Govt. Exams
Entrance Exams
In elastic collision between identical objects where one is at rest, they exchange velocities. The moving ball comes to rest and the stationary ball moves with initial velocity of the first.
Max acceleration occurs at maximum extension. F = kx = 100×0.2 = 20 N. Using F = ma: 20 = m×50, m = 0.4 kg
Range = u²sin(2θ)/g = (40)²sin(90°)/10 = 1600×1/10 = 160 m
Since 5² + 12² = 25 + 144 = 169 = 13², these form a right triangle. The angle between 5 N and 12 N is 90°
Component down = mg sin30° = 100×0.5 = 50 N. Max static friction = μN = 0.5×100cos30° = 43.3 N. Since 50 > 43.3, it should slide. Let me recalculate: friction = 0.5×86.6 = 43.3 N. Actually 50 N > 43.3 N, so it WILL slide.
a = dv/dt = 6t - 6. Setting a = 0: 6t - 6 = 0, t = 1 s
Net force = √(3² + 4²) = 5 N. Using F = ma, a = 5/2 = 2.5 m/s²
Max height = u²sin²θ/(2g) = (20)²(sin45°)²/(2×10) = 400×0.5/20 = 10 m
Using a = (m₂ - m₁)g/(m₁ + m₂) = (5-3)×10/(5+3) = 20/8 = 2.5 m/s²
For friction-free circular motion: tan θ = v²/(rg) = (10)²/(50 × 10) = 100/500 = 0.2. Therefore θ = tan⁻¹(0.2)
