The centre of mass (CM) of a two-particle system is the mass-weighted average position of the two particles. NCERT Class 11 Physics Chapter 6 (System of Particles and Rotational Motion), page 93, defines it as the point where the entire mass of the system can be considered concentrated for describing translational motion.
The trap that costs marks: When two unequal masses sit on a rod, many aspirants default to placing the CM at the midpoint (L/2). This is correct only when the masses are equal. For unequal masses, the CM shifts toward the heavier particle.
The formula. For two particles of masses m₁ and m₂ separated by distance L, with the origin at m₁:
x_cm = (m₁ × 0 + m₂ × L) / (m₁ + m₂) = m₂L / (m₁ + m₂)
Equivalently, the distance of the CM from m₁ is d₁ = m₂L/(m₁ + m₂), and from m₂ is d₂ = m₁L/(m₁ + m₂). Notice the cross-relationship: the distance from each mass is proportional to the other mass.
Key property: m₁d₁ = m₂d₂. The CM is the balance point — the torques about it are equal and opposite.
NEET context. This topic appears as a direct-application question (estimated ~0.4 questions per year, medium weight in the chapter). The standard format: two masses on a massless rod, find distance of CM from one end. The distractor that catches aspirants is L/2 — the equal-distribution default.
Watch out: Always check which end the question asks the distance from. "Distance of CM from the heavier mass" gives a smaller number than "distance from the lighter mass." Misreading which mass the answer refers to is an easy −1.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The centre of mass of a system of particles is defined as the point where:
Answer: A. By definition (NCERT Class 11 Physics Chapter 6, page 95), the centre of mass is the point at which the entire mass of the system can be thought of as concentrated for analyzing its translational motion.
Why B is wrong: B describes an equilibrium condition, not the definition of centre of mass. The CM exists regardless of whether external forces are zero.
Why C is wrong: C is incorrect. The CM is defined by mass distribution, not by any energy extremum condition.
Why D is wrong: D describes a rigid-body condition (all points having equal velocity only in pure translation), not the definition of the CM.
For a two-particle system with masses m₁ and m₂ separated by distance L, the position of the centre of mass from m₁ is:
Answer: C. Placing the origin at m₁, x_cm = (m₁ × 0 + m₂ × L)/(m₁ + m₂) = m₂L/(m₁ + m₂). This follows directly from the CM formula (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A gives L/2, which is valid only when m₁ = m₂. This is the equal-distribution default trap — the CM shifts toward the heavier mass for unequal masses.
Why B is wrong: B has the masses swapped: m₁ appears in the numerator instead of m₂. The distance from m₁ is proportional to the other mass (m₂), not to m₁ itself.
Why D is wrong: D gives a distance greater than L for any positive masses, which is physically impossible — the CM must lie between the two particles.
In a two-particle system, the distances d₁ and d₂ of the centre of mass from masses m₁ and m₂ respectively satisfy:
Answer: C. Since d₁ = m₂L/(m₁ + m₂) and d₂ = m₁L/(m₁ + m₂), multiplying gives m₁d₁ = m₁m₂L/(m₁ + m₂) = m₂d₂. This is the balance-point (lever) property of the CM (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A is true only when m₁ = m₂. For unequal masses the CM is closer to the heavier particle, so d₁ ≠ d₂.
Why B is wrong: B has the subscripts swapped. The correct relation is m₁d₁ = m₂d₂, meaning each mass times its own distance from the CM is equal — not each mass times the other's distance.
Why D is wrong: D would require one distance to be negative, but d₁ and d₂ are magnitudes (both positive). The vector form uses signed positions, but the question specifies distances.
Two particles of masses 3 kg and 5 kg are placed at the ends of a 2.0 m long massless rod. The distance of the centre of mass from the 3 kg mass is:
Answer: A. d₁ = m₂L/(m₁ + m₂) = 5 × 2.0/(3 + 5) = 10.0/8 = 1.25 m. The CM is closer to the heavier (5 kg) mass, so the distance from the lighter (3 kg) mass is the larger value (NCERT Class 11 Physics Chapter 6, page 95).
Why B is wrong: B gives m₁L/(m₁ + m₂) = 3 × 2.0/8 = 0.75 m, which is the distance of the CM from the 5 kg mass, not from the 3 kg mass. The numerator should use the other mass. (trap: mass-weighting inversion)
Why C is wrong: C gives L/2 = 1.00 m, which assumes equal masses. With m₁ = 3 kg and m₂ = 5 kg the masses are unequal, so the CM is not at the midpoint. (trap: equal-distribution default)
Why D is wrong: D gives 1.50 m, which would leave the CM 0.50 m from the 5 kg mass; then m₁d₁ = 3 × 1.50 = 4.5 does not balance m₂d₂ = 5 × 0.50 = 2.5, so it is not the balance point.
Two particles of masses 2 kg and 6 kg are separated by 4.0 m. How far is the centre of mass from the 6 kg particle?
Answer: B. Distance of CM from the 6 kg mass: d₂ = m₁L/(m₁ + m₂) = 2 × 4.0/(2 + 6) = 8.0/8 = 1.0 m. The CM is much closer to the heavier mass, as expected.
Why A is wrong: A gives m₂L/(m₁ + m₂) = 6 × 4.0/8 = 3.0 m, which is the distance from the 2 kg mass. The question asks for distance from the 6 kg mass — reading which end is referenced is critical. (trap: misidentifying which mass the distance is measured from)
Why C is wrong: C gives L/2 = 2.0 m, the midpoint. This ignores the 3:1 mass ratio. (trap: equal-distribution default)
Why D is wrong: D does not correspond to any correct application of the CM formula for these values.
Two particles of masses m and 3m are joined by a rigid massless rod of length L. The centre of mass divides the rod in the ratio (from mass m to mass 3m):
Answer: D. Distance from m: d₁ = 3m × L/(m + 3m) = 3L/4. Distance from 3m: d₂ = m × L/(m + 3m) = L/4. Ratio d₁ : d₂ = 3 : 1. The CM is 3 times farther from the lighter mass than from the heavier mass (NCERT Class 11 Physics Chapter 6, page 95).
Why A is wrong: A gives 1 : 3, which inverts the ratio. The CM is closer to the heavier mass (3m), so the distance from the lighter mass (m) must be the larger number, giving 3 : 1, not 1 : 3. (trap: mass-weighting inversion)
Why B is wrong: B gives 2 : 1, which would correspond to a mass ratio of 1 : 2, not the given 1 : 3.
Why C is wrong: C gives 1 : 1 (midpoint), valid only for equal masses. Here the mass ratio is 1 : 3, so the CM cannot be equidistant. (trap: equal-distribution default)
Two particles of masses 4 kg and 8 kg are placed at coordinates (1.0, 0) m and (4.0, 0) m respectively on the x-axis. The x-coordinate of the centre of mass is:
Answer: D. x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂) = (4 × 1.0 + 8 × 4.0)/(4 + 8) = (4.0 + 32.0)/12 = 36.0/12 = 3.0 m. This is a coordinate-based CM calculation requiring substitution into the general formula rather than the simplified distance form.
Why A is wrong: A gives (x₁ + x₂)/2 = (1.0 + 4.0)/2 = 2.5 m, which is the geometric midpoint. This ignores the mass weighting entirely. (trap: equal-distribution default)
Why B is wrong: B gives 3.5 m, which would place the CM beyond the midpoint toward the 8 kg mass by too much. The correct weighted average yields 3.0 m.
Why C is wrong: C gives 2.0 m, which comes from swapping the masses between the positions: (8 × 1.0 + 4 × 4.0)/12 = 24/12 = 2.0 m. It puts the CM nearer the lighter mass.
Two particles of masses 1.0 kg and 4.0 kg are separated by 1.00 m. If the 1.0 kg mass is moved 0.50 m closer to the 4.0 kg mass (separation becomes 0.50 m), how far does the centre of mass shift?
Answer: B. Place the origin at the 4.0 kg mass. Initial CM from origin: x_cm1 = 1.0 × 1.00/(1.0 + 4.0) = 0.200 m. After moving the 1.0 kg mass to 0.50 m away: x_cm2 = 1.0 × 0.50/5.0 = 0.100 m. Shift = 0.200 − 0.100 = 0.10 m toward the 4.0 kg mass (the CM moved closer to the origin, which is at the 4.0 kg mass).
Why A is wrong: A has the correct magnitude (0.10 m) but the wrong direction. Since the 1.0 kg mass moved toward the 4.0 kg mass, the CM also moves toward the 4.0 kg mass, not away from it.
Why C is wrong: C assumes the CM shifts by the same amount as the displaced particle. The CM shift is mass-weighted: only 1.0/(1.0 + 4.0) = 1/5 of the particle's displacement. (trap: assuming CM moves the same distance as the moved particle)
Why D is wrong: D uses the 4.0 kg mass fraction (4/5 × 0.50 = 0.40) instead of the 1.0 kg mass fraction. The CM shift due to moving m₁ equals m₁/(m₁ + m₂) times the displacement, not m₂/(m₁ + m₂). (trap: mass-weighting inversion)
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Given
• m₁ = 2.0 kg• m₂ = 3.0 kg• L = 0.50 m (separation)
Required
Distance of centre of mass from m₁ (the 2.0 kg particle).
Concept
The centre of mass of a two-particle system lies at the mass-weighted average position. Placing the origin at m₁, the CM position equals m₂L/(m₁ + m₂). The CM is closer to the heavier particle (NCERT Class 11 Physics Chapter 6, page 95).
Formula
d₁ = m₂ × L / (m₁ + m₂)
Substitution
d₁ = 3.0 × 0.50 / (2.0 + 3.0)
Calculation
d₁ = 1.50 / 5.0 = 0.30 m
Final answer
The centre of mass is 0.30 m from the 2.0 kg particle (and 0.20 m from the 3.0 kg particle).
Note on significant figures: the masses and length are each given to 2 significant figures. The integers in the formula (the addition and division) are exact counting operations and do not limit significant figures. The answer 0.30 m retains 2 significant figures.
Common trap
Answering 0.25 m (= L/2) by assuming equal distribution. The mass ratio is 2 : 3, so the CM is not at the midpoint. Always apply the formula — the midpoint shortcut works only when masses are equal.
Similar NEET-style question
"A system consists of two point masses, 5.0 kg and 15.0 kg, separated by 1.00 m. Find the distance of the centre of mass from the 15.0 kg mass." [Answer: d₂ = 5.0 × 1.00/(5.0 + 15.0) = 0.25 m. Note: from the heavier mass, so the answer is the smaller distance.]
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The centre of mass of a system of particles is the unique point where the entire mass of the system may be considered to be concentrated for purposes of describing translational motion. For a system of n particles: R_cm = (Σ m_i r_i) / (Σ m_i).
-- NCERT Class 11 Physics, Ch. 6, p. 96The position of the centre of mass equals the mass-weighted average of particle positions. For continuous bodies use integral form.
| Symbol | Quantity | SI Unit |
|---|---|---|
| R_cm | CoM position | m |
| m_i | mass of i-th particle | kg |
| r_i | position of i-th particle | m |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Overthinking
Student answers L/2 for two-particle CoM regardless of mass ratio.
Question gives two masses on rigid rod and asks for CoM distance.
R_cm from m1 = m2*L/(m1+m2). Heavier mass pulls CoM closer to it.
More in System of Particles and Rotational Motion: 6 exam traps and mistakes · 7 formulas · 3 question patterns from its other lessons.
All 11 past-paper questions from System of Particles and Rotational Motion →
uses equal distribution
Default to L/2 regardless of mass ratio
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