PhysicsISC Class 11 · System of Particles and Rotational Motion

Centre of Mass: Position, Motion and Conservation of Momentum

Updated 17 October 2026 · Fact-checked

The centre of mass of particles has coordinates x_cm = Σmx/Σm. It moves as if the whole mass were concentrated there and acted on by the net external force, M a_cm = F_ext. Internal forces do not change its motion; if F_ext = 0, the centre of mass stays at rest or moves uniformly.

The centre of mass sits on the line joining the masses, nearer the heavier one, so that m₁d₁ = m₂d₂ (here d₁ is three times d₂).

Illustration drawn for QuestionBuddy.

Understand Centre of Mass and Its Motion

Position. For particles of masses m₁, m₂, ... at positions x₁, x₂, ..., x_cm = (m₁x₁ + m₂x₂ + ...)/(m₁ + m₂ + ...), and likewise for y and z. For two particles, the centre of mass lies on the line joining them, nearer the heavier one: its distances from the masses are in the inverse ratio of the masses.

Rigid bodies. For a uniform body, the centre of mass is at its geometrical centre if the body is symmetrical (rod, ring, disc, sphere). For a lamina with a hole, treat the hole as a negative mass at its own centre: x_cm = (A₁x₁ − A₂x₂)/(A₁ − A₂) for a uniform sheet, where areas replace masses.

Motion. Velocity v_cm = Σmv/Σm, so the total momentum is P = M v_cm. Newton's second law for the system is F_ext = M a_cm. Internal forces (explosions, collisions, a person walking in a boat) cancel in pairs and cannot move the centre of mass.

Isolated system. If F_ext = 0, v_cm is constant. In "boat and man" problems on still water, the centre of mass of the pair stays fixed, so when the man walks, the boat moves the opposite way until the centre of mass is back to its original position.

Explosion. A shell bursting in flight leaves its centre of mass following the original parabola, whatever the fragments do, until the first piece strikes the ground.

Centre of Mass and Its Motion: formulas and key facts

Centre of mass: $x_{c} m = Σ m_{i} x_{i} /Σ m_{i}$; Similarly for y and z.
Velocity and momentum: $v_{c} m = Σ m_{i} v_{i} / M ∣ P = M v_{c} m$; M = total mass.
Newton's law for a system: $F_{e} x t = M a_{c} m$; Internal forces do not contribute.
Body with a hole: $x_{c} m = (A_{1} x_{1} - A_{2} x_{2}) / (A_{1} - A_{2})$; Uniform sheet; A₂ is the removed area.

Centre of Mass and Its Motion in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A boy walking in a boat

Suppose a 40 kg boy stands at one end of a 60 kg boat, 5 m long, floating at rest on still water (water resistance neglected). He walks to the other end.

Find how far the boat moves.

Try it first, then show the solutionSolution

The centre of mass of boy + boat does not move (no external horizontal force).
Let the boat move x backwards; the boy then moves (5 − x) forward relative to the ground.
40(5 − x) = 60x → 200 = 100x → x = 2 m.

Answer: The boat moves 2 m backwards; the boy moves 3 m forward relative to the ground.

Memory hook: The lighter body moves farther: displacements are in the inverse ratio of the masses.

Tricks to remember Centre of Mass and Its Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

Nearer the heavier
The centre of mass of two particles lies nearer to the larger mass.
Use it when: Quick check of a two-particle answer.
Hole = negative mass
Subtract the removed part's moment; the denominator is the remaining area or mass.
Use it when: Lamina and disc with hole questions.

Centre of Mass and Its Motion: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Two particles · 1 mark · MCQBoard level

Two particles of masses 2 kg and 3 kg lie on the x-axis at x = 0 and x = 5 m. The centre of mass is at:
(a) 2 m
(b) 2.5 m
(c) 3 m
(d) 3.5 m

Try it first, then show the solutionSolution

x_cm = (2 × 0 + 3 × 5)/5 = 3 m.

Answer: (c) 3 m

Example 2: Three particles in a plane · 3 marks · Short answerBoard level

Masses 1 kg, 2 kg and 3 kg are at (0, 0), (4, 0) and (0, 6) metres respectively. Find the coordinates of the centre of mass.

Try it first, then show the solutionSolution

M = 6 kg. (0.5 mark)
x_cm = (1 × 0 + 2 × 4 + 3 × 0)/6 = 8/6 = 1.33 m. (1.25 marks)
y_cm = (1 × 0 + 2 × 0 + 3 × 6)/6 = 3 m. (1.25 marks)

Answer: (1.33 m, 3 m)

Example 3: Man on a trolley · 4 marks · Short answerHarder board level

A 40 kg trolley stands on smooth rails with a 60 kg man at one end. The man walks along the trolley at 2 m s⁻¹ relative to it. Find the speed of the trolley and the speed of the man relative to the ground.

Try it first, then show the solutionSolution

Total momentum is zero throughout. Let the trolley move at v backwards; the man moves at (2 − v) forwards. (1 mark)
60(2 − v) = 40v → 120 = 100v. (1 mark)
v = 1.2 m s⁻¹ for the trolley. (1 mark)
The man's speed relative to the ground = 2 − 1.2 = 0.8 m s⁻¹. (1 mark)

Answer: Trolley 1.2 m s⁻¹ backwards; man 0.8 m s⁻¹ forwards (relative to the ground).

Example 4: Exploding shell · 4 marks · Short answerHarder board level

At the top of its flight, 45 m above the ground, a 6 kg shell moving horizontally at 20 m s⁻¹ breaks into 2 kg and 4 kg pieces. The 2 kg piece is brought to rest and falls vertically (g = 10 m s⁻²). Find where each piece lands relative to the point below the explosion, and verify with the centre of mass.

Try it first, then show the solutionSolution

Momentum: 6 × 20 = 4 × v → v = 30 m s⁻¹ for the 4 kg piece. (1 mark)
Fall time = √(2 × 45/10) = 3 s. (1 mark)
The 2 kg piece lands at 0 m; the 4 kg piece lands 30 × 3 = 90 m away. (1 mark)
Centre of mass: (2 × 0 + 4 × 90)/6 = 60 m, equal to 20 × 3 m for the undisturbed shell. (1 mark)

Answer: 2 kg piece lands directly below (0 m); 4 kg piece at 90 m; the centre of mass lands at 60 m.

Example 5: Two pulling bodies · 5 marks · Long answerHOTS / competency

Two bodies of masses 2 kg and 4 kg lie at rest on a smooth horizontal plane 6 m apart and attract each other with internal forces. (i) Where do they meet? Then a constant external force of 12 N is applied on the system (on the 4 kg body) along the line of motion, instead. (ii) Find the acceleration of the centre of mass, and its velocity and displacement after 3 s from rest.

Try it first, then show the solutionSolution

(i) The centre of mass is fixed: its distance from the 2 kg body is 4 × 6/6 = 4 m. They meet there, the 2 kg body moving 4 m and the 4 kg body 2 m. (2 marks)
(ii) a_cm = F_ext/M = 12/6 = 2 m s⁻² (internal forces do not matter). (1 mark)
v_cm = a t = 2 × 3 = 6 m s⁻¹. (1 mark)
s_cm = ½ × 2 × 9 = 9 m. (1 mark)

Answer: They meet 4 m from the lighter body's start; a_cm = 2 m s⁻², v_cm = 6 m s⁻¹, s_cm = 9 m.

Where students lose marks in Centre of Mass and Its Motion

The slips examiners see most often, and what to write instead.

Mistake: Using distances, not signed coordinates
Forgetting a negative coordinate or the minus sign for a hole.
Do this: Use signed coordinates and treat the hole as a negative mass.
Usually costs: 1 to 2 marks.
Mistake: Thinking internal forces move the centre of mass
Claiming the centre of mass of an exploding shell changes its path.
Do this: Only external forces change the motion of the centre of mass.
Usually costs: 1 to 2 marks.
Mistake: Measuring boat movement from the wrong frame
Using the boy's displacement relative to the boat as the ground displacement.
Do this: Ground displacement = relative displacement minus the boat's displacement.
Usually costs: 1 mark.

Exam tips

State the origin and the positive direction before using x_cm.
In boat or trolley problems, begin with "no external horizontal force, so the centre of mass is fixed".
Verify with the inverse mass ratio rule.
For laminas, use areas as masses.

Practice questions on Centre of Mass and Its Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A shell moving in a parabola explodes in mid-air without any external force other than gravity. The centre of mass of the fragments:
(a) moves vertically downwards
(b) continues on the original parabola
(c) stops
(d) moves in a straight line
Hint
Only gravity acts externally.
Answer
(b) continues on the original parabola
Q2 · 2 marks · Short answerBoard level
Two bricks of masses 3 kg and 5 kg have their centres 8 m apart on a line. Find the distance of their common centre of mass from the 3 kg brick.
Hint
Distances are in the inverse ratio of the masses.
Answer
d = 5 × 8/(3 + 5) = 5 m from the 3 kg brick.
Q3 · 3 marks · Short answerHarder board level
A uniform square plate of side 2 m has a square of side 1 m cut out at one corner. Find the centre of mass of the remaining plate relative to the centre of the original square.
Hint
Take the origin at the centre; the cut piece's centre is at (0.5, 0.5).
Answer
Areas: 4 and 1. x_cm = (4 × 0 − 1 × 0.5)/3 = −1/6 m; similarly y_cm = −1/6 m. It is 0.236 m from the centre, away from the cut corner.
Q4 · 3 marks · Short answerHarder board level
A 60 kg man at the end of a 120 kg raft (length 6 m) on still water walks to the other end. How far does the raft move?
Hint
The raft moves x, the man moves 6 − x relative to ground.
Answer
60(6 − x) = 120x → 360 = 180x → x = 2 m.
Q5 · 3 marks · Short answerHarder board level
A 2 kg body moves at 3 m s⁻¹ and a 3 kg body moves at 2 m s⁻¹ in the opposite direction. Find the velocity of the centre of mass and the total kinetic energy in the centre-of-mass frame.
Hint
Take the first direction as positive.
Answer
v_cm = (6 − 6)/5 = 0, so the centre of mass is at rest and the lab frame is the CM frame. K = ½ × 2 × 9 + ½ × 3 × 4 = 9 + 6 = 15 J.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a 70 kg man and a 30 kg child stand 10 m apart on frozen, frictionless ice and pull on a light rope joining them.
(i) How far from the man is the centre of mass? (1 mark)
(ii) Where do they meet? (1 mark)
(iii) If the man moves at 0.3 m s⁻¹, how fast does the child move? (1 mark)
(iv) How long do they take to meet? (1 mark)
Hint
The rope force is internal, so momentum stays zero.
Answer
(i) 30 × 10/100 = 3 m. (ii) At the centre of mass, 3 m from the man's starting point (he moves 3 m, the child 7 m). (iii) 70 × 0.3 = 30 v → v = 0.7 m s⁻¹. (iv) t = 3/0.3 = 10 s (also 7/0.7 = 10 s).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write x_cm for two particles.Show answer
(m₁x₁ + m₂x₂)/(m₁ + m₂).
2.What does F_ext = M a_cm tell us?Show answer
The centre of mass moves as if all the mass and external force were at that point.
3.Where is the centre of mass of a uniform rod?Show answer
At its middle point.
4.Can internal forces change the centre of mass motion?Show answer
No.

Centre of Mass and Its Motion: frequently asked questions

Does the centre of mass have to lie inside the body?

No. For a ring, the centre of mass is at its centre, where there is no matter, and for a crescent or L-shaped body it can also lie outside.

Why does the shell's centre of mass keep following the parabola after an explosion?

The explosion forces are internal. The only external force is gravity, which acts on the total mass as before.

When would the centre of mass and the point where weight acts differ?

Only for a body so large that g varies noticeably across it. For every body in ISC problems the two points coincide.