Laws of Motion — Important Questions

2v/5

Using conservation of momentum: Initial momentum = 2m × v = 2mv. Final momentum = 5m × v' (all five trolleys move together). Setting them equal: 2mv = 5mv' → v' = 2v/5. Option A (v/2) suggests only 4 trolleys move together. Option C (2v/3) suggests only 3 total. Option D (v/5) incorrectly halves the numerator.

1 m/s²

Kinetic friction: f_k = μ_k × mg = 0.2 × 10 × 10 = 20 N. Net force = F_applied − f_k = 30 − 20 = 10 N. Acceleration: a = F_net/m = 10/10 = 1 m/s². Option A (3 m/s²) ignores friction entirely. Option C (2 m/s²) uses incorrect friction value. Option D (0.5 m/s²) doubles the mass in the denominator.

No net external force is acting on the object.

Newton's First Law states that a body in uniform motion continues to do so unless a net external force acts on it. Constant velocity means zero acceleration, which by F = ma means net force = 0. Option A is wrong — a forward net force would accelerate the object. Option C is wrong — gravity acts but is balanced by the normal force. Option D is wrong — momentum = mv = 2 × 5 = 10 kg m/s, which is non-zero.

30 kg m/s

Change in momentum = Impulse = F × Δt = 150 N × 0.2 s = 30 kg m/s. This is a direct application of the Impulse-Momentum theorem. Option A (15 kg m/s) uses Δt = 0.1 s instead of 0.2 s. Option C (300 kg m/s) forgets to multiply by time (or uses Δt = 2 s). Option D (3 kg m/s) uses wrong arithmetic.