Physics — Mechanics, Heat, Waves & Oscillations — Practice

Question 1

A ball is projected at 30° to the horizontal with an initial velocity of 20 m/s. Find the maximum height reached. (g = 10 m/s²)

Accepted Answer

Vertical component of initial velocity: uᵧ = u sin 30° = 20 × 0.5 = 10 m/s. At maximum height, vertical velocity = 0. Using v² = u² − 2gH: 0 = (10)² − 2 × 10 × H. H = 100/20 = 5 m. Alternatively, using formula: H = u² sin²θ / 2g = (20)² × (0.5)² / (2 × 10) = 400 × 0.25 / 20 = 100/20 = 5 m. Maximum height = 5 m.

Question 2

A mass of 0.5 kg is attached to a spring of spring constant k = 200 N/m. Find (a) the angular frequency ω, (b) the time period T, and (c) the maximum velocity if amplitude A = 0.1 m.

Accepted Answer

(a) Angular frequency: ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s. (b) Time period: T = 2π/ω = 2π/20 = π/10 ≈ 0.314 s. Alternatively, T = 2π√(m/k) = 2π√(0.5/200) = 2π × 0.05 = 0.314 s. (c) Maximum velocity occurs at the equilibrium position (x = 0): v_max = Aω = 0.1 × 20 = 2 m/s. At equilibrium, all energy is kinetic: KE_max = ½mv_max² = ½ × 0.5 × 4 = 1 J = Total Energy = ½kA² = ½ × 200 × 0.01 = 1 J ✓

Question 3

Derive the expression for the escape velocity of a body from the surface of the Earth. Calculate its value. (G = 6.67 × 10⁻¹¹ Nm²/kg², M = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m)

Accepted Answer

DERIVATION: Escape velocity is the minimum velocity needed to escape Earth's gravitational pull (reaching infinity with zero kinetic energy). Using energy conservation: KE at surface = PE to be overcome. ½mv² = GMm/R (energy needed to escape from r=R to r=∞). v² = 2GM/R. Therefore vₑ = √(2GM/R). Since g = GM/R²: vₑ = √(2gR). CALCULATION: vₑ = √(2 × 6.67 × 10⁻¹¹ × 6 × 10²⁴ / 6.4 × 10⁶). = √(2 × 6.67 × 6 × 10⁻¹¹⁺²⁴⁻⁶) = √(80.04 × 10⁷) = √(8.004 × 10⁸) = 2.83 × 10⁴ m/s ≈ 11.2 km/s. This is why rockets need to achieve ~11.2 km/s to leave Earth.

Practice — Physics — Mechanics, Heat, Waves & Oscillations

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