Physics — Important Derivations
Step-by-step derivations for TS Inter 1st Year Physics — the ones that appear as LAQs in TGBIE exams. Based on the 14 chapters in the official 2025-26 annual plan.
Equations of Motion (Ch. 3)
Very frequent LAQ- 1Consider a body with initial velocity u, uniform acceleration a, final velocity v, time t, displacement s.
- 21st equation: v = u + at → from definition of acceleration: a = (v−u)/t, rearrange to get v = u + at.
- 32nd equation: s = ut + ½at² → s = average velocity × time = ((u+v)/2)×t = ((u+u+at)/2)×t = ut + ½at².
- 43rd equation: v² = u² + 2as → from (1): t = (v−u)/a; substitute in (2): s = u(v−u)/a + ½a(v−u)²/a² → simplify to get v² = u² + 2as.
- 5Result: three kinematic equations, valid for uniform acceleration in a straight line.
Projectile Motion — Range and Max Height (Ch. 4)
Very frequent LAQ- 1A projectile is launched at angle θ with speed u. Horizontal: no acceleration; Vertical: acceleration = −g.
- 2Horizontal: x = u·cosθ·t; Vertical: y = u·sinθ·t − ½gt².
- 3Time of flight: when projectile returns to ground, y = 0 → t(u·sinθ − ½gt) = 0 → T = 2u·sinθ/g.
- 4Range R: R = u·cosθ × T = u·cosθ × (2u·sinθ/g) = u²·sin2θ/g.
- 5Maximum height H: at highest point, vertical velocity = 0 → 0 = (u·sinθ)² − 2gH → H = u²·sin²θ/(2g).
- 6Maximum range: R_max when sin2θ = 1 → θ = 45° → R_max = u²/g.
Work-Energy Theorem (Ch. 6)
Frequent- 1Consider a body of mass m acted upon by a net force F, displacing it by ds along the direction of force.
- 2Work done: dW = F·ds = ma·ds (using Newton's 2nd law)
- 3Using v·dv = a·ds: dW = m·v·dv
- 4Total work: W = ∫(v₁ to v₂) mv·dv = ½mv₂² − ½mv₁²
- 5W_net = ΔKE — this is the work-energy theorem.
- 6For conservative forces only: W = −ΔPE, so total mechanical energy (KE + PE) is conserved.
Torque and Angular Momentum (Ch. 7)
Frequent- 1Torque τ = r × F (vector product); magnitude: τ = rF·sinθ.
- 2Angular momentum L = r × p = r × mv; magnitude: L = mvr·sinθ.
- 3Relation: τ = dL/dt (analogous to F = dp/dt in linear motion).
- 4For a rigid body: L = Iω; τ = Iα where I is moment of inertia and α is angular acceleration.
- 5Conservation of angular momentum: if τ_ext = 0, then L = Iω = constant (e.g., figure skater pulling arms in).
Orbital Velocity and Escape Velocity (Ch. 9)
Every year- 1For a satellite in circular orbit of radius r: gravitational force provides centripetal force.
- 2GMm/r² = mv²/r → v_o² = GM/r → v_o = √(GM/r).
- 3At Earth's surface (r = R): v_o = √(gR) ≈ 7.9 km/s.
- 4Escape velocity: a body must have KE ≥ |gravitational PE|.
- 5½mv_e² = GMm/R → v_e = √(2GM/R) = √(2gR) ≈ 11.2 km/s.
- 6Note: v_e = √2 × v_o (escape velocity is √2 times orbital velocity at surface).
Carnot Engine Efficiency (Ch. 13)
Very frequent LAQ- 1A Carnot engine operates between source at temperature T₁ and sink at T₂ (T₁ > T₂).
- 2Process: isothermal expansion at T₁ (heat absorbed Q₁), adiabatic expansion, isothermal compression at T₂ (heat rejected Q₂), adiabatic compression.
- 3For ideal Carnot cycle: Q₁/Q₂ = T₁/T₂.
- 4Work done: W = Q₁ − Q₂.
- 5Efficiency: η = W/Q₁ = (Q₁−Q₂)/Q₁ = 1 − Q₂/Q₁ = 1 − T₂/T₁.
- 6η = (T₁−T₂)/T₁. Maximum possible efficiency for a heat engine between T₁ and T₂.
Kinetic Theory — Pressure of a Gas (Ch. 14)
Frequent- 1Consider N molecules in a cubical box of side L, mass m each.
- 2A molecule with x-velocity vₓ hits a wall: momentum change = 2mvₓ per collision.
- 3Force on wall from one molecule: F = 2mvₓ × (vₓ/2L) = mvₓ²/L.
- 4Total force from N molecules: F = m(v₁ₓ² + v₂ₓ² + ... + vₙₓ²)/L = mN<vₓ²>/L.
- 5By symmetry: <vₓ²> = <vy²> = <vz²> = <v²>/3 where <v²> is mean square speed.
- 6Pressure: P = F/L² = mN<v²>/(3L³) = (1/3)ρ<v²> = (1/3)(nm)<v²>.
- 7Also: P = nkT (ideal gas) → <v²> = 3kT/m → v_rms = √(3kT/m).