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Mechanics Formulas

Complete chapter-wise formula sheet for Mechanics (Physics) covering key equations, parameters, traps, and concept diagrams for JEE Main & Advanced.

11
Chapters
83
Formulas
5,102+
DB PYQs

Chapter Directory (24 Chapters)

Click any chapter to view dedicated formula sheet & concept diagrams
Mechanics462 Qs

Units, Dimensions & Errors

Dimensional analysis, dimensional formulas of physical quantities, propagation of errors, vernier callipers and screw gauge.

8 Formulas
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Mechanics560 Qs

Kinematics & Motion in 1D & 2D

Uniformly accelerated motion, projectile motion, trajectory parabola, circular kinematics, and relative velocity.

8 Formulas Diagram
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Mechanics348 Qs

Laws of Motion & Friction

Newton's laws of motion, free body diagrams, static and kinetic friction, banking of roads, and pseudo-force.

7 Formulas
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Mechanics343 Qs

Work, Power & Energy

Work-energy theorem, conservative forces, potential energy, elastic collisions in 1D/2D, and power.

7 Formulas
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Mechanics271 Qs

Center of Mass & Momentum Conservation

Center of mass coordinates, continuous mass distribution, momentum conservation, and explosion of bodies.

7 Formulas
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Mechanics665 Qs

Rotational Motion & Rigid Bodies

Moment of inertia, torque, angular momentum conservation, rolling motion without slipping, and rotational kinetic energy.

8 Formulas Diagram
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Mechanics519 Qs

Gravitation & Satellite Motion

Newton's law, variation of g with height/depth/latitude, gravitational potential energy, orbital and escape speed.

8 Formulas
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Mechanics413 Qs

Properties of Matter & Elasticity

Hooke's law, Young's modulus, bulk modulus, shear modulus, elastic potential energy, and thermal stress.

8 Formulas
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Mechanics368 Qs

Fluid Mechanics & Surface Tension

Pascal's law, Archimedes principle, continuity equation, Bernoulli's theorem, viscosity (Stokes law, Poiseuille), and surface tension.

7 Formulas
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Mechanics511 Qs

Simple Harmonic Motion (SHM)

Kinematics and dynamics of SHM, spring-mass systems, simple pendulum, energy conservation, and superpositions.

8 Formulas Diagram
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Mechanics642 Qs

Wave Motion & Sound Waves

Wave equation, speed of sound in gases, standing waves in strings and organ pipes, beats, and Doppler effect.

7 Formulas
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Formulas Reference (Mechanics)

11 Chapters match
MechanicsHigh Yield (1-2 Qs in JEE)462 PYQs

Units, Dimensions & Errors

Dimensional analysis, dimensional formulas of physical quantities, propagation of errors, vernier callipers and screw gauge.

Propagation of Errors (Relative & Percentage)

Z=ApBqCr  ⟹  ΔZZ=pΔAA+qΔBB+rΔCCZ = \frac{A^p B^q}{C^r} \implies \frac{\Delta Z}{Z} = p \frac{\Delta A}{A} + q \frac{\Delta B}{B} + r \frac{\Delta C}{C}

Fractional and percentage error estimation in derived quantities.

Errors always add up; never subtract error terms even if quantity is in the denominator.

Vernier Calliper & Screw Gauge Least Count

LCvernier=1 MSD−1 VSD=1 MSDN,LCscrew gauge=PitchTotal Circular Scale Divisions\text{LC}_{\text{vernier}} = 1 \text{ MSD} - 1 \text{ VSD} = \frac{1 \text{ MSD}}{N}, \quad \text{LC}_{\text{screw gauge}} = \frac{\text{Pitch}}{\text{Total Circular Scale Divisions}}

Instrument precision formulas for laboratory physics where N divisions of vernier scale coincide with (N - 1) main scale divisions.

Pitch is distance advanced in one complete rotation of the thimble (usually 0.5 mm or 1.0 mm).

Zero Error & True Reading Correction

True Reading=Observed Reading−(Zero Error),Observed Reading=MSR+(VSR×LC)\text{True Reading} = \text{Observed Reading} - (\text{Zero Error}), \quad \text{Observed Reading} = \text{MSR} + (\text{VSR} \times \text{LC})

Correction for instrumental zero offset. If zero is to the right/above, error is positive; if to the left/below, error is negative.

Positive zero error is SUBTRACTED; negative zero error is ADDED (since -(-e) = +e).

Important Dimensional Formulas

[G]=[M−1L3T−2],[h]=[ML2T−1],[ε0]=[M−1L−3T4A2],[μ0]=[MLT−2A−2][G] = [M^{-1}L^3T^{-2}], \quad [h] = [ML^2T^{-1}], \quad [\varepsilon_0] = [M^{-1}L^{-3}T^4A^2], \quad [\mu_0] = [MLT^{-2}A^{-2}]

Fundamental universal constants tested in JEE Main dimensional matching.

Speed of light relation: c = 1 / √(μ₀ ε₀) has dimension [LT⁻¹].

Basic Mathematics Used in Physics (Approximations)

For θ≪1 rad:sin⁡θ≈θ,tan⁡θ≈θ,cos⁡θ≈1−θ22;(1+x)n≈1+nx(∣x∣≪1)\text{For } \theta \ll 1 \text{ rad}: \sin\theta \approx \theta, \quad \tan\theta \approx \theta, \quad \cos\theta \approx 1 - \frac{\theta^2}{2}; \quad (1 + x)^n \approx 1 + nx \quad (|x| \ll 1)

Essential mathematical approximations for small oscillations, optics angles, thermal expansions, and gravitation height variations.

Angle θ must be converted to radians before applying small angle trigonometric approximations.

Significant Figures & Calculation Rules

Add/Sub: Least decimal places;Mult/Div: Least significant figures;σmean=σN\text{Add/Sub: Least decimal places}; \quad \text{Mult/Div: Least significant figures}; \quad \sigma_{\text{mean}} = \frac{\sigma}{\sqrt{N}}

Precision rules for experimental physics observations and standard deviation of sample mean.

All non-zero digits are significant; zeroes between non-zero digits are significant; trailing zeroes after decimal are significant.

Propagation of Experimental Errors

Z=ApBqCr  ⟹  ΔZZ=p(ΔAA)+q(ΔBB)+r(ΔCC)Z = \frac{A^p B^q}{C^r} \implies \frac{\Delta Z}{Z} = p\left(\frac{\Delta A}{A}\right) + q\left(\frac{\Delta B}{B}\right) + r\left(\frac{\Delta C}{C}\right)

Logarithmic differentiation error combination. Fractional and percentage errors add up in quadrature or worst-case linear bound.

In addition or subtraction (Z = A ± B), absolute errors always add: ΔZ = ΔA + ΔB (never subtract errors).

Spherometer Radius of Curvature Formula

R=l26h+h2,Pitch=Distance movedRotations,LC=PitchNR = \frac{l^2}{6h} + \frac{h}{2}, \quad \text{Pitch} = \frac{\text{Distance moved}}{\text{Rotations}}, \quad \text{LC} = \frac{\text{Pitch}}{N}

Laboratory measurement of radius of curvature of convex/concave optical lenses where l is mean leg separation and h is central sagitta elevation.

For very small sagitta (h << l), R ≈ l² / (6h).
MechanicsHigh Yield (2-3 Qs in JEE)560 PYQs

Kinematics & Motion in 1D & 2D

Uniformly accelerated motion, projectile motion, trajectory parabola, circular kinematics, and relative velocity.

Chapter

Equations of Motion (Constant Acceleration)

v=u+at,s=ut+12at2,v2=u2+2as,sn=u+a2(2n−1)v = u + at, \quad s = ut + \frac{1}{2}at^2, \quad v^2 = u^2 + 2as, \quad s_n = u + \frac{a}{2}(2n - 1)

Standard 1D and vector kinematic equations.

s_n is the displacement specifically in the n-th second.

Projectile Motion Core Parameters

T=2usin⁡θg,H=u2sin⁡2θ2g,R=u2sin⁡2θgT = \frac{2u\sin\theta}{g}, \quad H = \frac{u^2\sin^2\theta}{2g}, \quad R = \frac{u^2\sin 2\theta}{g}

Time of flight, maximum height, and horizontal range on flat ground.

Range is identical for complementary launch angles θ and (90° - θ).

Equation of Projectile Trajectory

y=xtan⁡θ−gx22u2cos⁡2θ=xtan⁡θ(1−xR)y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta} = x\tan\theta \left(1 - \frac{x}{R}\right)

Parabolic trajectory equation linking horizontal coordinate x to vertical coordinate y.

The form y = x tan θ (1 - x/R) saves massive calculation time when Range R is known.

Centripetal & Tangential Acceleration

ac=v2r=ω2r,at=dvdt=αr,atotal=ac2+at2a_c = \frac{v^2}{r} = \omega^2 r, \quad a_t = \frac{dv}{dt} = \alpha r, \quad a_{total} = \sqrt{a_c^2 + a_t^2}

Kinematics of circular motion with radial and tangential acceleration components.

Uniform circular motion has a_t = 0 (constant speed, changing velocity direction).

Relative Velocity & River-Boat Crossing

v⃗AB=v⃗A−v⃗B,tmin=dvbr,Zero Drift Condition: sin⁡θ=vrvbr  (vbr>vr)\vec{v}_{AB} = \vec{v}_A - \vec{v}_B, \quad t_{min} = \frac{d}{v_{br}}, \quad \text{Zero Drift Condition: } \sin\theta = \frac{v_r}{v_{br}} \; (v_{br} > v_r)

Relative velocity in 2D and crossing a river of width d.

For shortest time, boat must steer perpendicular to the river bank.

Projectile on an Inclined Plane

T=2usin⁡(α−β)gcos⁡β,R=u2gcos⁡2β[sin⁡(2α−β)−sin⁡β],Rmax=u2g(1+sin⁡β)T = \frac{2u\sin(\alpha - \beta)}{g\cos\beta}, \quad R = \frac{u^2}{g\cos^2\beta}[\sin(2\alpha - \beta) - \sin\beta], \quad R_{\text{max}} = \frac{u^2}{g(1 + \sin\beta)}

Launch angle α from horizontal onto plane inclined at angle β. Replace g with components g cos β and g sin β.

Maximum range up the plane occurs when α = π/4 + β/2; down the plane replace β with -β.

Radius of Curvature of Trajectory

ρ=v2a⊥=[1+(dy/dx)2]3/2∣d2y/dx2∣,ρtop=u2cos⁡2θg\rho = \frac{v^2}{a_\perp} = \frac{[1 + (dy/dx)^2]^{3/2}}{|d^2y/dx^2|}, \quad \rho_{\text{top}} = \frac{u^2\cos^2\theta}{g}

Radius of the osculating circle at any point along a curved path where a_⊥ is normal acceleration.

At the highest point of standard projectile motion, speed is u cos θ and normal acceleration is g.

Rain-Man & Wind-Air Relative Motion

v⃗r/m=v⃗r−v⃗m,tan⁡θ=∣v⃗m∣∣v⃗r∣\vec{v}_{r/m} = \vec{v}_r - \vec{v}_m, \quad \tan\theta = \frac{|\vec{v}_m|}{|\vec{v}_r|}

Apparent direction of rainfall relative to a moving observer.

To protect from rain, hold umbrella in the direction of apparent velocity v_{r/m}.
MechanicsHigh Yield (2 Qs in JEE)348 PYQs

Laws of Motion & Friction

Newton's laws of motion, free body diagrams, static and kinetic friction, banking of roads, and pseudo-force.

Newton's Second Law & Momentum

F⃗net=dp⃗dt=ma⃗,J⃗=∫F⃗ dt=Δp⃗\vec{F}_{net} = \frac{d\vec{p}}{dt} = m\vec{a}, \quad \vec{J} = \int \vec{F} \, dt = \Delta \vec{p}

Net force equals rate of change of linear momentum; Impulse J equals change in momentum.

Area under Force vs Time (F-t) graph gives impulse.

Static & Kinetic Friction

fs≤μsN,fk=μkN(μs>μk)f_s \le \mu_s N, \quad f_k = \mu_k N \quad (\mu_s > \mu_k)

Limiting static friction and constant sliding kinetic friction.

Angle of repose θ equals angle of friction: tan θ = μ_s.

Banking of Roads — v_max, v_min & Optimum Speed

vopt=rgtan⁡θ,vmax=rg⋅tan⁡θ+μs1−μstan⁡θ,vmin=rg⋅tan⁡θ−μs1+μstan⁡θv_{\text{opt}} = \sqrt{rg\tan\theta}, \quad v_{\text{max}} = \sqrt{rg\cdot\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}}, \quad v_{\text{min}} = \sqrt{rg\cdot\frac{\tan\theta - \mu_s}{1 + \mu_s\tan\theta}}

On a banked road of radius r and bank angle θ: at optimum speed v_opt lateral friction is zero. Between v_min and v_max the car negotiates the bend safely. Below v_min, car slides inward (friction acts up the bank); above v_max, car skids outward (friction acts down the bank).

v_min exists only when tan θ > μ_s. For tan θ ≤ μ_s, even at rest the car won't slide inward. On unbanked road (θ = 0): v_max = √(μ_s rg) and v_min = 0.

Connected Bodies & Pulley Systems

a=Net driving forceTotal mass,T=2m1m2m1+m2g  (Atwood machine)a = \frac{\text{Net driving force}}{\text{Total mass}}, \quad T = \frac{2 m_1 m_2}{m_1 + m_2} g \; (\text{Atwood machine})

Standard acceleration and string tension formulas.

Tension in light inextensible string is uniform throughout unless passing over massive pulley.

Two-Block System & Slipping Threshold

acommon=Fm1+m2,Fmax=μm1g(1+m2m1)=μ(m1+m2)ga_{\text{common}} = \frac{F}{m_1 + m_2}, \quad F_{\text{max}} = \mu m_1 g \left(1 + \frac{m_2}{m_1}\right) = \mu(m_1 + m_2)g

Maximum external pulling force F applied on bottom block before top block m₁ starts to slip relative to m₂.

If applied force F ≤ F_max, both blocks accelerate together as a single combined unit.

Pseudo-Force & Apparent Weight in Elevator

F⃗pseudo=−ma⃗frame,N=m(g±a)(+ for upward a,  − for downward a)\vec{F}_{\text{pseudo}} = -m \vec{a}_{\text{frame}}, \quad N = m(g \pm a) \quad (+ \text{ for upward } a, \; - \text{ for downward } a)

Inertial force applied to masses when observed from a non-inertial frame accelerating at a_frame.

In free fall (a = g downwards), apparent weight N = 0 (weightlessness).

Virtual Work & String Constraint Relations

∑T⃗⋅x⃗=0  ⟹  ∑T⃗⋅v⃗=0  ⟹  ∑T⃗⋅a⃗=0\sum \vec{T} \cdot \vec{x} = 0 \implies \sum \vec{T} \cdot \vec{v} = 0 \implies \sum \vec{T} \cdot \vec{a} = 0

Total virtual work done by tension in an inextensible, massless string system is identically zero.

Wedge constraint: Relative velocity components perpendicular to the common inclined contact face must be equal.
MechanicsHigh Yield (2 Qs in JEE)343 PYQs

Work, Power & Energy

Work-energy theorem, conservative forces, potential energy, elastic collisions in 1D/2D, and power.

Work-Energy Theorem

Wnet=Wconservative+Wnon-conservative+Wexternal=ΔK=Kf−KiW_{\text{net}} = W_{\text{conservative}} + W_{\text{non-conservative}} + W_{\text{external}} = \Delta K = K_f - K_i

Net work done by all forces on a particle equals the change in its kinetic energy.

Most reliable problem-solving tool in mechanics.

Conservative Force & Potential Energy

F⃗=−∇⃗U=−(∂U∂xi^+∂U∂yj^+∂U∂zk^),ΔU=−Wcons\vec{F} = -\vec{\nabla} U = -\left(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k}\right), \quad \Delta U = -W_{\text{cons}}

Force derived from scalar potential energy gradient.

Stable equilibrium occurs where dU/dx = 0 and d²U/dx² > 0 (minimum potential energy).

1D Elastic Collision Velocities

v1=(m1−m2m1+m2)u1+(2m2m1+m2)u2,e=v2−v1u1−u2=1v_1 = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2, \quad e = \frac{v_2 - v_1}{u_1 - u_2} = 1

Final velocities after head-on perfectly elastic collision.

If m₁ = m₂ in elastic collision, velocities are completely exchanged: v₁ = u₂ and v₂ = u₁.

Instantaneous & Average Power

P=dWdt=F⃗⋅v⃗,Pavg=WtotalΔtP = \frac{dW}{dt} = \vec{F} \cdot \vec{v}, \quad P_{\text{avg}} = \frac{W_{\text{total}}}{\Delta t}

Rate of doing work (1 Watt = 1 Joule/second = 1 kg m² s⁻³).

1 horsepower (hp) ≈ 746 Watts.

Vertical Circular Motion (String & Light Rod)

vbottom≥5gR,vtop≥gR,Tbottom−Ttop=6mg,T(θ)=mv2R+mgcos⁡θv_{\text{bottom}} \ge \sqrt{5gR}, \quad v_{\text{top}} \ge \sqrt{gR}, \quad T_{\text{bottom}} - T_{\text{top}} = 6mg, \quad T(\theta) = \frac{mv^2}{R} + mg\cos\theta

Critical thresholds for looping the vertical circle without slackening of string.

For a massless rigid rod, speed at top can be 0, so critical bottom speed is only √(4gR) = 2√(gR).

Inelastic Collision & Kinetic Energy Loss

ΔK=12(m1m2m1+m2)(1−e2)(u1−u2)2,v1=m1u1+m2u2−em2(u1−u2)m1+m2\Delta K = \frac{1}{2} \left(\frac{m_1 m_2}{m_1 + m_2}\right) (1 - e^2) (u_1 - u_2)^2, \quad v_1 = \frac{m_1 u_1 + m_2 u_2 - e m_2(u_1 - u_2)}{m_1 + m_2}

Loss of mechanical kinetic energy dissipated as heat and sound during collision with coefficient of restitution e.

For perfectly inelastic collision (e = 0), both bodies stick together with maximum possible KE loss.

Spring Potential Energy & Work

Us=12kx2,Wspring=−ΔUs=12k(xi2−xf2)U_s = \frac{1}{2} k x^2, \quad W_{\text{spring}} = -\Delta U_s = \frac{1}{2} k (x_i^2 - x_f^2)

Elastic potential energy stored in an ideal spring deformed by displacement x.

Effective spring constant for parallel springs: k_p = k₁ + k₂; for series springs: 1/k_s = 1/k₁ + 1/k₂.
MechanicsHigh Yield (1-2 Qs in JEE)271 PYQs

Center of Mass & Momentum Conservation

Center of mass coordinates, continuous mass distribution, momentum conservation, and explosion of bodies.

Center of Mass Coordinates

R⃗cm=∑mir⃗i∑mi=∫r⃗ dm∫dm\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i} = \frac{\int \vec{r} \, dm}{\int dm}

Mass-weighted average position vector of a multi-particle or continuous system.

If net external force is zero, center of mass moves with constant velocity.

CM of Standard Uniform Symmetrical Bodies

ycm=2Rπ  (Semicircular ring),ycm=4R3π  (Semicircular disc),ycm=3R8  (Solid hemisphere)y_{cm} = \frac{2R}{\pi} \; (\text{Semicircular ring}), \quad y_{cm} = \frac{4R}{3\pi} \; (\text{Semicircular disc}), \quad y_{cm} = \frac{3R}{8} \; (\text{Solid hemisphere})

Standard heights of center of mass measured from the diametric base.

Hollow hemisphere has y_{cm} = R / 2.

Conservation of Linear Momentum

F⃗ext=0  ⟹  P⃗system=Mv⃗cm=constant\vec{F}_{ext} = 0 \implies \vec{P}_{system} = M \vec{v}_{cm} = \text{constant}

Total linear momentum remains constant in the absence of external forces.

Explosions and rifle recoil conserve total momentum.

CM of Cones & Hemispherical Shells

ycm=h4  (Solid Cone),ycm=h3  (Hollow Cone),ycm=R2  (Hemispherical Shell)y_{cm} = \frac{h}{4} \; (\text{Solid Cone}), \quad y_{cm} = \frac{h}{3} \; (\text{Hollow Cone}), \quad y_{cm} = \frac{R}{2} \; (\text{Hemispherical Shell})

Height of center of mass measured from flat circular base along axis of symmetry.

Memory mnemonic from base: Shell (R/2) > Solid hemisphere (3R/8); Hollow cone (h/3) > Solid cone (h/4).

Cavity Problems (Negative Mass Concept)

R⃗cm=Morigr⃗1−Mremovedr⃗2Morig−Mremoved\vec{R}_{cm} = \frac{M_{\text{orig}}\vec{r}_1 - M_{\text{removed}}\vec{r}_2}{M_{\text{orig}} - M_{\text{removed}}}

Calculates the shift in center of mass when a symmetrical portion or hole is carved out from a uniform body.

For circular disc with circular hole of radius r at distance d: x_{cm} = - (r² d) / (R² - r²).

Reduced Mass (Two-Body Connected System)

μ=m1m2m1+m2,T=2πμk\mu = \frac{m_1 m_2}{m_1 + m_2}, \quad T = 2\pi \sqrt{\frac{\mu}{k}}

Reduces two interacting bodies connected by a spring into an equivalent single particle problem.

When a spring connects m₁ and m₂ on a frictionless surface, natural frequency is ω = √(k / μ).

Variable Mass & Rocket Propulsion

Fthrust=vrel(−dmdt),v=v0+vrelln⁡(m0m)−gtF_{\text{thrust}} = v_{\text{rel}} \left(-\frac{dm}{dt}\right), \quad v = v_0 + v_{\text{rel}} \ln\left(\frac{m_0}{m}\right) - gt

Tsiolkovsky rocket equation for mass ejection rate dm/dt with exhaust velocity v_rel.

For upward acceleration just after liftoff: a = (v_rel / m₀)(|dm/dt|) - g.
MechanicsTop Priority (3-4 Qs in JEE)665 PYQs

Rotational Motion & Rigid Bodies

Moment of inertia, torque, angular momentum conservation, rolling motion without slipping, and rotational kinetic energy.

Chapter

Moment of Inertia of Standard Bodies

Iring=MR2,  Idisc=12MR2,  Isolid sphere=25MR2,  Irod=112ML2I_{\text{ring}} = MR^2, \; I_{\text{disc}} = \frac{1}{2}MR^2, \; I_{\text{solid sphere}} = \frac{2}{5}MR^2, \; I_{\text{rod}} = \frac{1}{12}ML^2

Moments of inertia about central symmetry axes.

Hollow sphere has I = (2/3)MR².

Parallel & Perpendicular Axis Theorems

I=Icm+Md2  (Parallel),Iz=Ix+Iy  (Perpendicular, 2D laminas)I = I_{cm} + Md^2 \; (\text{Parallel}), \quad I_z = I_x + I_y \; (\text{Perpendicular, 2D laminas})

Translating and rotating axes of inertia.

Parallel axis theorem requires I_{cm} through center of mass, NOT any random point.

Torque & Angular Momentum

τ⃗=r⃗×F⃗=Iα⃗,L⃗=r⃗×p⃗=Iω⃗,τ⃗ext=dL⃗dt\vec{\tau} = \vec{r} \times \vec{F} = I\vec{\alpha}, \quad \vec{L} = \vec{r} \times \vec{p} = I\vec{\omega}, \quad \vec{\tau}_{ext} = \frac{d\vec{L}}{dt}

Rotational analogs of Newton's second law.

If net external torque is zero, angular momentum is conserved (I₁ω₁ = I₂ω₂).

Pure Rolling Motion Kinetic Energy

Ktotal=Ktrans+Krot=12Mvcm2(1+k2R2)K_{total} = K_{trans} + K_{rot} = \frac{1}{2}Mv_{cm}^2 \left(1 + \frac{k^2}{R^2}\right)

Total kinetic energy of body rolling without slipping with radius of gyration k.

For pure rolling: v_{cm} = Rω and a_{cm} = Rα.

Acceleration on an Inclined Plane

a=gsin⁡θ1+k2R2,μmin=tan⁡θ1+R2k2a = \frac{g\sin\theta}{1 + \frac{k^2}{R^2}}, \quad \mu_{min} = \frac{\tan\theta}{1 + \frac{R^2}{k^2}}

Linear acceleration of a rolling body down an incline of angle θ.

Body with smallest k²/R² reaches the bottom first (Solid Sphere > Disc > Hollow Sphere > Ring).

Angular Momentum in Combined Motion (Translation + Rotation)

L⃗O=r⃗cm×Mv⃗cm+Icmω⃗,K=12IIAOR ω2\vec{L}_O = \vec{r}_{cm} \times M\vec{v}_{cm} + I_{cm}\vec{\omega}, \quad K = \frac{1}{2}I_{\text{IAOR}}\,\omega^2

Total angular momentum about origin O decomposes into orbital CM term plus spin intrinsic term.

Pure rolling can be treated as pure rotation about the Instantaneous Axis of Rotation (IAOR) passing through the contact point.

Rolling Kinematics & Point Velocity on Wheel

vP=2vcmsin⁡(θ2)  ⟹  vtop=2vcm,vbottom=0,acontact=vcm2R  (towards center)v_P = 2 v_{cm} \sin\left(\frac{\theta}{2}\right) \implies v_{\text{top}} = 2 v_{cm}, \quad v_{\text{bottom}} = 0, \quad a_{\text{contact}} = \frac{v_{cm}^2}{R} \; (\text{towards center})

Speed of any point on perimeter of rolling cylinder at angle θ from bottom contact point.

The bottom contact point is instantaneously at rest (v = 0), but its instantaneous acceleration is non-zero (directed toward the center).

Toppling vs Sliding of a Block

Toppling before sliding condition: μ>bh,Ftopple=mgb2h\text{Toppling before sliding condition: } \mu > \frac{b}{h}, \quad F_{\text{topple}} = \frac{mg b}{2h}

Critical threshold for block of base width b and height h pushed by horizontal force at top edge.

Normal reaction shifts forward by distance x = (F·h)/(mg) as horizontal force F increases, reaching the front corner (x = b/2) at the verge of toppling.
MechanicsHigh Yield (2 Qs in JEE)519 PYQs

Gravitation & Satellite Motion

Newton's law, variation of g with height/depth/latitude, gravitational potential energy, orbital and escape speed.

Newton's Gravitation Law & Potential Energy

F=Gm1m2r2,U=−GMmr,V=−GMrF = G \frac{m_1 m_2}{r^2}, \quad U = -G \frac{M m}{r}, \quad V = -\frac{GM}{r}

Universal inverse-square gravitational force and potential.

Potential V is negative everywhere, approaching 0 at infinity.

Variation of g with Altitude & Depth

gh=g(1+hR)2≈g(1−2hR),gd=g(1−dR)g_h = \frac{g}{\left(1 + \frac{h}{R}\right)^2} \approx g\left(1 - \frac{2h}{R}\right), \quad g_d = g\left(1 - \frac{d}{R}\right)

Decrease of gravitational acceleration above and below Earth's surface.

At Earth's center (d = R), g = 0.

Escape Velocity & Orbital Velocity

ve=2GMR=2gR≈11.2 km/s,vo=GMR=gR≈7.92 km/sv_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} \approx 11.2 \text{ km/s}, \quad v_o = \sqrt{\frac{GM}{R}} = \sqrt{gR} \approx 7.92 \text{ km/s}

Speeds required to escape Earth and maintain circular low-orbit.

Escape speed is exactly √2 times orbital speed: v_e = √2 v_o.

Kepler's Third Law (Harmonic Law)

T2=(4π2GM)r3  ⟺  T2∝r3T^2 = \left(\frac{4\pi^2}{GM}\right) r^3 \iff T^2 \propto r^3

Square of satellite orbital period is proportional to cube of semi-major axis.

Geostationary satellite: T = 24 hours, altitude h ≈ 36,000 km.

Satellite Total Energy & Binding Energy

Etotal=−GMm2r=−K=12U,Binding Energy=GMm2rE_{\text{total}} = -\frac{GMm}{2r} = -K = \frac{1}{2}U, \quad \text{Binding Energy} = \frac{GMm}{2r}

Mechanical energy partition for stable circular orbit.

Negative total energy signifies a bound gravitational state.

Gravitational Field & Potential Inside Solid Sphere & Shell

Shell (r<R):I=0,  V=−GMR;Solid Sphere (r≤R):I=−GMrR3,  V(r)=−GM2R3(3R2−r2)\text{Shell } (r < R): I = 0, \; V = -\frac{GM}{R}; \quad \text{Solid Sphere } (r \le R): I = -\frac{GMr}{R^3}, \; V(r) = -\frac{GM}{2R^3}(3R^2 - r^2)

Internal field and potential for uniform spherical distributions. Field inside hollow shell is zero everywhere.

Potential at the center of solid planet is 1.5 times the surface potential: V_center = -1.5 GM/R.

Variation of g with Earth's Rotation & Latitude

g′=g−ω2Rcos⁡2λ  ⟹  gequator=g−ω2R,gpole=gg' = g - \omega^2 R \cos^2\lambda \implies g_{\text{equator}} = g - \omega^2 R, \quad g_{\text{pole}} = g

Apparent gravitational acceleration considering Earth's diurnal rotation at latitude angle λ.

If Earth stopped rotating, effective weight would increase at the equator by m·ω²R, but would remain unchanged at the poles.

Orbital Shift Energy & Hyperbolic Escape Speed

ΔE=GMm2(1r1−1r2),v∞=v2−ve2(v>ve)\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right), \quad v_\infty = \sqrt{v^2 - v_e^2} \quad (v > v_e)

Energy required to boost a satellite between circular orbits of radii r₁ and r₂, and residual interstellar speed at infinity.

For elliptic orbit of semi-major axis a: total energy is E = -GMm / (2a).
MechanicsHigh Yield (1-2 Qs in JEE)413 PYQs

Properties of Matter & Elasticity

Hooke's law, Young's modulus, bulk modulus, shear modulus, elastic potential energy, and thermal stress.

Hooke's Law & Moduli of Elasticity

Y=StressStrain=F/AΔL/L,B=−ΔPΔV/V,η=F/AθY = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta L / L}, \quad B = -\frac{\Delta P}{\Delta V / V}, \quad \eta = \frac{F / A}{\theta}

Young's, Bulk, and Shear moduli.

Compressibility K = 1 / B.

Elastic Potential Energy Stored in Wire

U=12×Force×ΔL,u=UVolume=12×Stress×Strain=12Y(Strain)2U = \frac{1}{2} \times \text{Force} \times \Delta L, \quad u = \frac{U}{\text{Volume}} = \frac{1}{2} \times \text{Stress} \times \text{Strain} = \frac{1}{2} Y (\text{Strain})^2

Work done in stretching a wire stored as elastic energy.

Area under Stress-Strain curve up to elastic limit gives energy density.

Thermal Stress in Clamped Wire

Stress=YαΔT,F=YAαΔT\text{Stress} = Y \alpha \Delta T, \quad F = Y A \alpha \Delta T

Force developed when temperature of clamped wire changes by ΔT.

α is the linear coefficient of thermal expansion.

Poisson's Ratio & Relations Between Elastic Moduli

σ=−Δr/rΔL/L,Y=3B(1−2σ)=2η(1+σ),9Y=1B+3η\sigma = -\frac{\Delta r / r}{\Delta L / L}, \quad Y = 3B(1 - 2\sigma) = 2\eta(1 + \sigma), \quad \frac{9}{Y} = \frac{1}{B} + \frac{3}{\eta}

Lateral contraction strain to longitudinal extension strain ratio, and master algebraic interrelations among Y, B, η, and σ.

Theoretical limits of σ are -1 to 0.5; practical engineering materials lie between 0 and 0.5.

Elongation of Wire Under Self-Weight

ΔLself=MgL2AY=ρgL22Y\Delta L_{\text{self}} = \frac{M g L}{2AY} = \frac{\rho g L^2}{2Y}

Elongation produced purely by the wire's own gravitational weight (mass M, density ρ, length L).

Elongation under self-weight is exactly HALF the elongation produced by hanging a mass M at the bottom end.

Depression of a Loaded Beam

δ=WL348YIg=WL34bd3Y(Rectangular bar with breadth b, depth d)\delta = \frac{W L^3}{48 Y I_g} = \frac{W L^3}{4 b d^3 Y} \quad (\text{Rectangular bar with breadth } b, \text{ depth } d)

Sagging depression δ at the center of a beam of length L supported at both ends loaded with weight W.

To minimize sagging, depth d is made large compared to breadth b (I-shaped girders in construction).

Bulk Modulus, Compressibility & Fluid Density with Depth

B=−VdPdV,K=1B,ρ(h)=ρ0(1+ρ0ghB)B = -V\frac{dP}{dV}, \quad K = \frac{1}{B}, \quad \rho(h) = \rho_0\left(1 + \frac{\rho_0 g h}{B}\right)

Volumetric elasticity and fractional increase in liquid density at depth h under hydrostatic pressure.

Minus sign indicates volume decreases as pressure increases.

Torsion of a Cylinder (Twisting Couple)

C=πηr42L,W=12Cθ2=πηr4θ24LC = \frac{\pi \eta r^4}{2L}, \quad W = \frac{1}{2}C\theta^2 = \frac{\pi \eta r^4 \theta^2}{4L}

Restoring couple C per unit angle of twist in a circular shaft of shear modulus η, radius r, and length L.

A hollow cylinder is much more resistant to twisting than a solid cylinder of identical mass and length.
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Fluid Mechanics & Surface Tension

Pascal's law, Archimedes principle, continuity equation, Bernoulli's theorem, viscosity (Stokes law, Poiseuille), and surface tension.

Continuity Equation & Bernoulli's Principle

A1v1=A2v2,P+12ρv2+ρgh=constantA_1 v_1 = A_2 v_2, \quad P + \frac{1}{2}\rho v^2 + \rho g h = \text{constant}

Mass and energy conservation in ideal streamline fluid flow.

Torricelli's speed of efflux: v = √(2gh).

Stokes' Law & Terminal Velocity

Fv=6πηrv,vt=29r2(ρ−σ)gηF_v = 6\pi\eta r v, \quad v_t = \frac{2}{9}\frac{r^2(\rho - \sigma)g}{\eta}

Viscous drag force on sphere and steady terminal fall speed.

Terminal velocity is proportional to the square of sphere radius (v_t ∝ r²).

Excess Pressure in Drops & Bubbles

ΔPdrop=2TR,ΔPsoap bubble=4TR\Delta P_{\text{drop}} = \frac{2T}{R}, \quad \Delta P_{\text{soap bubble}} = \frac{4T}{R}

Pressure difference across curved liquid interface due to surface tension T.

Soap bubble has 2 free surfaces, so excess pressure is doubled.

Capillary Rise (Jurin's Law)

h=2Tcos⁡θρgrh = \frac{2T\cos\theta}{\rho g r}

Height of liquid column climbing a capillary tube of radius r with contact angle θ.

For pure water in clean glass, contact angle θ ≈ 0° (cos θ = 1).

Torricelli's Law & Efflux Range

v=2gh,R=2h(H−h),Rmax=H  (at h=H/2),tempty=Aa2Hgv = \sqrt{2gh}, \quad R = 2\sqrt{h(H - h)}, \quad R_{\text{max}} = H \; (\text{at } h = H/2), \quad t_{\text{empty}} = \frac{A}{a}\sqrt{\frac{2H}{g}}

Speed of efflux from orifice at depth h below surface, horizontal range on ground, and total emptying time of tank.

Two holes at depths h and (H - h) yield identical horizontal ranges.

Poiseuille's Viscous Flow & Reynolds Number

Q=πΔPr48ηL,Re=ρvDηQ = \frac{\pi \Delta P r^4}{8\eta L}, \quad R_e = \frac{\rho v D}{\eta}

Volume flow rate Q through a circular capillary pipe and dimensionless Reynolds number determining flow stability.

Flow is laminar for R_e < 2000, unstable for 2000 < R_e < 3000, and turbulent for R_e > 3000.

Surface Energy & Droplet Coalescence

W=TΔA,R=n1/3r,ΔUreleased=4πTr2(n−n2/3),ΔT=3Tsρ(1r−1R)W = T \Delta A, \quad R = n^{1/3}r, \quad \Delta U_{\text{released}} = 4\pi T r^2 (n - n^{2/3}), \quad \Delta T = \frac{3T}{s \rho}\left(\frac{1}{r} - \frac{1}{R}\right)

Work done in changing surface area, and heat released when n identical droplets coalesce into a single large drop.

When droplets coalesce, surface area decreases, releasing energy and warming the liquid.
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Simple Harmonic Motion (SHM)

Kinematics and dynamics of SHM, spring-mass systems, simple pendulum, energy conservation, and superpositions.

Chapter

Kinematics of SHM

x=Asin⁡(ωt+ϕ),v=ωA2−x2,a=−ω2xx = A\sin(\omega t + \phi), \quad v = \omega\sqrt{A^2 - x^2}, \quad a = -\omega^2 x

Displacement, velocity, and restoring acceleration in simple harmonic motion.

Maximum velocity v_max = ωA; maximum acceleration a_max = ω²A.

Energy in Simple Harmonic Motion

K=12mω2(A2−x2),U=12mω2x2,Etotal=12mω2A2=constantK = \frac{1}{2}m\omega^2(A^2 - x^2), \quad U = \frac{1}{2}m\omega^2 x^2, \quad E_{total} = \frac{1}{2}m\omega^2 A^2 = \text{constant}

Continuous interchange between kinetic and potential energy.

At x = A/√2, kinetic energy equals potential energy (K = U = E/2).

Time Period of Spring-Mass System

T=2πmk,kseries=k1k2k1+k2,kparallel=k1+k2T = 2\pi\sqrt{\frac{m}{k}}, \quad k_{\text{series}} = \frac{k_1 k_2}{k_1 + k_2}, \quad k_{\text{parallel}} = k_1 + k_2

Oscillation period of spring with mass m.

Cutting a spring into n equal parts increases stiffness of each part to n*k.

Simple Pendulum Time Period

T=2πLgT = 2\pi\sqrt{\frac{L}{g}}

Small-angle oscillation period of ideal pendulum.

Seconds pendulum has T = 2 s and length L ≈ 1 meter.

Superposition of Two Collinear SHMs

Ares=A12+A22+2A1A2cos⁡δ,tan⁡θ=A2sin⁡δA1+A2cos⁡δA_{\text{res}} = \sqrt{A_1^2 + A_2^2 + 2 A_1 A_2 \cos\delta}, \quad \tan\theta = \frac{A_2\sin\delta}{A_1 + A_2\cos\delta}

Vector phasor addition of two SHMs of identical frequency with phase difference δ.

Constructive interference at δ = 0 gives A = A₁ + A₂.

Physical (Compound) Pendulum & Minimum Time Period

T=2πImgd=2πk2+d2gd  ⟹  Tmin=2π2kg(when d=k)T = 2\pi\sqrt{\frac{I}{mgd}} = 2\pi\sqrt{\frac{k^2 + d^2}{gd}} \implies T_{\text{min}} = 2\pi\sqrt{\frac{2k}{g}} \quad (\text{when } d = k)

Rigid body swinging about horizontal suspension axis at distance d from CM (k is radius of gyration about CM).

Points of suspension and oscillation are interchangeable: interchanging d and k²/d yields the exact same time period.

Damped Harmonic Oscillations & Quality Factor

A(t)=A0e−bt2m,ω′=km−b24m2,E(t)=E0e−btm,Q=ω0mbA(t) = A_0 e^{-\frac{b t}{2m}}, \quad \omega' = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}}, \quad E(t) = E_0 e^{-\frac{b t}{m}}, \quad Q = \frac{\omega_0 m}{b}

Oscillations with viscous damping force F_d = -b v. Amplitude decays exponentially with damping constant b/(2m).

Critical damping occurs when b² = 4mk (no oscillations occur; body returns to equilibrium most rapidly).

Liquid Column in U-Tube & Floating Cylinder

TU-tube=2πhg=2πL2g,Tcylinder=2πmAρg=2πhsubmergedgT_{\text{U-tube}} = 2\pi\sqrt{\frac{h}{g}} = 2\pi\sqrt{\frac{L}{2g}}, \quad T_{\text{cylinder}} = 2\pi\sqrt{\frac{m}{A\rho g}} = 2\pi\sqrt{\frac{h_{\text{submerged}}}{g}}

Restoring buoyant and gravitational head pressures creating natural simple harmonic oscillations.

Both systems oscillate with effective length equal to the equilibrium submerged/arm depth h.
MechanicsHigh Yield (2 Qs in JEE)642 PYQs

Wave Motion & Sound Waves

Wave equation, speed of sound in gases, standing waves in strings and organ pipes, beats, and Doppler effect.

General Wave Equation & Wave Speed

y(x,t)=Asin⁡(kx±ωt+ϕ),v=ωk=fλ=Tμ  (String)y(x, t) = A\sin(kx \pm \omega t + \phi), \quad v = \frac{\omega}{k} = f\lambda = \sqrt{\frac{T}{\mu}} \; (\text{String})

Wave traveling with wavenumber k = 2π/λ and angular frequency ω = 2πf.

Minus sign (-ωt) signifies propagation in positive x-direction.

Speed of Sound (Laplace Correction)

v=γPρ=γRTMv = \sqrt{\frac{\gamma P}{\rho}} = \sqrt{\frac{\gamma R T}{M}}

Speed of acoustic waves in ideal gas under adiabatic compressions.

Speed of sound increases with temperature: v ∝ √T (in Kelvin).

Standing Waves in Organ Pipes

fn=nv2L  (n=1,2,3…,Open Pipe),fn=(2n−1)v4L  (n=1,2,3…,Closed Pipe)f_n = n \frac{v}{2L} \; (n=1,2,3\dots, \text{Open Pipe}), \quad f_n = (2n - 1) \frac{v}{4L} \; (n=1,2,3\dots, \text{Closed Pipe})

Harmonics in resonant air columns.

Fundamental frequency of open pipe is twice that of closed pipe of identical length.

Beats Frequency & Doppler Effect

fbeat=∣f1−f2∣,f′=f0(v±vov∓vs)f_{\text{beat}} = |f_1 - f_2|, \quad f' = f_0 \left( \frac{v \pm v_o}{v \mp v_s} \right)

Periodic amplitude modulation from interfering sound waves, and frequency shift due to relative motion.

Use upper signs when observer and source approach each other.

Sound Intensity & Loudness Level (Decibels)

I=12ρvω2A2=2π2ρvf2A2,β=10log⁡10(II0) dB  (I0=10−12 W/m2)I = \frac{1}{2}\rho v \omega^2 A^2 = 2\pi^2 \rho v f^2 A^2, \quad \beta = 10\log_{10}\left(\frac{I}{I_0}\right) \text{ dB} \; (I_0 = 10^{-12} \text{ W/m}^2)

Energy transmission per unit area per second and logarithmic decibel scale.

A 10-fold increase in intensity corresponds to a +10 dB loudness increase.

Resonance Tube & End Correction

e=0.6r,λ=2(L2−L1),v=2f(L2−L1)e = 0.6 r, \quad \lambda = 2(L_2 - L_1), \quad v = 2f(L_2 - L_1)

Acoustic end correction for organ pipe of inner radius r. Eliminates end correction error by subtracting first and second resonance lengths.

End correction applies at every open end: open pipe has 2e, closed pipe has 1e.

Standing Waves on Stretched Fixed String

fn=n2LTμ=nf1(n=1,2,3… )f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}} = n f_1 \quad (n = 1, 2, 3\dots)

Natural vibration frequencies of string clamped at both ends under tension T with linear mass density μ.

Nodes occur at fixed ends; anti-nodes occur at centers of vibration loops.