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JEE Main (Engineering) · Physics (JEE & NEET)

Center of Mass and System of Particles

Center of mass location and motion for particle systems, rigid bodies, explosions, impulse-momentum links and variable-mass framing.

A standalone systems topic: locate the COM, then use it as the one point whose motion ignores all internal pushes and pulls.

  • JEE Main (Engineering)
  • Hard level
  • 5 concepts
  • 5 practice questions

1Locating the center of mass

The center of mass is the mass-weighted average position. For two particles it lies on the joining line, closer to the heavier particle. For a rigid body with symmetry, the integral has already been done by geometry: the COM of a uniform rod is its midpoint, and that of a uniform disc is its centre.

Figure. The centre of mass lies on the line between the masses and closer to the heavier 3 kg block.

How it works

  1. Choose originCOM coordinates depend on the coordinate origin, so set a convenient one.
  2. Weight by massMultiply each coordinate by its mass before adding.
  3. Divide by total massThe denominator is total mass, not number of particles.

Two masses on a line

Masses 2\text{ kg} and 3\text{ kg} are at 0 and 10\text{ m}.

  • x_{cm}=(2\times0+3\times10)/(2+3)6\text{ m}
  • Distances from masses6\text{ m} from 2 kg, 4\text{ m} from 3 kg

Pro tip. The COM being nearer the 3 kg mass is a sanity check, not a coincidence.

Two equal masses are at x=0 and x=8\text{ m}. Their COM is at
  1. 2\text{ m}
  2. 4\text{ m}
  3. 8\text{ m}

Equal masses put the COM at the midpoint.

2COM motion only listens to external force

For a system of particles, internal forces cancel in pairs when you sum over the whole system. What remains is M\vec a_{cm}=\vec F_{ext}. A person walking inside a boat can shift relative positions, but without an external horizontal force the horizontal COM of person plus boat cannot accelerate.

Figure. Internal pushes cancel in pairs. With no horizontal external force the system COM stays put: the person steps right and the boat slides left so the weighted average does not move.

How it works

  1. Define the systemInclude every body whose internal forces you want to cancel.
  2. Sum all equationsInternal action-reaction pairs cancel in the total.
  3. Use external forceOnly the remaining external force changes COM acceleration.
An isolated system explodes internally. Its COM acceleration during the explosion is
  1. Zero
  2. Huge
  3. Opposite to the lighter fragment

With no external force, M a_{cm}=F_{ext}=0.

3Momentum, impulse and the COM

Total momentum is the mass of the system times COM velocity: \vec P=M\vec V_{cm}. Therefore an external impulse changes the COM motion, while internal impulses only redistribute momentum among parts. This is the clean link between center of mass and collisions.

Figure. Total momentum is just the sum of the pieces, and that sum equals M V_cm. An external impulse changes the COM bar; internal impulses only reshuffle the parts.

How it works

  1. Compute total momentumAdd vector momenta of all particles.
  2. Divide by massV_{cm}=P/M.
  3. Apply impulseExternal impulse equals change in total momentum.

COM velocity from momenta

Masses 2\text{ kg} and 3\text{ kg} move right at 4\text{ m s}^{-1} and left at 1\text{ m s}^{-1}.

  • P=2\times4+3\times(-1)5\text{ kg m s}^{-1}
  • V_{cm}=P/M=5/51\text{ m s}^{-1} right

Pro tip. Momentum signs are vector signs. Do not average speeds.

If total momentum of a 10\text{ kg} system is 30\text{ kg m s}^{-1} east, V_{cm} is
  1. 3\text{ m s}^{-1} east
  2. 30\text{ m s}^{-1} east
  3. 300\text{ m s}^{-1} east

V_{cm}=P/M=30/10=3\text{ m s}^{-1}.

4Explosions and recoil keep COM motion fixed

An explosion can create kinetic energy because internal energy is released, but it cannot create net momentum. If the system starts from rest and external impulse is negligible, fragments must leave with momenta adding to zero. Recoil of a gun, a person jumping from a boat and spring-separated blocks are the same COM statement.

Figure. An explosion can create kinetic energy from internal stores, but it cannot create net momentum. Fragments fly opposite ways so their momentum vectors cancel.

How it works

  1. Before eventWrite total momentum or COM velocity just before separation.
  2. During eventTreat internal forces as unable to change total momentum.
  3. After eventSet vector sum of final momenta equal to the initial total.

Two-fragment explosion

A body at rest splits into 1 kg and 3 kg parts; the 1 kg part moves right at 12 m/s.

  • Initial total momentum0
  • 1\times12+3v0
  • v-4\text{ m s}^{-1}

Pro tip. Kinetic energy increased, so energy conservation would be the wrong equation; momentum conservation is the COM statement.

A gun recoils because
  1. Momentum is conserved for gun plus bullet
  2. Kinetic energy is conserved exactly
  3. The bullet pushes air backward only

The bullet momentum forward is balanced by gun momentum backward when external impulse is negligible.

5Variable mass needs a system boundary

A rocket is not a fixed-mass body: exhaust leaves the system carrying momentum relative to the rocket. The safe idea is to write momentum balance across a chosen boundary. With external forces neglected and exhaust speed u relative to the rocket, the ideal result is v-v_0=u\ln(M_0/M). The logarithm is a boundary result, not a new force law.

Figure. Choose the rocket as the system; exhaust crosses the rear boundary carrying relative momentum. The thrust term is v_rel dm/dt, not an ordinary contact force on a fixed-mass body.

How it works

  1. Choose bodySay whether the system includes expelled mass or only the rocket.
  2. Account exhaustMass leaving carries momentum relative to the rocket.
  3. Then simplifyOnly under ideal assumptions does the rocket equation follow.
In the ideal rocket equation, increasing exhaust speed u while mass ratio is fixed
  1. Increases final speed gain
  2. Decreases final speed gain
  3. Has no effect

\Delta v=u\ln(M_0/M), so the speed gain is proportional to exhaust speed.

Notes

  • For discrete particles, \vec R_{cm}=\dfrac{\sum m_i\vec r_i}{\sum m_i}. For a continuous body, replace the sum by \dfrac{1}{M}\int \vec r\,dm. Symmetry places the COM immediately for uniform rods, discs and spheres.
  • The COM moves as if the total external force acts on total mass: M\vec a_{cm}=\vec F_{ext}. Internal forces can rearrange particles but cannot accelerate the COM of an isolated system.
  • Total momentum is \vec P=M\vec V_{cm}. An external impulse changes COM momentum; without external impulse, the COM velocity stays constant through explosions and collisions.
  • For two particles on a line, the COM divides the separation inversely in the ratio of masses. The heavier particle lies closer to the COM.
  • Variable-mass examples such as rockets need a system boundary and relative exhaust speed; naive F=ma on the changing body alone usually misses the momentum carried away.

Formulas

  • \vec R_{cm}=\dfrac{\sum m_i\vec r_i}{\sum m_i}
  • \vec V_{cm}=\dfrac{\sum m_i\vec v_i}{M}, \vec P=M\vec V_{cm}
  • M\vec a_{cm}=\vec F_{ext}
  • Two masses on x-axis: x_{cm}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}
  • External impulse: \vec J_{ext}=\Delta\vec P=M\Delta\vec V_{cm}
  • Rocket ideal relation: v-v_0=u\ln(M_0/M) when external forces are neglected

Exam traps & shortcuts

  • If no external horizontal force acts, the horizontal COM position or velocity cannot change no matter what happens internally.
  • For two masses, COM is closer to the heavier mass; this catches numerator swaps.
  • In explosions from rest, momenta of fragments sum to zero even when kinetic energy increases.

Reference tables

COM problem cues
CueUseWatch for
Location of systemR_{cm}=\sum mr/Mheavier mass pulls COM closer
No external forcea_{cm}=0internal motion can still occur
Explosion/recoilP_{initial}=P_{final}kinetic energy may change
External impulseJ_{ext}=\Delta Pimpulse changes COM velocity
Rocketmomentum with mass flowdefine the system boundary

Recap

Read only this before a system-of-particles problem.

Location
COM is a mass-weighted average, not a geometric midpoint unless symmetry or equal masses says so.
Motion
M a_{cm}=F_{ext}; internal forces cannot move the COM of an isolated system.
Momentum
P=M V_{cm}, so external impulse changes COM momentum.
Explosions
Momentum is conserved; kinetic energy can be created from internal energy.
Rockets
Variable mass problems are momentum-balance problems with a system boundary.

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