, not true weight.

Question 3

For a body moving in a vertical circle of radius $R$, the minimum speed at the bottom to complete the full circle is:

Minimum speed at the top = $\sqrt{gR}$ (so string remains taut: T = 0). Using energy conservation from bottom to top: $\tfrac{1}{2}mv^2_{\text{bottom}} = \tfrac{1}{2}mv^2_{\text{top}} + mg(2R) \to v^2_{\text{bottom}} = gR + 4gR = 5gR \to v_{\text{bottom}} = \sqrt{5gR}$

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Work, Energy & Power

Energy methods — often faster than Newton's laws

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1. Work Done by a Force

Work is done when a force causes displacement along its direction. If force $F$ acts at angle $\theta$ to displacement $d$:

$W = Fd\cos\theta$
$\theta = 0°$ → maximum positive work (force and displacement same direction)
$\theta = 90°$ → zero work (normal force on horizontal motion does no work)
$\theta = 180°$ → negative work (friction opposes motion)
For variable force: $W = \int F\cdot dx$ $=$ area under $F$-$x$ graph

2. Kinetic Energy and Work-Energy Theorem

The work-energy theorem is the most powerful shortcut in JEE Mechanics — it bypasses the need to find acceleration when you only care about speeds.

$KE = \tfrac{1}{2}mv^2$
$W_{net} = \Delta KE = KE_f - KE_i$ (net work by ALL forces $=$ change in KE)
This includes work by normal, gravity, friction, tension — every force

3. Potential Energy and Conservative Forces

Force	Conservative?	Potential Energy
Gravity	✅ Yes	$U = mgh$
Spring (Hooke's law)	✅ Yes	$U = \tfrac{1}{2}kx^2$
Friction	❌ No	No PE defined; energy lost as heat
Normal force	—	Does zero work (always ⊥ to motion)

For conservative forces only: $W = -\Delta U$. Energy is conserved: $KE + PE = \text{constant}$.

4. Collisions

Type	Momentum conserved?	KE conserved?	e
Elastic	✅ Yes	✅ Yes	$e = 1$
Inelastic	✅ Yes	❌ No (some lost)	$0 < e < 1$
Perfectly inelastic	✅ Yes	❌ No (max loss)	$e = 0$

Coefficient of restitution: $e = \dfrac{v_2 - v_1}{u_1 - u_2}$ $=$ relative speed after / relative speed before.

5. Power

Average power: $P = \dfrac{W}{t}$
Instantaneous power: $P = \vec F\cdot\vec v = Fv\cos\theta$
Unit: Watt (W) $=$ J/s; 1 hp $= 746$ W

💡 JEE Tip: When a problem asks "find the speed at point B," use energy conservation — it is almost always faster than Newton's laws. Only use F=ma when you need the actual force or acceleration at a specific instant.

⚠ Spring PE: $\tfrac{1}{2}kx^2$ where $x$ is measured from the natural (unstretched) length, not from equilibrium in a hanging spring. Don't confuse $x$ (extension) with the equilibrium position shift.

💥

Work, Energy & Power — Formulas

Work & Energy

Work (constant force)

$$ W = Fd \cos\theta $$

$\theta$ = angle between $F$ and displacement; $W < 0$ if $\theta > 90°$

Work (variable force)

$$ W = \int F dx $$

Area under $F$-$x$ graph; use for spring, friction with varying $N$

Kinetic Energy

$$ KE = \tfrac{1}{2}mv^{2} $$

Always $\ge 0$; doubles when $v$ increases by $\sqrt 2$

Work-Energy Theorem

$$ W_{net} = \Delta KE = \tfrac{1}{2}m(v^{2}-u^{2}) $$

Net work by all forces $=$ change in kinetic energy

Gravitational PE

$$ U_{g} = mgh $$

$h$ measured from chosen reference; can be negative

Spring PE

$$ U_{s} = \tfrac{1}{2}kx^{2} $$

$x$ = extension/compression from natural length; always positive

Mechanical Energy Conservation

$$ KE_{1} + PE_{1} = KE_{2} + PE_{2} $$

Valid only when no non-conservative forces act

Work by friction

$$ W_{f} = -\mu mg \cos\theta \cdot d $$

Always negative; equals heat generated $= \mu mg\cdot d$ (horizontal)

Collisions

Elastic Collision — v₁'

$$ v_{1}' = (m_{1}-m_{2})u_{1} + 2m_{2}u_{2}) / (m_{1}+m_{2}) $$

Both momentum and KE conserved

m/s

Coeff. of Restitution

$$ e = \frac{v_{2}-v_{1}}{u_{1}-u_{2}} $$

$e=1$ elastic, $e=0$ perfectly inelastic, $0<e<1$ inelastic

—

KE Loss in Collision

$$ \Delta KE = \tfrac{1}{2} \cdot m_{1}m_{2}/(m_{1}+m_{2}) \cdot (u_{1}-u_{2})^{2}(1-e^{2}) $$

Maximum loss when $e = 0$ (perfectly inelastic)

Power

$$ P = \frac{W}{t} = F\cdot v \cos\theta $$

Instantaneous: $P = Fv$ when $\vec F \parallel \vec v$

💥

Work, Energy & Power — Solved Examples

A 2 kg block slides 5 m down a frictionless 30° incline. Find: (a) work done by gravity, (b) speed at the bottom if it starts from rest. [$g = 10$ m/s²]

Approach hint

Work by gravity $= mgh = mg \times (5\sin 30°)$. Then use work-energy theorem: $W_{net} = \Delta KE$ since no friction.

Height dropped: $h = 5 \times \sin 30° = 5 \times 0.5 = 2.5$ m
Work by gravity: $W = mgh = 2 \times 10 \times 2.5 =$ 50 J
Work by normal force $= 0$ (perpendicular to motion)
$W_{net} = \Delta KE \to 50 = \tfrac{1}{2}\times 2\times v^2 - 0 \to v^2 = 50 \to v =$ $5\sqrt 2 \approx 7.07$ m/s

✓ $W_{gravity} = 50$ J | $v$ at bottom $= 5\sqrt 2$ m/s $\approx 7.07$ m/s

A ball of mass 2 kg moving at 6 m/s collides elastically with a 4 kg ball at rest. Find the velocities of both balls after the collision.

Approach hint

Elastic collision: use the formula directly. Alternatively, write two equations — conservation of momentum and conservation of KE — and solve simultaneously.

$m_1 = 2$ kg, $u_1 = 6$ m/s; $m_2 = 4$ kg, $u_2 = 0$
$v_1 = \dfrac{(m_1-m_2)u_1 + 2m_2 u_2}{m_1+m_2} = \dfrac{(2-4)\times 6 + 0}{6} = -\dfrac{12}{6} =$ $-2$ m/s (bounces back)
$v_2 = \dfrac{2m_1 u_1 + (m_2-m_1)u_2}{m_1+m_2} = \dfrac{2\times 2\times 6 + 0}{6} = \dfrac{24}{6} =$ $+4$ m/s
Check KE: initial $= \tfrac{1}{2}\times 2\times 36 = 36$ J; final $= \tfrac{1}{2}\times 2\times 4 + \tfrac{1}{2}\times 4\times 16 = 4 + 32 = 36$ J ✓

✓ $v_1 = -2$ m/s (reverses) | $v_2 = +4$ m/s

A block of mass $m$ is released from the top of a smooth hemispherical bowl of radius $R$. Find: (a) speed at the bottom, (b) normal force at the bottom, (c) the angle from vertical at which the block leaves the surface if the bowl is inverted (convex side up).

Approach hint

Part (a)+(b): energy conservation + Newton's 2nd law at bottom (circular motion). Part (c): on convex surface, block leaves when $N = 0$. Set up energy equation and centripetal condition simultaneously.

Speed at bottom (concave bowl): $mgh = \tfrac{1}{2}mv^2 \to v = \sqrt{2gR}$
Normal at bottom: $N - mg = \dfrac{mv^2}{R} = \dfrac{m(2gR)}{R} = 2mg \to N =$ $3mg$
Convex surface — leaving angle: Let block be at angle $\theta$ from top. Height fallen $= R - R\cos\theta = R(1-\cos\theta)$. Speed: $v^2 = 2gR(1-\cos\theta)$
At leaving point, $N = 0$: $mg\cos\theta = \dfrac{mv^2}{R} \to g\cos\theta = \dfrac{v^2}{R} = 2g(1-\cos\theta)$
$\cos\theta = 2 - 2\cos\theta \to 3\cos\theta = 2 \to \cos\theta = \dfrac{2}{3} \to$ $\theta = \cos^{-1}(\dfrac{2}{3}) \approx 48.2°$

✓ $v_{\text{bottom}} = \sqrt{2gR}$ | $N_{bottom} = 3mg$ | Leaves at $\cos\theta = \dfrac{2}{3}$

💥

Work, Energy & Power — Diagrams

Click to zoom

Energy conservation on frictionless incline: all PE converts to KE at the bottom

Spring potential energy: $U = \tfrac{1}{2}kx^2$ where $x$ is measured from natural (unstretched) length

Elastic vs perfectly inelastic collision: momentum conserved in both; KE conserved only in elastic

💥

Work, Energy & Power — Quick Quiz

Question 1

A body of mass 2 kg is dropped from a height of 10 m. Just before hitting the ground, its kinetic energy is (ignore air resistance, $g = 10$ m/s²):

All PE converts to KE: $KE = mgh = 2 \times 10 \times 10 = 200$ J ✓

Question 2

A 3 kg block moving at 4 m/s collides perfectly inelastically with a 1 kg stationary block. The loss in kinetic energy is:

Common velocity: $v = 3\times \dfrac{4}{3+1} = 3$ m/s. Initial KE $= \tfrac{1}{2}\times 3\times 16 = 24$ J. Final KE $= \tfrac{1}{2}\times 4\times 9 = 18$ J. Loss $= 24-18 = 6$ J. Using formula: loss $= \tfrac{1}{2}\times (m_1 m_2)/(m_1+m_2)\times (u_1-u_2)^2 = \tfrac{1}{2}\dfrac{\times(3)}{4}\times 16 = 6$ J. Answer is B (6 J).

Question 2 (corrected)

The work done by a spring force when it is compressed by $x$ from natural length is:

Spring force opposes compression: $F = -kx$ (restoring). Work done BY spring on block $= \int F\, dx = -\tfrac{1}{2}kx^2$. The block does positive work on spring (stores PE $= +\tfrac{1}{2}kx^2$). Work by spring on block $= -\tfrac{1}{2}kx^2$. ✓

🔄

Rotational Motion

Angular analogues of linear motion + moment of inertia

○ Not Started

1. Angular Kinematic Analogies

Linear	Angular	Relation
Displacement $s$	Angle $\theta$	$s = r\theta$
Velocity $v$	Angular velocity $\omega$	$v = r\omega$
Acceleration $a$	Angular acceleration $\alpha$	$a_{tangential} = r\alpha$
Mass $m$	Moment of inertia $I$	$I = \Sigma mr^2$
Force $F$	Torque $\tau$	$\tau = rF\sin\phi$
$F = ma$	$\tau = I\alpha$	—
Momentum $p = mv$	Angular momentum $L = I\omega$	—
$KE = \tfrac{1}{2}mv^2$	$KE_{rot} = \tfrac{1}{2}I\omega^2$	—

2. Moment of Inertia of Common Bodies

Body	Axis	I
Solid sphere	Through centre	$\dfrac{2MR^2}{5}$
Hollow sphere	Through centre	$\dfrac{2MR^2}{3}$
Solid cylinder / disk	Along axis	$\dfrac{MR^2}{2}$
Hollow cylinder / ring	Along axis	$MR^2$
Thin rod	Through centre $\perp$ rod	$\dfrac{ML^2}{12}$
Thin rod	Through one end $\perp$ rod	$\dfrac{ML^2}{3}$
Rectangular plate ($a\times b$)	Through centre	$\dfrac{M(a^2+b^2)}{12}$

3. Theorems on MOI

Parallel axis theorem: $I = I_{cm} + Md^2$ — $I$ about any axis $= I_{cm} +$ mass $\times$ (distance between axes)$^2$
Perpendicular axis theorem (planar bodies only): $I_z = I_x + I_y$ — $I$ about axis $\perp$ to plane $=$ sum of $I$ about two $\perp$ axes in the plane

4. Rolling Motion (Pure Rolling)

For a body rolling without slipping on a surface:

Velocity of centre of mass: $v_{cm} = R\omega$
Point of contact has zero velocity (no slipping)
Total KE $= KE_{trans} + KE_{rot} = \tfrac{1}{2}mv^2 + \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}mv^2(1 + \dfrac{k^2}{R^2})$ where $I = mk^2$
Acceleration down an incline: $a = g\sin\theta/(1 + \dfrac{k^2}{R^2})$
Solid sphere rolls faster than hollow sphere (smaller $\dfrac{k^2}{R^2} = \dfrac{2}{5}$ vs $\dfrac{2}{3}$)

5. Conservation of Angular Momentum

When net external torque $= 0$, angular momentum $L = I\omega$ is conserved.

Spinning skater pulls arms in: $I$ decreases → $\omega$ increases ($L = I\omega = \text{const}$)
Planets in elliptical orbit: faster near perihelion (Kepler's 2nd law)
Diver tucking: reduces $I$ → increases $\omega$ for faster rotation

💡 JEE Tip: For rolling problems, always split KE into translational + rotational. The fraction of KE that is rotational depends only on the body shape: disk = 1/3, solid sphere = 2/7, ring = 1/2 of total KE.

⚠ Torque direction: $\vec\tau = \vec r \times \vec F$ (cross product). Use the right-hand rule. Torque is zero when force passes through the pivot ($\vec r$ and $\vec F$ are parallel/anti-parallel, $\sin\phi = 0$).

🔄

Rotational Motion — Formulas

Rotational Kinematics

Angular velocity

$$ \omega = d\frac{\theta}{dt} \quad|\quad v = r\omega $$

$\omega$ in rad/s; $v$ = tangential speed at radius $r$

rad/s

Angular equations

$$ \omega = \omega_0 + \alpha t \quad|\quad \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2 $$

Direct analogues of SUVAT for constant $\alpha$

rad

Torque

$$ \tau = rF \sin\phi = I\alpha $$

$\phi$ = angle between $\vec r$ and $\vec F$; $\vec\tau = \vec r \times \vec F$ (vector)

N·m

Angular Momentum

$L = I\omega = mvr$ (for point mass)

Conserved when $\tau_{ext} = 0$

kg·m²/s

Moment of Inertia & Theorems

Parallel Axis

$$ I = I_{cm} + Md^{2} $$

$d$ = distance from cm to new axis; $I$ always increases

kg·m²

Perpendicular Axis

$$ I_{z} = I_{x} + I_{y} $$

For planar (2D) bodies only; $z$ is axis $\perp$ to the plane

kg·m²

Solid Sphere

$$ I = \frac{2MR^{2}}{5} $$

About diameter; $k^2 = \dfrac{2R^2}{5}$

kg·m²

Solid Cylinder

$$ I = \frac{MR^{2}}{2} $$

About central axis; disk has same formula

kg·m²

Rolling Motion

Total KE (rolling)

$$ KE = \tfrac{1}{2}mv^{2}(1 + \frac{k^{2}}{R^{2}}) $$

$k^2 = \dfrac{I}{m}$ (radius of gyration squared)

Acceleration on incline

$$ a = g \sin \theta / (1 + \frac{k^2}{R^2}) $$

Smaller $\dfrac{k^2}{R^2}$ → faster rolling (solid sphere fastest)

m/s²

Friction in rolling

$$ f = Mk^{2}g \sin\theta / (R^{2} + k^{2}) $$

Static friction provides torque for rotation; no energy loss

Rotational KE

$$ KE_{rot} = \tfrac{1}{2}I\omega^{2} $$

For disk: $\dfrac{1}{3}$ of total KE; sphere: $\dfrac{2}{7}$ of total KE

🔄

Rotational Motion — Solved Examples

A solid sphere of mass 2 kg and radius 0.1 m rolls without slipping with a speed of 4 m/s. Find: (a) translational KE, (b) rotational KE, (c) total KE.

Approach hint

$KE_{trans} = \tfrac{1}{2}mv^2, KE_{rot} = \tfrac{1}{2}I\omega^2$ where $I = \dfrac{2MR^2}{5}$ and $\omega = \dfrac{v}{R}$ for pure rolling.

$KE_{trans} = \tfrac{1}{2}\times 2\times 16 =$ 16 J
$I = \dfrac{2MR^2}{5} = 2\times 2\times \dfrac{0.01}{5} = 0.008$ kg·m²
$\omega = \dfrac{v}{R} = \dfrac{4}{0.1} = 40$ rad/s
$KE_{rot} = \tfrac{1}{2}\times 0.008\times 1600 =$ 6.4 J
Total KE $= 16 + 6.4 =$ 22.4 J $= \tfrac{7}{5}\times \tfrac{1}{2}mv^2$ ✓

✓ $KE_{trans} = 16$ J | $KE_{rot} = 6.4$ J | Total $= 22.4$ J

A disc of mass $M$ and radius $R$ rotates about its central axis at $\omega_1 = 10$ rad/s. A ring of mass $\dfrac{M}{2}$ and radius $R$ is gently placed coaxially on the disc. Find the new angular velocity. (All surfaces smooth.)

Approach hint

No external torque acts on the disc-ring system (surfaces smooth → no friction torque from outside). Conserve angular momentum: $L_i = L_f$.

$I_{disc} = \dfrac{MR^2}{2}$
$I_{ring} = (\dfrac{M}{2})R^2 = \dfrac{MR^2}{2}$
$L_i = I_{disc}\times \omega_1 = \dfrac{MR^2}{2} \times 10 = 5MR^2$
$L_f = (I_{disc} + I_{ring})\times \omega_2 = (\dfrac{MR^2}{2} + \dfrac{MR^2}{2})\times \omega_2 = MR^2\,\omega_2$
$5MR^2 = MR^2\omega_2 \to \omega_2 =$ 5 rad/s

✓ New angular velocity $= 5$ rad/s (halved because total $I$ doubled)

A solid cylinder of mass $M$ and radius $R$ rolls without slipping down an incline of angle $\theta$ and length $L$. Starting from rest, find: (a) the acceleration, (b) the speed at the bottom, (c) the friction force.

Approach hint

Use the rolling-on-incline formula. Alternatively, set up Newton's 2nd law ($F=Ma$ for cm) and torque equation ($\tau = I\alpha$) with rolling constraint $a = R\alpha$, then solve simultaneously for $a$ and friction $f$.

For cylinder: $k^{2} = \dfrac{R^{2}}{2}, so 1 + \dfrac{k^{2}}{R^{2}} = \dfrac{3}{2}$
Acceleration: a = g sinθ / (3/2) = 2g sinθ / 3
Speed at bottom using v² = 2aL: $v = \sqrt{4gL\sin\theta/3}$
From Newton's law: $Mg \sin\theta - f = Ma \to f = Mg \sin\theta - M(2g \sin\theta/3) =$ Mg sinθ / 3
Check: τ = fR = Iα = (MR²/2)(a/R) = MRa/2 = MR(2g sinθ/3)/2 = MRg sinθ/3 = fR ✓

✓ a = 2g sinθ/3 | $v = \sqrt{4gL\sin\theta/3}$ | f = Mg sinθ/3

🔄

Rotational Motion — Diagrams

Click to zoom

Rolling cylinder on incline: contact point has zero velocity; static friction provides rotational torque

MOI comparison: hollow bodies have larger MOI than solid ones of same mass and radius

Conservation of angular momentum: spinning skater pulls arms in → I decreases → ω increases

🔄

Rotational Motion — Quick Quiz

Question 1

A solid sphere and a hollow sphere of the same mass and radius start rolling from rest down an incline. Which reaches the bottom first?

$a = g\sin\theta/(1 + \dfrac{k^2}{R^2})$. Solid sphere: $\dfrac{k^2}{R^2} = \dfrac{2}{5}$, so $a = 5g\sin\theta/7$. Hollow sphere: $\dfrac{k^2}{R^2} = \dfrac{2}{3}$, so $a = 3g\sin\theta/5$. Since $\dfrac{5}{7} > \dfrac{3}{5}$, solid sphere has larger acceleration → reaches bottom first. ✓

Question 2

The moment of inertia of a thin rod of mass $M$ and length $L$ about an axis through one end and perpendicular to the rod is:

$I_{cm} = \dfrac{ML^2}{12}$ (about centre). Using parallel axis: $I_{end} = \dfrac{ML^2}{12} + M(\dfrac{L}{2})^2 = \dfrac{ML^2}{12} + \dfrac{ML^2}{4} = \dfrac{ML^2}{3}$. ✓

Question 3

A disc rotating at 10 rad/s has its angular momentum doubled by applying a constant torque for 4 s. The applied torque is ($I = 2$ kg·m²):

$\Delta L = \tau \times t$. Initial $L = I\omega = 2\times 10 = 20$ kg·m²/s. Final $L = 40$. $\Delta L = 20$. $\tau = \Delta \dfrac{L}{t} = \dfrac{20}{4} = 5$ N·m ✓

🌍

Gravitation

Newton's law of gravitation, satellites, Kepler's laws

○ Not Started

1. Universal Law of Gravitation

Every mass attracts every other mass with force: $F = Gm_1 \dfrac{m_2}{r^2}$

$G = 6.67 \times 10^{-11}$ N·m²/kg² (universal gravitational constant)
Force is always attractive, acts along the line joining the two masses
Follows inverse-square law: doubling distance → force becomes $\dfrac{1}{4}$

2. Variation of g

Condition	Formula	Effect
Surface	$g = \dfrac{GM}{R^2}$	Standard $9.8$ m/s²
Altitude $h$ ($h \ll R$)	$g' \approx g(1 - \dfrac{2h}{R})$	Decreases with height
Altitude $h$ (any)	$g' = \dfrac{gR^2}{(R+h)^2}$	Exact formula
Depth $d$	$g'' = g(1 - \dfrac{d}{R})$	Decreases with depth; zero at centre
Latitude $\phi$	$g_\phi = g - R\omega^2\cos^2\phi$	Max at poles, min at equator

3. Kepler's Laws

1st Law (Ellipse): Planets move in elliptical orbits with the Sun at one focus
2nd Law (Equal areas): A planet sweeps equal areas in equal time intervals → moves faster near perihelion (angular momentum conserved)
3rd Law ($T^2 \propto r^3$): Square of period is proportional to cube of semi-major axis: $T^2 = (4\dfrac{\pi^2}{GM})r^3$

4. Satellites

Orbital velocity: $v_0 = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{gR^2}{r}}$ — independent of satellite mass
Time period: $T = 2\pi \dfrac{r}{v_0} = 2\pi\sqrt{\dfrac{r^3}{GM}}$
Total energy: $E = KE + PE = -\dfrac{GMm}{2r}$ (negative → bound orbit)
Geostationary orbit: $T = 24$ h, $r \approx 42{,}000$ km from centre, directly above equator
Escape velocity: $v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}$ $\approx 11.2$ km/s from Earth
Note: $v_e = \sqrt 2 \times v_0$ at surface (escape speed is $\sqrt 2$ times orbital speed at surface)

5. Gravitational Potential Energy

$U = -\dfrac{GMm}{r}$ (negative; zero at infinity)
Binding energy $=$ energy needed to remove satellite to infinity $= +\dfrac{GMm}{2r}$
For height $h$ above surface: $\Delta U = mgh$ (only valid for $h \ll R$)

💡 JEE Tip: The most tested formula is $T^2 \propto r^3$. If two satellites have orbital radii in ratio 1:4, their periods are in ratio 1:8 (since $4^{3/2} = 8$). Also, $g$ at height $R$ above surface $= \dfrac{g}{4}$ (not $\dfrac{g}{2}$ — a classic trap).

⚠ Common Mistake: $g$ decreases linearly with depth but as inverse-square with height. At depth $\dfrac{R}{2}$, $g = \dfrac{g}{2}$. At height $R$, $g = \dfrac{g}{4}$. Students often confuse these two cases.

🌍

Gravitation — Formulas

Gravitational Force & Field

Newton's Law

$$ F = \frac{Gm_{1}m_{2}}{r^{2}} $$

$G = 6.67\times 10^{-11}$ N·m²/kg²; always attractive

Surface gravity

$$ g = \frac{GM}{R^{2}} $$

$g \approx 9.8$ m/s² at Earth's surface; depends on $M$ and $R$

m/s²

g at altitude h

$$ g' = \frac{gR^{2}}{(R+h)^{2}} $$

Exact; for $h \ll R$: $g' \approx g(1-\dfrac{2h}{R})$

m/s²

g at depth d

$$ g'' = g(1-\frac{d}{R}) $$

Linear decrease; zero at Earth's centre

m/s²

Orbital Mechanics

Orbital velocity

$$ v_{0} = \sqrt{\frac{GM}{r}} $$

$r$ = distance from Earth's centre; independent of satellite mass

m/s

Orbital period

$$ T = 2\pi\sqrt{\frac{r^{3}}{GM}} $$

$T^2 \propto r^3$ (Kepler's 3rd law)

Escape velocity

$$ v_{e} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} $$

$\approx 11.2$ km/s from Earth; $v_e = \sqrt 2\cdot v_0$ at surface

m/s

Satellite total energy

$$ E = -\frac{GMm}{2r} $$

Negative (bound); $KE = \dfrac{GMm}{2r}$; $PE = -\dfrac{GMm}{r}$

Gravitational PE & Kepler

Gravitational PE

$$ U = -\frac{GMm}{r} $$

Zero at infinity; negative for bound systems

Kepler's 3rd Law

$$ T^{2} = (4\frac{\pi^{2}}{GM})a^{3} $$

$a$ = semi-major axis; $T_1^2/T_2^2 = \dfrac{a_1^3}{a_2^3}$

—

🌍

Gravitation — Solved Examples

At what height above Earth's surface does $g$ become $\dfrac{g}{4}$? [$R = 6400$ km]

Approach hint

Use the exact formula $g' = \dfrac{gR^2}{(R+h)^2}$. Set $g' = \dfrac{g}{4}$ and solve for $h$.

$\dfrac{g}{4} = \dfrac{gR^2}{(R+h)^2} \to (R+h)^2 = 4R^2 \to R+h = 2R$
$h = R =$ 6400 km

✓ At height $h = R = 6400$ km above surface, $g$ reduces to $\dfrac{g}{4}$

A satellite orbits Earth at height 3600 km above the surface. Find its orbital speed and period. [$R = 6400$ km, $g = 10$ m/s²]

Approach hint

$r = R + h$. Use $v_0 = \sqrt{\dfrac{gR^2}{r}}$ and $T = 2\pi \dfrac{r}{v_0}$.

$r = 6400 + 3600 = 10{,}000$ km $= 10^7$ m
$v_0 = \sqrt{\dfrac{gR^2}{r}} = \sqrt{10 \times (6.4\times 10^6)^2 / 10^7}$
$= \sqrt{10 \times 4.096\times 10^{13} / 10^7} = \sqrt{4.096\times 10^7} \approx$ 6400 m/s $= 6.4$ km/s
$T = 2\pi \dfrac{r}{v_0} = 2\pi \times \dfrac{10^7}{6400} \approx$ 9817 s $\approx 2.73$ hours

✓ $v_0 \approx 6.4$ km/s | $T \approx 2.73$ hours

Two satellites A and B orbit Earth in circular orbits of radii $R$ and $4R$ respectively. Compare: (a) orbital speeds, (b) periods, (c) total mechanical energies (satellite mass $m$ each).

Approach hint

$v \propto \dfrac{1}{\sqrt r}, T \propto r^{3/2}$, $E = -\dfrac{GMm}{2r} \propto -\dfrac{1}{r}$. Express ratios directly.

Speeds: $\dfrac{v_A}{v_B} = \sqrt{\dfrac{4R}{R}} = 2$ → A is twice as fast as B
Periods: $T_A/T_B = (\dfrac{R}{4R})^{3/2} = (\dfrac{1}{4})^{3/2} = \dfrac{1}{8} \to T_B = 8T_A$
Total energy: $\dfrac{E_A}{E_B} = (\dfrac{4R}{R}) = 4 \to |E_A| = 4|E_B|$; A has more negative energy (more bound)

✓ $v_A = 2v_B$ | $T_B = 8T_A$ | $E_A = 4E_B$ (A is more tightly bound)

🌍

Gravitation — Diagrams

Click to zoom

g vs distance from Earth's centre: linear inside (depth), inverse-square outside (altitude)

Circular satellite orbit: gravitational force provides centripetal force; total energy = −GMm/2r

🌍

Gravitation — Quick Quiz

Question 1

The ratio of escape velocity to orbital velocity of a satellite at Earth's surface is:

$v_e = \sqrt{2gR}$; $v_0 = \sqrt{gR}$. Ratio $= \dfrac{\sqrt{2gR}}{\sqrt{gR}} = \sqrt 2$ ✓

Question 2

Two planets have the same average density but different radii $R_1 = 2R_2$. The ratio of their escape velocities $\dfrac{v_{e1}}{v_{e2}}$ is:

$M \propto \rho R^3$, so $v_e = \sqrt{\dfrac{2GM}{R}} \propto \sqrt{R^2} = R$ (for same density). $\dfrac{v_{e1}}{v_{e2}} = \dfrac{R_1}{R_2} = 2$ ✓

Question 3

A satellite is moved from orbit radius $r$ to $4r$. Its kinetic energy becomes:

$KE = \dfrac{GMm}{2r} \propto \dfrac{1}{r}$. When $r \to 4r$, $KE \to \dfrac{KE}{4}$. So KE becomes $\dfrac{1}{4}$ of original. ✓

🔬

Properties of Matter

Elasticity, fluid statics, surface tension, viscosity, Bernoulli

○ Not Started

1. Elasticity

Modulus	Stress	Strain	Formula
Young's ($Y$)	$F/A$ (longitudinal)	$\Delta L/L$	$Y = \dfrac{FL}{A\,\Delta L}$
Bulk ($B$)	$-\Delta P$ (hydrostatic)	$\Delta V/V$	$B = -\Delta \dfrac{P}{\Delta V/V}$
Shear ($\dfrac{G}{\eta}$)	$F/A$ (tangential)	$\phi$ (shear angle)	$G = \dfrac{F/A}{\phi}$

Poisson's ratio: $\sigma = -\dfrac{\text{lateral strain}}{\text{longitudinal strain}}$; range: 0 to 0.5.

2. Fluid Statics

Pressure at depth $h$: $P = P_0 + \rho gh$ (same at all points at same horizontal level)
Pascal's Law: Pressure applied to enclosed fluid transmits equally in all directions → hydraulic lift: $F_2/F_1 = A_2/A_1$
Archimedes' Principle: Buoyancy force $=$ weight of fluid displaced $= \rho_{fluid}\,V g$
Body floats when $\rho_{body} \le \rho_{fluid}$; fraction submerged $= \dfrac{\rho_{body}}{\rho_{fluid}}$

3. Surface Tension

Surface tension $T = \dfrac{\text{Force}}{\text{Length}}$ (N/m); or Surface energy $= T \times \Delta A$
Excess pressure inside a drop: $\Delta P = \dfrac{2T}{r}$
Excess pressure inside a soap bubble (two surfaces): $\Delta P = \dfrac{4T}{r}$
Capillary rise: $h = 2T\cos\theta/(\rho gr)$; rises in tubes it wets ($\theta < 90°$)
Angle of contact: water-glass $\approx 0°$ (wets), mercury-glass $\approx 140°$ (doesn't wet)

4. Viscosity & Fluid Flow

Stokes' drag: $F = 6\pi\eta rv$ (sphere of radius $r$, speed $v$, viscosity $\eta$)
Terminal velocity: $v_T = 2r^2(\rho - \sigma)\dfrac{g}{9\eta}$ ($\rho$ = sphere density, $\sigma$ = fluid density)
Continuity equation: $A_1 v_1 = A_2 v_2$ (mass conservation for incompressible flow)
Bernoulli's theorem: $P + \tfrac{1}{2}\rho v^2 + \rho gh = \text{constant}$ (energy conservation per unit volume)
Applications: venturimeter, pitot tube, aerofoil lift, atomiser

💡 JEE Tip: Always distinguish drop (one surface, $\Delta P = \dfrac{2T}{r}$) from bubble (two surfaces, $\Delta P = \dfrac{4T}{r}$). In Bernoulli problems: where pipe is narrow ($A$ small), velocity is high (from continuity) → pressure is low (from Bernoulli).

⚠ Bernoulli ≠ always applicable: It holds for streamline, non-viscous, incompressible flow. Don't apply it to turbulent flow or highly viscous fluids without qualification.

🔬

Properties of Matter — Formulas

Elasticity

Young's Modulus

$$ Y = (F/A) / (\Delta L/L) = \frac{FL}{A\Delta L} $$

Steel: ~200 GPa; rubber: ~0.01 GPa

Bulk Modulus

$$ B = -\Delta P / (\Delta V/V) $$

Compressibility k = 1/B; water has very large B

Elastic PE in wire

$$ U = \tfrac{1}{2} \times (\text{stress}) \times (\text{strain}) \times \text{volume} $$

$= \tfrac{1}{2}F\cdot\Delta L = F^2 \dfrac{L}{2AY}$

Thermal stress

$$ \text{Thermal stress} = Y\alpha\,\Delta T $$

$\alpha$ = coefficient of linear expansion; occurs when expansion is prevented

Fluid Statics & Surface Tension

Gauge pressure

$$ P = \rho gh $$

$h$ = depth below free surface; independent of cross-section shape

Buoyant force

$$ F_{b} = \rho_{fluid}Vg $$

$V$ = volume submerged; equals weight of displaced fluid

Excess pressure (drop)

$$ \Delta P = \frac{2T}{r} $$

One liquid surface; e.g. oil drop in water

Excess pressure (bubble)

$$ \Delta P = \frac{4T}{r} $$

Two surfaces (inner + outer); soap bubble in air

Capillary rise

$$ h = 2T \cos\theta / (\rho gr) $$

$r$ = tube radius; $\theta$ = contact angle; rises if $\theta < 90°$

Terminal velocity

$$ v_{T} = 2r^{2}(\rho-\sigma)g / (9\eta) $$

$\rho$ = sphere, $\sigma$ = fluid density; $\eta$ = viscosity

m/s

Bernoulli's Theorem

$$ P + \tfrac{1}{2} \rho v^2 + \rho gh = const $$

Energy per unit volume; for non-viscous, streamline flow

Continuity

$$ A_{1}v_{1} = A_{2}v_{2} $$

Incompressible fluid; flow rate $Q = Av$ = constant

m³/s

🔬

Properties of Matter — Solved Examples

A steel wire of length 2 m and cross-section 2 mm² is stretched by a force of 400 N. Find the extension. [$Y_{steel} = 2\times 10^{11}$ Pa]

Approach hint

$Y = \dfrac{FL}{A\Delta L} \to \Delta L = \dfrac{FL}{AY}$. Convert mm² to m².

$A = 2$ mm² $= 2\times 10^{-6}$ m²
$\Delta L = \dfrac{FL}{AY} = \dfrac{400\times 2}{2\times 10^{-6}\times 2\times 10^{11}}$
$= \dfrac{800}{4\times 10^5} =$ $2\times 10^{-3}$ m $= 2$ mm

✓ Extension $= 2$ mm

Water flows through a horizontal pipe that narrows from diameter 10 cm to 5 cm. If the speed in the wider section is 1 m/s, find: (a) speed in the narrow section, (b) pressure difference. [$\rho_{water} = 1000$ kg/m³]

Approach hint

Use continuity ($A_1 v_1 = A_2 v_2$) to get $v_2$, then Bernoulli ($P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$ for horizontal pipe with $h_1 = h_2$).

$A_1 = \pi(0.05)^2 = 7.85\times 10^{-3}$ m²; $A_2 = \pi(0.025)^2 = 1.96\times 10^{-3}$ m²
By continuity: $v_2 = A_1 \dfrac{v_1}{A_2} = (\dfrac{D_1}{D_2})^2 v_1 = (\dfrac{10}{5})^2 \times 1 =$ 4 m/s
Bernoulli (horizontal): $P_1 + \tfrac{1}{2}\rho v_1^2 = P_2 + \tfrac{1}{2}\rho v_2^2$
$P_1 - P_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2) = \tfrac{1}{2}\times 1000\times (16-1) =$ 7500 Pa

✓ $v_{narrow} = 4$ m/s | $P_{wide} - P_{narrow} = 7500$ Pa

A spherical ball of density 8000 kg/m³ and radius 5 mm falls through glycerine of density 1200 kg/m³ and viscosity $\eta = 0.83$ Pa·s. Find the terminal velocity.

Approach hint

At terminal velocity, drag (Stokes) + buoyancy = weight. Use $v_T = 2r^2(\rho-\sigma)\dfrac{g}{9\eta}$ directly.

$r = 5\times 10^{-3}$ m, $\rho = 8000$ kg/m³, $\sigma = 1200$ kg/m³, $\eta = 0.83$ Pa·s
$v_T = 2r^2(\rho-\sigma)\dfrac{g}{9\eta} = 2\times (5\times 10^{-3})^2 \times (8000-1200) \times 10 / (9 \times 0.83)$
$= 2 \times 25\times 10^{-6}\times 6800 \times \dfrac{10}{7.47}$
$= \dfrac{3.4}{7.47} \approx$ 0.455 m/s $\approx 45.5$ cm/s

✓ Terminal velocity $\approx 0.455$ m/s ($45.5$ cm/s)

🔬

Properties of Matter — Diagrams

Click to zoom

Stress-strain curve: elastic region (Hooke's law), yield point, plastic region, fracture

Bernoulli + continuity: narrower section → faster flow → lower pressure (Venturimeter principle)

🔬

Properties of Matter — Quick Quiz

Question 1

The excess pressure inside a soap bubble of radius $r$ and surface tension $T$ is:

A soap bubble has TWO surfaces (inner and outer). Each surface contributes $\dfrac{2T}{r}$ of excess pressure, and there are two surfaces → $\Delta P = \dfrac{4T}{r}$ ✓

Question 2

A wooden block of density 600 kg/m³ floats in water ($\rho = 1000$ kg/m³). The fraction of the block above the water surface is:

Fraction submerged $= \dfrac{\rho_{block}}{\rho_{water}} = \dfrac{600}{1000} = 0.6$. Fraction above $= 1 - 0.6 = 0.4$ ✓

Question 3

Terminal velocity of a ball falling in a viscous liquid depends on the radius $r$ as:

$v_T = 2r^2(\rho-\sigma)\dfrac{g}{9\eta} \propto r^2$. Doubling the radius quadruples the terminal velocity. ✓

Linear	Angular	Relation
Displacement \(s\)	Angle \(\theta\)	\(s = r\theta\)
Velocity \(v\)	Angular velocity \(\omega\)	\(v = r\omega\)
Acceleration \(a\)	Angular acceleration \(\alpha\)	\(a_{tangential} = r\alpha\)
Mass \(m\)	Moment of inertia \(I\)	\(I = \Sigma mr^2\)
Force \(F\)	Torque \(\tau\)	\(\tau = rF\sin\phi\)
\(F = ma\)	\(\tau = I\alpha\)	—
Momentum \(p = mv\)	Angular momentum \(L = I\omega\)	—
\(KE = \tfrac{1}{2}mv^2\)	\(KE_{rot} = \tfrac{1}{2}I\omega^2\)	—

Condition	Formula	Effect
Surface	\(g = \dfrac{GM}{R^2}\)	Standard \(9.8\) m/s²
Altitude \(h\) (\(h \ll R\))	\(g' \approx g(1 - \dfrac{2h}{R})\)	Decreases with height
Altitude \(h\) (any)	\(g' = \dfrac{gR^2}{(R+h)^2}\)	Exact formula
Depth \(d\)	\(g'' = g(1 - \dfrac{d}{R})\)	Decreases with depth; zero at centre
Latitude \(\phi\)	\(g_\phi = g - R\omega^2\cos^2\phi\)	Max at poles, min at equator

Modulus	Stress	Strain	Formula
Young's (\(Y\))	\(F/A\) (longitudinal)	\(\Delta L/L\)	\(Y = \dfrac{FL}{A\,\Delta L}\)
Bulk (\(B\))	\(-\Delta P\) (hydrostatic)	\(\Delta V/V\)	\(B = -\Delta \dfrac{P}{\Delta V/V}\)
Shear (\(\dfrac{G}{\eta}\))	\(F/A\) (tangential)	\(\phi\) (shear angle)	\(G = \dfrac{F/A}{\phi}\)

Law	Statement	Mathematical Form
1st (Inertia)	A body stays at rest or uniform motion unless acted on by a net external force	If \(F_{net} = 0\), then \(a = 0\)
2nd (F=ma)	Net force equals rate of change of momentum	\(F = ma = \dfrac{dp}{dt}\)
3rd (Action-Reaction)	Every action has an equal and opposite reaction — on different bodies	\(\vec F_{AB} = -\vec F_{BA}\)

Type	Formula	When?	Note
Static	\(f_s \le \mu_s N\)	Body at rest; value adjusts to prevent motion	Can be less than \(\mu_s N\)
Kinetic	\(f_k = \mu_k N\)	Body in motion relative to surface	Always \(\mu_k < \mu_s\)
Rolling	\(f_r = \mu_r N\)	Rolling contact	\(\mu_r \ll \mu_k\)

Force	Conservative?	Potential Energy
Gravity	✅ Yes	\(U = mgh\)
Spring (Hooke's law)	✅ Yes	\(U = \tfrac{1}{2}kx^2\)
Friction	❌ No	No PE defined; energy lost as heat
Normal force	—	Does zero work (always ⊥ to motion)

Body	Axis	I
Solid sphere	Through centre	\(\dfrac{2MR^2}{5}\)
Hollow sphere	Through centre	\(\dfrac{2MR^2}{3}\)
Solid cylinder / disk	Along axis	\(\dfrac{MR^2}{2}\)
Hollow cylinder / ring	Along axis	\(MR^2\)
Thin rod	Through centre \(\perp\) rod	\(\dfrac{ML^2}{12}\)
Thin rod	Through one end \(\perp\) rod	\(\dfrac{ML^2}{3}\)
Rectangular plate (\(a\times b\))	Through centre	\(\dfrac{M(a^2+b^2)}{12}\)