Govt. Exams
Acceleration a = dv/dt = 6t + 2. At t=2s: a = 6(2) + 2 = 14 m/s². Force F = ma = 2 × 14 = 28 N
For Atwood-like system with one block horizontal: a = (m_B × g)/(m_A + m_B) = (2 × 10)/(3 + 2) = 20/5 = 4 m/s²
Using s = ut + ½gt² for complete journey (s=0): 0 = 30t - 5t². Solving: t(30 - 5t) = 0, giving t = 6 s (non-zero solution)
Resultant force = √(8² + 6²) = √(64 + 36) = √100 = 10 N. Acceleration = F/m = 10/6 = 1.67 m/s²
Net force = 20 - 12 = 8 N. Work done by net force = 8 × 5 = 40 J. This equals KE gained = 40 J
Stability increases with larger base area and lower center of gravity. The 20×30 cm face down provides the largest base area, ensuring maximum stability.
Initial PE = mgh. With 20% loss, KE = 0.8mgh. ½mv² = 0.8mgh. v = √(1.6gh) = √(1.6×10×5) = √80 ≈ 8.94 m/s. Closest is A or B; B is more precise.
For circular motion: μₛmg ≥ mv²/r. μₛ ≥ v²/(rg) = 400/(100×10) = 0.4
In elastic collision between equal masses where one is at rest, the moving ball comes to rest and transfers all momentum and energy to the stationary ball.
Time to fall: h = ½gt². 20 = ½(10)t². t² = 4. t = 2 s. Horizontal displacement = vₓ × t = 15 × 2 = 30 m
