Govt. Exams
Entrance Exams
By impulse-momentum theorem, Δp = F·Δt = 10 × 5 = 50 kg·m/s
Range = u²sin(2θ)/g = (400 × sin(90°))/10 = 400/10 = 40 m
At h/2, PE = mg(h/2), Total energy = mgh. KE = mgh - mg(h/2) = mg(h/2). Fraction = (h/2)/h = 1/2
In elastic collision between equal masses, if one is at rest, they exchange velocities completely. This satisfies both momentum and energy conservation.
At maximum angle, tan(θ) = μs = 0.5, so θ = tan⁻¹(0.5) ≈ 26.6°
Using v² = u² - 2as, where v = 0, u = 20 m/s, a = 5 m/s². Therefore, s = u²/(2a) = 400/10 = 40 m
Impulse = Δ(momentum) = m(v₂ - v₁) = 5(10 - 4) = 30 N·s
Total mass = m + 2m + 3m = 6m. Using F = Ma: a = F/(6m)
v = ωr = 10 × 0.2 = 2 m/s
Efficiency = (Work output/Work input) × 100. 80 = (8000/W) × 100 → W = 10000 J