Govt. Exams
Entrance Exams
Total acceleration = F/(m₁+m₂) = 10/5 = 2 m/s². Tension = F - m₁a = 10 - 2×2 = 6 N (or T = m₂a = 3×2 = 6 N)
a = g(sinθ - μcosθ) = 10(sin30° - 0.1×cos30°) = 10(0.5 - 0.1×0.866) = 10(0.5 - 0.0866) = 4.134 ≈ 4.3 m/s²
At maximum height, vertical component vᵧ = 0. Horizontal component vₓ remains 20 m/s. Total velocity = 20 m/s
Orbital velocity v = √(GM/r). If r → 2r, then v' = √(GM/2r) = v/√2
Tangential acceleration at = r(dω/dt) = 5 × 2 = 10 m/s² (constant as dω/dt = 2)
By momentum conservation: 4(5) - 6(3) = (4+6)v → 20 - 18 = 10v → v = 0.2 m/s
e = √(h₂/h₁) = √(h/2 ÷ h) = √(1/2) ≈ 0.707
For SHM, ωmax = ω × A where ω = √(k/m) = √(50/2) = 5 rad/s. vmax = 5 × 0.1 = 0.5 m/s
Deceleration a = g(sinθ + μcosθ) = 10(0.5 + 0.2×0.866) = 10(0.5 + 0.173) = 6.73 m/s². Using v² = u² - 2as: 0 = 100 - 2(6.73)s, s ≈ 3.2 m
Both objects have constant acceleration -g. If thrown with same initial velocity but at different times, they will have the same velocity at different times. For same velocity at same instant: vA = u - gt₁ and vB = u - gt₂. Since t₁ ≠ t₂, velocities won't be equal at the same instant.