PN01 — Mechanics

In uniform circular motion, the speed (magnitude of velocity) is constant, but the direction of velocity continuously changes. Since velocity is a vector, it is NOT constant. Acceleration (centripetal, directed toward centre) also changes direction. Linear momentum changes direction with velocity.

R = u²sin2θ/g; H = u²sin²θ/2g. Setting R = H: 2sinθcosθ = sin²θ/2 → 4cosθ = sinθ → tanθ = 4 → θ ≈ 76°. This is a standard NDA mechanics derivation requiring manipulation of projectile equations.

After perfectly inelastic collision: combined velocity = v/2 (momentum conservation: mv = 2m × v_f). Initial KE = mv²/2. Final KE = ½(2m)(v/2)² = mv²/4. Loss = mv²/2 - mv²/4 = mv²/4. Exactly half the initial KE is lost in a perfectly inelastic collision between equal masses.

Static friction acts on a body that does not move relative to the surface. Since there is no relative displacement of the contact point, static friction does zero work. This is different from kinetic friction (which does negative work on a sliding body). Walking: static friction at the foot does positive work on the body.

The rotational analogue of Newton's Second Law: τ_net = dL/dt. Just as F_net = dp/dt for translational motion. If τ_net = 0, then dL/dt = 0 → L = constant (conservation of angular momentum). This explains why a spinning ice skater pulls arms in to spin faster (L conserved, I decreases → ω increases).

I = MR²/2 for a uniform disc (solid cylinder) about its central axis perpendicular to the disc. For a ring: I = MR². For a solid sphere: I = 2MR²/5. For a hollow sphere: I = 2MR²/3. The disc formula is derived by integrating dm × r² from 0 to R.

Defining property of conservative forces: work done in any closed path (returning to starting point) is zero. This is equivalent to saying work done is path-independent (depends only on endpoints). Consequence: a potential energy function can be defined. Gravity, spring force, and electrostatic force are conservative.

Newton's second law for a system: F_ext = M_total × a_cm. The centre of mass of a system accelerates as if all the external force acts on the total mass concentrated at the CM. Internal forces (Newton's 3rd Law pairs) cancel and do not affect CM motion.

For rolling without slipping: v = ωR. KE_rot = ½Iω² = ½(2MR²/5)(v/R)² = Mv²/5. KE_trans = Mv²/2. Ratio = (Mv²/5)/(Mv²/2) = 2/5. So KE_rot : KE_trans = 2:5. Total KE = 7Mv²/10. Rolling objects arrive later at the bottom of an incline than sliding ones due to this rotational energy.

Angular momentum L = Iω = (mr²)ω = mr²ω. Alternatively: L = mvr = m(ωr)r = mr²ω. Angular momentum is a vector directed along the axis of rotation (right-hand rule). For circular motion, L = mr²ω = mrv (since v = ωr).

At the highest point: the vertical component of velocity = 0 (momentarily). The horizontal component remains constant throughout (no air resistance): v_x = u cos θ. So the velocity at the highest point = u cos θ (directed horizontally). For a vertical throw (θ=90°): v = 0 at top.

Translational equilibrium: ΣF = 0 (vector sum of all forces = 0). The body either remains at rest or moves with constant velocity. For complete equilibrium (translational + rotational): BOTH ΣF = 0 AND Στ = 0. A spinning top may have rotational equilibrium but not translational (if precessing).

Power P = F · v = Fv cosθ (dot product), where θ is angle between F and v. Maximum power when F parallel to v (θ=0): P = Fv. Zero power when F perpendicular to v (θ=90°, e.g., centripetal force). P = dW/dt = F · (ds/dt) = F · v.

ve = √(2gR). New ve = √(2 × 2g × R/2) = √(2gR) = same. The changes cancel exactly. This illustrates that escape velocity is determined by the product gR, not by g or R individually. Since g = GM/R², we have gR = GM/R, and ve = √(2GM/R).

Horizontal projection: vertical motion starts from rest. h = ½gt² → t = √(2h/g). This is independent of horizontal speed u. The horizontal speed determines range (x = u√(2h/g)) but not time of flight. All objects projected horizontally from the same height take the same time to reach the ground.

For near-surface situations, we take the Earth surface as the reference level where PE = 0. GPE = mgh increases with height h. For large heights: U = -GMm/r (negative, with infinity as reference). Both conventions are valid; CDS/NDA problems use mgh for near-surface unless otherwise specified.

For a uniform ladder of length L, mass m: torque balance gives tan θ_min = 1/(2μ), where μ = coefficient of friction at the floor. The minimum angle depends only on μ. The mass and length cancel in the equilibrium equations. This is a classic NDA statics problem.

For a conical pendulum: vertical equilibrium gives T cosθ = mg; horizontal gives T sinθ = mω²r = mω²L sinθ. So T = 2π/ω = 2π√(L cosθ/g). Note: when θ → 0, it reduces to T = 2π√(L/g) (simple pendulum). Larger θ → shorter period.

v_cm = constant when net external force = 0 (by Newton's second law: F_ext = Ma_cm = 0 → a_cm = 0 → v_cm constant). Internal forces cannot change v_cm. In an explosion, v_cm remains at the original velocity of the object even as pieces fly apart — used to find positions of fragments.

v = dr/dt = 2ti + 2j. At t=2: v = 4i + 2j. Speed = |v| = √(4² + 2²) = √(16+4) = √20 = 2√5 m/s. This requires differentiating the position vector and then finding the magnitude. Option C shows the calculation; D is the simplified form (both correct, D is simplest).