Centre of Mass — Important Questions

The CM falls freely under gravity as if no explosion had occurred.

Step 1: Before the explosion, the bomb is at rest at 50 m height. The only external force is gravity (F_ext = Mg downward). Step 2: The explosion is caused by internal chemical forces — these are internal to the system. Step 3: Since internal forces cancel (Newton's third law), they do not change the motion of the CM. Step 4: The CM continues to experience only gravity and falls freely, exactly as the bomb would have if it had not exploded. Option A is wrong — the bomb was already subject to gravity and would fall. Option C is wrong — internal energy cannot move the CM upward against gravity.

10 m

Step 1: Since the system starts at rest and no external force acts, the CM remains fixed throughout. Step 2: The particles will meet at the position of their CM. Step 3: Let d₁ = distance of CM from 3 kg mass. Using the CM formula: d₁ = m₂ × d / (m₁ + m₂). Step 4: d₁ = 5 × 16 / (3 + 5) = 80 / 8 = 10 m. So the 3 kg particle travels 10 m and the 5 kg particle travels 6 m to meet at the CM. Key insight: r ∝ 1/m, meaning the lighter particle travels more distance. Option B (8 m) is the midpoint, valid only for equal masses. Option A confuses d₁ and d₂.

CM lies at the midpoint of the line joining the two particles.

Step 1: If m₁ = m₂ = m, the formula becomes r_cm = (m·r₁ + m·r₂)/(m + m) = (r₁ + r₂)/2. Step 2: (r₁ + r₂)/2 is exactly the average of the two position vectors, which geometrically is the midpoint. Step 3: The CM divides the line in the ratio m₂:m₁ from m₁. If m₁ = m₂, this ratio is 1:1, i.e., the midpoint. Option A is incorrect — CM at a particle's location is only possible if the other mass is zero. Option C (1/3 position) is wrong. Option D is always wrong — CM always lies on the line joining the two particles.

0 m/s

Step 1: Assign signs — east is positive. v₁ = +3 m/s (4 kg), v₂ = -6 m/s (2 kg). Step 2: v_cm = (m₁v₁ + m₂v₂) / (m₁ + m₂) Step 3: v_cm = (4×3 + 2×(-6)) / (4 + 2) = (12 - 12) / 6 = 0/6 = 0 m/s. Step 4: The CM is at rest! The total momentum of the system is zero (p = Mv_cm = 6×0 = 0). Physical insight: This is also verifiable by total momentum: p = 4×3 + 2×(-6) = 12 - 12 = 0 N·s. Option A is wrong (ignores direction). Option C is an arithmetic error. Option D is incorrect in both magnitude and direction.