Motion of centre of mass

Fire a firework shell into the night sky and, at the peak of its arc, it bursts into a hundred glowing sparks that scatter every which way. Each spark seems to obey its own will — some shoot sideways, some stall, some tumble back down early. Yet if you could paint a single dot exactly at the average position of all that flying debris, weighted by mass, that dot would trace a perfectly smooth, boring parabola — the exact same parabola the unexploded shell would have followed. The explosion, however violent, could not touch it. That dot is the centre of mass, and the rule that keeps it so stubbornly predictable is the subject of this article. The same idea quietly runs a hundred everyday scenes: two skaters push off each other and glide apart, a high jumper's body folds through an arch over the bar while their centre of mass barely clears it, a spinning wrench thrown across a workshop wobbles wildly while one point on it sails in a clean straight line. In every case, a system made of many parts — parts that push, pull, explode, and collide against each other in complicated ways — has one point that answers only to the outside world. Internal forces, no matter how large, are invisible to it. This article turns that observation into a single vector equation, $M\vec A_{cm} = \vec F_{ext}$ , and mines it for two big payoffs: a clean proof of why momentum conservation works the way it does, and a fast, board-exam-ready method for explosion and recoil problems that would otherwise need particle-by-particle bookkeeping. We reuse and extend the two skater and shell-explosion examples already worked out for the centre of mass itself, so the numbers should feel familiar even as the reasoning gets sharper.

Key takeaways

The centre of mass (CM) of a system of particles moves exactly as a single particle of mass $M$ (the total mass) would move under the net external force alone: $M\vec A_{cm} = \vec F_{ext}$ .
Internal forces — pushes, explosions, collisions between parts of the same system — occur in Newton's-third-law pairs and cancel completely when summed over the system; they cannot change $\vec V_{cm}$ or $\vec A_{cm}$ .
If $\vec F_{ext} = 0$ , then $\vec A_{cm} = 0$ and $\vec V_{cm}$ stays constant — the centre of mass moves with uniform velocity, which is the seed of momentum conservation for the whole system.
An exploding projectile's centre of mass keeps following the same parabolic trajectory the intact projectile would have traced, even as individual fragments scatter wildly.
Two skaters pushing off each other on frictionless ice leave their common centre of mass exactly where it started, because the only external forces (gravity and the normal reaction) are vertical and cancel.
$\vec V_{cm} = \dfrac{\sum m_i\vec v_i}{M}$ is the same weighted average used for position, just applied to velocities — differentiate the CM position formula once for velocity, twice for acceleration.
Key idea: sum Newton's second law over every particle in a system and the internal action-reaction pairs cancel exactly, leaving $M\vec A_{cm} = \vec F_{ext}$ — so only external forces can change how the centre of mass moves, however violently the system rearranges itself internally.

The core insight

Newton's second law applied to a whole system of particles collapses to $M\vec A_{cm} = \vec F_{ext}$ because internal forces cancel in equal-and-opposite pairs, so the centre of mass moves exactly like a single particle of the total mass driven only by the net external force.

Where it fits in the syllabus

This topic sits immediately after the definition of the centre of mass in the CBSE Class 11 chapter "System of Particles and Rotational Motion," and it is squarely inside the prescribed scope: the two-particle centre of mass $\vec R_{cm} = \dfrac{m_1\vec r_1+m_2\vec r_2}{m_1+m_2}$ , extended to a general system, together with momentum conservation and CM motion. It is the bridge concept: once you can differentiate $\vec R_{cm}$ with respect to time and apply Newton's second law to the whole system, you are one short step from the "linear momentum of a system of particles," the very next topic in the chapter. Nothing here needs calculus of variations, torque, or moment of inertia — this article stays entirely inside plain differentiation and vector addition, exactly as the board expects.

The physical picture

Every particle in a system obeys Newton's second law individually, $m_i\vec a_i = \vec F_i$ , where $\vec F_i$ is the total force on particle $i$ — part external (gravity, applied pushes, friction from outside the system), part internal (forces from the other particles of the same system, such as a chemical bond, a collision impulse, or a muscular push). Before summing anything, it helps to see the centre of mass velocity and acceleration as nothing more than the time-derivatives of the position formula you already know:

\vec V_{cm} = \frac{d\vec R_{cm}}{dt} = \frac{\sum m_i \vec v_i}{M}, \qquad \vec A_{cm} = \frac{d\vec V_{cm}}{dt} = \frac{\sum m_i \vec a_i}{M}

where: $\vec V_{cm}$ and $\vec A_{cm}$ are the velocity and acceleration of the centre of mass, $m_i$ and $\vec v_i,\vec a_i$ are the mass, velocity and acceleration of the $i$ -th particle, and $M=\sum m_i$ is the total mass of the system. Physically, this says the centre of mass is a mass-weighted average that moves and accelerates by averaging every particle's own velocity and acceleration — nothing mysterious, just the same weighting rule applied one derivative deeper each time. The picture below shows why summing the individual equations of motion is so powerful: it splits neatly into an internal part that always vanishes and an external part that survives.

Figure

Two particles connected by an internal push arrow that cancels itself, with an external force arrow acting on the whole system and the centre of mass shown moving along a smooth path

Derivation ANewton's second law for a system: M times A of the centre of mass equals the net external force

The goal is to show, starting only from Newton's second and third laws applied particle by particle, that summing the equations of motion of every particle in a system collapses into one clean statement about the centre of mass alone: $M\vec A_{cm} = \vec F_{ext}$ , with every internal force having vanished from the final result. This is the single most useful shortcut in the chapter — it lets you treat an exploding, colliding, or deforming system as if it were one point mass, provided you only ever ask about the motion of that one point.

Step 1. Write Newton's second law for each particle separately. For particle $i$ , the total force is the sum of an external part and an internal part contributed by every other particle $j$ of the same system: $m_i\vec a_i = \vec F_i^{ext} + \sum_{j\neq i} \vec F_{ij}$ , where $\vec F_{ij}$ is the force exerted on particle $i$ by particle $j$ .

Step 2. Sum this equation over every particle in the system, $i=1$ to $N$ . The left side becomes $\sum_i m_i\vec a_i$ . The right side splits into $\sum_i \vec F_i^{ext}$ , which is simply the net external force $\vec F_{ext}$ on the whole system, plus a double sum $\sum_i\sum_{j\neq i}\vec F_{ij}$ that collects every internal force pair.

Step 3. Show the internal double sum vanishes. By Newton's third law, the force of particle $j$ on particle $i$ is equal and opposite to the force of particle $i$ on particle $j$ : $\vec F_{ij} = -\vec F_{ji}$ . Every internal force in the double sum therefore appears exactly twice, once as $\vec F_{ij}$ and once as $\vec F_{ji}$ , and the two always add to zero. Summed over all pairs, the entire internal contribution is exactly zero, no matter how many particles or how complicated the internal forces are.

Step 4. Identify the left side as $M\vec A_{cm}$ . From the definition $\vec A_{cm} = \dfrac{\sum m_i\vec a_i}{M}$ , we have $\sum_i m_i\vec a_i = M\vec A_{cm}$ . Combining Steps 2 to 4, the internal forces having cancelled, gives the result: $M\vec A_{cm} = \vec F_{ext}$ .

Step 5. Read the physical content. The centre of mass of any system — however many particles it has, however they push, pull, collide, or explode against one another — moves exactly as a single particle of mass $M$ would move under the net external force alone. As a direct corollary, if $\vec F_{ext}=0$ , then $\vec A_{cm}=0$ , so $\vec V_{cm}$ is constant: this is the launching point for conservation of the total momentum of an isolated system, $\vec P = M\vec V_{cm} = \text{constant}$ .

Worked examples

Board level - two skaters push off on ice

Two skaters, of masses $60\ \text{kg}$ and $40\ \text{kg}$ , stand at rest on frictionless ice and then push off each other. Find the acceleration of their combined centre of mass while they push apart, and state what happens to $\vec V_{cm}$ before and after.

Step 1. Identify the system and classify the forces. Take both skaters together as one system of total mass $M = 60+40 = 100\ \text{kg}$ . The push between them is an internal force — it cancels in the sum by Newton's third law and cannot appear in $\vec F_{ext}$ .

Step 2. List the external forces. Only gravity (downward) and the normal reaction from the ice (upward) act from outside the system; the ice is frictionless and horizontal, so there is no horizontal external force, and the vertical ones cancel: $\vec F_{ext} = 0$ .

Step 3. Apply $M\vec A_{cm}=\vec F_{ext}$ . With $\vec F_{ext}=0$ , $\vec A_{cm} = \vec F_{ext}/M = 0$ . The centre of mass does not accelerate.

Step 4. Track $\vec V_{cm}$ before and after. Both skaters start at rest, so $\vec V_{cm}=0$ initially; since $\vec A_{cm}=0$ throughout the push, $\vec V_{cm}$ stays zero even as the skaters fly apart in opposite directions. The centre of mass remains fixed at its starting point for all time, consistent with the fact that the skaters' individual momenta must be equal and opposite at every instant.

Competitive level - shell explosion with a moving centre of mass afterwards

A shell of mass $10\ \text{kg}$ is fired and, at the top of its trajectory where its velocity is purely horizontal at $20\ \text{m/s}$ , it explodes into two fragments of masses $4\ \text{kg}$ and $6\ \text{kg}$ . The $4\ \text{kg}$ fragment moves off horizontally at $50\ \text{m/s}$ in the same direction the shell was travelling. Find the horizontal velocity of the $6\ \text{kg}$ fragment right after the explosion, and confirm that the centre of mass of the two fragments continues to move at the shell's original horizontal speed, $20\ \text{m/s}$ , immediately after the blast.

Step 1. Recognise the explosion as an internal event. The explosive forces act only between the two fragments, so they are internal to the shell-system and cannot change the system's total horizontal momentum during the brief blast (air resistance ignored, gravity is vertical and irrelevant to the horizontal component here).

Step 2. Conserve horizontal momentum across the explosion. $Mv = m_1v_1 + m_2v_2$ , so $10\times 20 = 4\times 50 + 6\,v_2$ , giving $200 = 200 + 6v_2$ , so $v_2 = 0\ \text{m/s}$ . The $6\ \text{kg}$ fragment has zero horizontal velocity immediately after the explosion and will fall under gravity alone from that point.

Step 3. Verify the centre-of-mass statement directly, as the competitive twist. Compute $V_{cm}$ of the fragments right after the blast using the CM velocity formula: $V_{cm} = \dfrac{m_1v_1+m_2v_2}{m_1+m_2} = \dfrac{4\times 50 + 6\times 0}{10} = \dfrac{200}{10} = 20\ \text{m/s}$ . This matches the shell's speed just before explosion exactly, confirming $M\vec A_{cm}=\vec F_{ext}$ (here, $\vec F_{ext}$ has no horizontal component, so $V_{cm}$ horizontal is unchanged by the internal blast). Timing note: under a minute and a half. The fast route is to skip fragment-by-fragment vector diagrams and go straight to horizontal-momentum conservation, since the explosion's internal forces are guaranteed to leave $MV_{cm}$ untouched — treat the shell's centre of mass as continuing along its undisturbed projectile parabola and only solve for the one unknown fragment velocity. The one trap is forgetting that "immediately after the explosion" freezes the calculation at that instant — gravity has not yet had time to act, so it is safe to ignore it for the horizontal-momentum step and only bring it back in if asked what happens next.

Common traps

Believing that internal forces (explosions, muscular pushes, collisions between parts of the same system) can change $\vec V_{cm}$ or $\vec A_{cm}$ — they cannot, by Newton's third law, however large or violent they are.
Forgetting to check whether gravity, friction, or a normal force counts as external in a given problem — gravity is almost always external and does affect $\vec A_{cm}$ unless the question states it is balanced or to be ignored.
Assuming every particle in the system moves like the centre of mass — individual particles can follow wildly different, even curved or reversing, paths while the centre of mass itself moves smoothly, such as along a single parabola for a projectile-type system.
Writing $M\vec A_{cm}=\vec F_{ext}$ without first correctly identifying and summing all the external forces on the system — a missed normal force or an unnoticed friction term throws off the whole answer.
Mixing up scalar and vector momentum conservation: in the shell example, only the horizontal component of momentum is conserved through the blast because gravity supplies a nonzero, though momentarily negligible-during-the-instant, vertical external force.

You should be able to…

State and apply $M\vec A_{cm} = \vec F_{ext}$ for a system of particles, explaining why the internal-force double sum vanishes by Newton's third law.
Explain, with a reason, why the centre of mass of an isolated system (zero net external force) moves with constant velocity.
Solve explosion, recoil, or internal-push problems by conserving the system's momentum instead of tracking every particle separately.
Correctly classify a given force in a problem as internal (cancels, ignore for $\vec A_{cm}$ ) or external (must be included in $\vec F_{ext}$ ).
Compute $\vec V_{cm}$ or $\vec A_{cm}$ for a two- or three-particle system directly from the mass-weighted average formulas.
Connect this force-based result to the momentum-based statement of momentum conservation used in the next topic.

Ask the tutor: Want to work through why an exploding object's centre of mass keeps tracing the same parabola, or need help sorting internal from external forces in a tricky recoil problem — ask the AI tutor to walk it through with you.

If the two skaters push off each other and one is much heavier, does the centre of mass move towards the lighter one since it moves faster?

No. Speed alone does not matter — momentum does, and the two skaters always have equal and opposite momenta right after the push (since they started at rest together). The centre of mass formula weights each skater's velocity by their mass, and those two momentum terms cancel exactly, however different their speeds are. So the centre of mass stays exactly where it started, for any mass ratio, as long as the ice really is frictionless and horizontal.

In the shell problem, why can we ignore gravity while doing the momentum conservation step?

Gravity is a real external force and it does act during the explosion, but the explosion itself is treated as happening over an extremely short time interval. Over that vanishingly small time, gravity's impulse (force times time) is negligible compared with the huge internal explosive forces, so the horizontal momentum just before and just after the blast is essentially unchanged. Right after the explosion, though, gravity resumes acting on each fragment separately, and their paths will curve downward from that moment on.

Does M times A of centre of mass equals F external mean the whole system literally behaves like one point particle in every way?

Only for describing where the centre of mass is and how it moves — not for anything else about the system. Individual particles can rotate, collide, deform, or fly off in any direction; energy can be lost to heat or sound in a collision; shapes can change completely. The one thing guaranteed is that the mass-weighted average position keeps obeying $M\vec A_{cm} = \vec F_{ext}$ regardless of all that internal chaos, because internal forces always cancel when summed over the system.

Is this the same physics as conservation of momentum, or a different topic?

They are two views of the same fact. If $\vec F_{ext}=0$ , then $M\vec A_{cm}=0$ means $\vec V_{cm}$ is constant, and since the system's total momentum is $\vec P = M\vec V_{cm}$ , a constant $\vec V_{cm}$ is exactly conservation of total momentum. This article derives the underlying "why" using forces; the next topic in the chapter, linear momentum of a system of particles, restates and extends the same idea directly in terms of momentum.