In short

The centre of mass (CM) of a system obeys \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} — it accelerates as if all the system's mass were concentrated at one point and all external forces acted there. Internal forces (explosions, collisions, springs) cannot change the CM's velocity. If no external force acts, the CM moves at constant velocity or stays at rest, no matter how the parts inside rearrange themselves.

A Diwali rocket soars upward, tracing a smooth arc across the night sky. At the top of its path it explodes — a burst of red and gold — and fragments scatter in every direction. Some fly up, some tumble down, some spin sideways. The beautiful symmetry of the original arc seems destroyed.

But it is not. If you could track the centre of mass of all those scattered fragments — every glowing ember, every cardboard shard — you would find it keeps moving along the exact same parabola the intact rocket was following. The explosion, no matter how violent, cannot push the centre of mass off course. Gravity is the only external force, and gravity alone decides where the CM goes.

This is the most powerful idea in the mechanics of many-body systems: the centre of mass moves as if the entire system were a single particle, subject only to external forces. Internal forces — the explosion, the spring, the collision — redistribute motion among the parts but cannot touch the CM.

Velocity of the centre of mass

The position of the centre of mass for two bodies of mass m_1 and m_2 is:

\vec{r}_{\text{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{M}

where M = m_1 + m_2 is the total mass. A point that sits still is not interesting. What happens when the bodies move?

Differentiate both sides with respect to time:

\vec{v}_{\text{cm}} = \frac{d\vec{r}_{\text{cm}}}{dt} = \frac{m_1 \dfrac{d\vec{r}_1}{dt} + m_2 \dfrac{d\vec{r}_2}{dt}}{M}
\boxed{\vec{v}_{\text{cm}} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{M}} \tag{1}

Why: the velocity of the CM is the mass-weighted average of the individual velocities. A heavier body pulls the CM velocity toward its own, just as a heavier body pulls the CM position toward its own location.

The numerator is the total linear momentum: m_1 \vec{v}_1 + m_2 \vec{v}_2 = \vec{p}_{\text{total}}. So:

\vec{v}_{\text{cm}} = \frac{\vec{p}_{\text{total}}}{M} \tag{2}

Why: the CM velocity is just the total momentum divided by the total mass. If you know the total momentum of a system and its total mass, you know how fast the CM is moving — without tracking each body individually.

This extends naturally to n bodies:

\vec{v}_{\text{cm}} = \frac{\sum_{i=1}^{n} m_i \vec{v}_i}{\sum_{i=1}^{n} m_i} = \frac{\vec{p}_{\text{total}}}{M}

Acceleration of the centre of mass — Newton's second law for systems

Differentiate equation (1) once more:

\vec{a}_{\text{cm}} = \frac{d\vec{v}_{\text{cm}}}{dt} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{M} \tag{3}

Why: same differentiation technique. The acceleration of the CM is the mass-weighted average of the individual accelerations.

Now apply Newton's second law to each body separately. Body 1 feels forces from body 2 (internal force \vec{f}_{12}) and from the outside world (external force \vec{F}_1^{\text{ext}}):

m_1 \vec{a}_1 = \vec{F}_1^{\text{ext}} + \vec{f}_{12}
m_2 \vec{a}_2 = \vec{F}_2^{\text{ext}} + \vec{f}_{21}

Why: each body obeys \vec{F}_{\text{net}} = m\vec{a} individually. The net force on each body is the sum of external forces and internal forces from the other body.

Add the two equations:

m_1 \vec{a}_1 + m_2 \vec{a}_2 = \left(\vec{F}_1^{\text{ext}} + \vec{F}_2^{\text{ext}}\right) + \left(\vec{f}_{12} + \vec{f}_{21}\right)

By Newton's third law, \vec{f}_{12} + \vec{f}_{21} = \vec{0}. The internal forces cancel in pairs — always, exactly, without exception.

m_1 \vec{a}_1 + m_2 \vec{a}_2 = \vec{F}_{\text{ext}} \tag{4}

where \vec{F}_{\text{ext}} = \vec{F}_1^{\text{ext}} + \vec{F}_2^{\text{ext}} is the total external force on the system.

Why: Newton's third law guarantees that every internal force has an equal and opposite partner. When you sum forces over the entire system, every internal pair adds to zero. Only external forces survive.

Substitute equation (4) into equation (3):

\vec{a}_{\text{cm}} = \frac{\vec{F}_{\text{ext}}}{M}
\boxed{\vec{F}_{\text{ext}} = M \vec{a}_{\text{cm}}} \tag{5}

Why: this is Newton's second law for the system as a whole. The centre of mass accelerates exactly as if the entire mass M were concentrated at one point and the total external force acted there. The internal forces — however complicated, however violent — have vanished from the equation.

Read equation (5) again. No matter how many bodies the system contains, no matter how they interact — collide, explode, attract, repel, connect with springs, break apart — the CM cares only about the total external force and the total mass. Internal forces are invisible to it.

For n bodies, the same result holds. Every internal force \vec{f}_{ij} on body i from body j is paired with \vec{f}_{ji} = -\vec{f}_{ij} on body j from body i. All \frac{n(n-1)}{2} internal pairs cancel, leaving:

\vec{F}_{\text{ext}} = M \vec{a}_{\text{cm}}

What this means — three consequences

1. No external force → CM velocity is constant

If \vec{F}_{\text{ext}} = \vec{0}, then \vec{a}_{\text{cm}} = \vec{0}, which means \vec{v}_{\text{cm}} is constant. The CM either stays at rest or moves in a straight line at constant speed — regardless of how the parts inside rearrange.

Picture two ice skaters standing face to face on a frictionless rink. They push off each other. Skater A flies left, skater B flies right. But the push is an internal force. There is no horizontal external force on the two-skater system. So the CM does not budge — it stays exactly where it was before the push.

A person walking inside a boat on a still lake is the same physics. The person moves forward; the boat slides backward. The person-plus-boat CM stays in place, because the walking force is internal to the system.

2. Under gravity alone, the CM follows a parabola

If gravity is the only external force, \vec{F}_{\text{ext}} = M\vec{g}, so \vec{a}_{\text{cm}} = \vec{g}. The CM follows the same trajectory as a single projectile — a parabola — regardless of what happens internally.

This is why the Diwali rocket's fragments, despite scattering wildly, have a CM that traces the original arc. Before the explosion, the rocket was a single projectile under gravity. After the explosion, the CM is still a system under gravity — same total mass, same external force, same trajectory.

Animated: a projectile explodes mid-flight, but its centre of mass continues on the original parabolic path A projectile launched at 20 m/s at 60 degrees explodes at t equals 1.5 seconds into two equal fragments. Fragment 1 flies forward faster and higher, fragment 2 slows and drops. A grey marker traces the centre of mass, which follows the original parabola throughout.
A projectile (launched at 20 m/s, 60° above horizontal) explodes at $t = 1.5$ s into two equal fragments. Fragment 1 (red) flies forward with extra speed. Fragment 2 (dark) slows and drops earlier. The centre of mass (grey dot) keeps tracing the original parabola as if no explosion occurred. Click replay to watch again.

3. Internal forces redistribute, external forces redirect

An explosion inside a system cannot change the CM velocity — it can only shuffle motion among the parts. The total momentum is conserved during the explosion (no external impulse), so \vec{v}_{\text{cm}} stays the same by equation (2).

An external force can change \vec{v}_{\text{cm}}. When a cricket ball hits the ground, the normal force from the pitch is an external force on the ball. That external force changes the CM velocity — the ball bounces. But the internal forces during a collision between two bodies within a system? Those leave the CM untouched.

This is why \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} is such a powerful shortcut. You do not need to know the details of the internal interactions — the explosions, the springs, the collisions. You only need the total external force and the total mass.

Explore: how mass affects the CM velocity

Two bodies move along a straight line. Body 1 (mass 2 kg) moves to the right at 5 m/s. Body 2 (mass m_2) moves to the left at 3 m/s. Drag the red point to change m_2 and watch the CM velocity respond.

Interactive: CM velocity as a function of the second body's mass A curve showing v_cm versus m2 for two bodies moving in opposite directions. As m2 increases past 10/3 kg, the CM velocity changes sign. A draggable point lets the reader explore different mass values. m₂ (kg) v_cm (m/s) 0 1 2 3 4 −1 2 4 6 8 v_cm drag the red dot along the axis
Body 1 (2 kg) moves right at 5 m/s. Body 2 ($m_2$ kg) moves left at 3 m/s. Drag the red dot to change $m_2$. When $m_2$ is small (below $10/3 \approx 3.3$ kg), body 1's rightward momentum dominates and the CM moves right. When $m_2$ exceeds $10/3$ kg, body 2's leftward momentum wins and the CM reverses direction. At $m_2 = 10/3$ kg exactly, the momenta balance and $v_{\text{cm}} = 0$.

The key point: once you set v_{\text{cm}} by choosing the masses and velocities, no internal force — no collision, no explosion, no spring — can change it. Only an external force can shift the red dot on this curve.

Worked examples

Example 1: Ice skaters pushing off

Two skaters stand face to face on a frictionless ice rink. Skater A has mass 50 kg and skater B has mass 70 kg. They push off each other and skater A slides backward at 2.1 m/s. Find skater B's velocity after the push, and verify that the centre of mass does not move.

Before and after diagram for two ice skaters pushing off Before: both skaters at rest, standing together. After: skater A moves left at 2.1 m/s, skater B moves right at 1.5 m/s. Before 50 kg 70 kg at rest After A 2.1 m/s B 1.5 m/s
Before: both skaters at rest. After: they push off — A slides left, B slides right. The frictionless ice means no horizontal external force acts on the system.

Step 1. Use conservation of momentum. Before the push, both skaters are at rest, so \vec{p}_{\text{total}} = 0.

m_A v_A + m_B v_B = 0
50 \times (-2.1) + 70 \times v_B = 0

Why: the push is an internal force — A pushes B and B pushes A (Newton's third law). No horizontal external force acts on the two-skater system, so total momentum stays zero.

Step 2. Solve for v_B.

v_B = \frac{50 \times 2.1}{70} = \frac{105}{70} = 1.5 \text{ m/s}

Why: v_B is positive (rightward), opposite to A's direction. The lighter skater moves faster: |v_A|/|v_B| = 2.1/1.5 = 1.4 = 70/50 = m_B/m_A. The speed ratio is the inverse of the mass ratio — a direct consequence of momentum conservation.

Step 3. Compute the CM velocity after the push.

v_{\text{cm}} = \frac{m_A v_A + m_B v_B}{M} = \frac{50 \times (-2.1) + 70 \times 1.5}{120} = \frac{-105 + 105}{120} = 0

Why: v_{\text{cm}} = 0 both before and after the push. The internal force changed the individual velocities dramatically — A went from 0 to -2.1 m/s, B from 0 to +1.5 m/s — but the CM velocity did not change at all. This is \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} at work: zero external force means zero CM acceleration.

Step 4. Verify the CM position after 2 seconds.

Place the origin at the initial position of the skaters. After 2 s: x_A = -2.1 \times 2 = -4.2 m, x_B = 1.5 \times 2 = 3.0 m.

x_{\text{cm}} = \frac{50 \times (-4.2) + 70 \times 3.0}{120} = \frac{-210 + 210}{120} = 0

Why: two seconds later, A is 4.2 m to the left and B is 3.0 m to the right, but the CM is still at the origin. The heavier skater moved less, exactly compensating for the lighter one moving more.

Result: Skater B moves at 1.5 m/s opposite to A. The CM velocity is zero both before and after — the push changed nothing about the CM.

What this shows: Internal forces (the push) send the skaters flying apart, but the CM does not budge. Zero external horizontal force means zero CM acceleration — and the numbers confirm it exactly.

Example 2: Diwali rocket exploding mid-flight

A Diwali rocket of mass 0.4 kg is launched at 30 m/s at 60° above the horizontal. At the highest point of its trajectory, it explodes into two equal fragments (0.2 kg each). One fragment has zero horizontal velocity after the explosion and falls vertically from the peak. Where does the other fragment land? (Use g = 9.8 m/s².)

Trajectory diagram for the exploding Diwali rocket A parabolic trajectory from launch to the original landing point at range R. At the peak, the rocket explodes. Fragment 1 falls vertically from the peak at R/2. Fragment 2 lands beyond R, at 3R/2. The CM lands at R. R explosion frag 1 R/2 frag 2 3R/2 CM at R launch
The rocket follows a parabola (dashed) until it explodes at the peak. Fragment 1 (red) drops straight down at $R/2$. Fragment 2 (dark) lands at $3R/2$. The CM (grey) lands exactly at $R$ — the original range.

Step 1. Find the original range R (as if no explosion happened).

R = \frac{u^2 \sin 2\theta}{g} = \frac{900 \times \sin 120°}{9.8} = \frac{900 \times 0.866}{9.8} = \frac{779.4}{9.8} \approx 79.5 \text{ m}

Why: even though the rocket explodes, its CM must land at this distance — because gravity is the only external force and the CM follows the original parabola.

Step 2. Locate the explosion point.

At the peak, the horizontal distance from the launch point is R/2 (for symmetric projectile motion with no air resistance):

x_{\text{peak}} = \frac{R}{2} \approx 39.75 \text{ m}

Fragment 1 falls vertically from the peak, so it lands at x_1 = 39.75 m.

Why: for a projectile launched at angle \theta over flat ground, the peak occurs at the midpoint of the range. Fragment 1 drops straight down from this point.

Step 3. Find the velocity at the peak.

At the highest point, the vertical component of velocity is zero. Only the horizontal component survives:

v_x = u \cos 60° = 30 \times 0.5 = 15 \text{ m/s}

Why: gravity acts only vertically, so it does not change the horizontal velocity. At the peak, v_y = 0 (the rocket momentarily stops climbing), leaving only v_x = 15 m/s.

Step 4. Apply conservation of momentum at the explosion.

Before the explosion, the horizontal momentum is p = Mv_x = 0.4 \times 15 = 6 kg·m/s. Fragment 1 has zero horizontal velocity after the explosion:

m_1 \times 0 + m_2 \times v_{2x} = 6
0.2 \times v_{2x} = 6 \implies v_{2x} = 30 \text{ m/s}

Why: the explosion is internal — it conserves total momentum. Fragment 1 carries zero horizontal momentum, so fragment 2 must carry all of it. Its horizontal velocity doubles from 15 to 30 m/s.

Step 5. Find where fragment 2 lands.

After the explosion, fragment 2 is at the peak height H with horizontal velocity 30 m/s and vertical velocity 0. It behaves as a projectile launched horizontally:

H = \frac{u^2 \sin^2\theta}{2g} = \frac{900 \times 0.75}{19.6} \approx 34.4 \text{ m}

Time to fall from height H:

t = \sqrt{\frac{2H}{g}} = \sqrt{\frac{68.8}{9.8}} \approx 2.65 \text{ s}

Horizontal distance from the explosion point:

\Delta x = 30 \times 2.65 = 79.5 \text{ m}

Why: this is not a coincidence — \Delta x equals R exactly. Fragment 2 travels the full original range from the peak, landing at x_2 = R/2 + R = 3R/2.

x_2 = 39.75 + 79.5 = 119.25 \text{ m}

Step 6. Verify the CM landing position.

x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{M} = \frac{0.2 \times 39.75 + 0.2 \times 119.25}{0.4} = \frac{7.95 + 23.85}{0.4} = \frac{31.8}{0.4} = 79.5 \text{ m} = R \; \checkmark

Why: the CM lands exactly at the original range. Fragment 1 at R/2, fragment 2 at 3R/2 — perfectly symmetric about R. The explosion was invisible to the CM.

Result: Fragment 2 lands at 3R/2 \approx 119.25 m from the launch point, about 79.5 m beyond the explosion.

What this shows: You can find where a fragment lands without knowing anything about the explosion forces. The CM must land at R. If one equal-mass fragment lands at R/2, the other must land at 3R/2. The internal explosion is completely invisible to the centre of mass.

Common confusions

  • "If no external force acts, nothing moves." Wrong. If no external force acts, the CM does not accelerate — but individual parts can move however they want. Two skaters push off and fly apart at high speed. A bomb explodes into a dozen fragments. The parts move; the CM does not (or keeps moving at constant velocity if it was already moving).

  • "Internal forces can move the centre of mass." Never. Internal forces always appear in Newton's-third-law pairs that sum to zero. An explosion inside a sealed box cannot change the CM velocity of the box-plus-contents system. A person walking inside a boat cannot change the person-plus-boat CM position — the boat slides backward by exactly the right amount.

  • "The CM always stays in the same spot." Only if the net external force is zero and the CM was initially at rest. Under gravity, the CM accelerates downward at g. Under a push from outside the system, the CM accelerates in the direction of the push. The rule is \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} — the CM is stationary only when \vec{F}_{\text{ext}} = 0 and \vec{v}_{\text{cm}} = 0.

  • "This works only for two bodies." It works for any number. For n bodies, all internal-force pairs cancel. There are n(n-1)/2 such pairs, and each sums to zero by Newton's third law. The result \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} holds regardless of how many interacting parts the system contains.

  • "The CM must be inside one of the bodies." The CM is a mathematical point, not a physical object. It can lie in empty space. For a hollow ring, the CM is at the geometric centre where there is no material. For two skaters standing apart, the CM is somewhere on the ice between them.

If you came here to understand how the centre of mass moves and to solve problems using \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}, you have everything you need. What follows is for readers who want the centre-of-mass reference frame and a first look at variable-mass systems.

The centre-of-mass reference frame

The CM velocity is \vec{v}_{\text{cm}} = \vec{p}_{\text{total}}/M. Jump into a frame that moves with the CM by subtracting \vec{v}_{\text{cm}} from every velocity:

\vec{v}_i' = \vec{v}_i - \vec{v}_{\text{cm}}

In this frame, the total momentum is always zero:

\sum m_i \vec{v}_i' = \sum m_i(\vec{v}_i - \vec{v}_{\text{cm}}) = \vec{p}_{\text{total}} - M\vec{v}_{\text{cm}} = \vec{0}

Why: subtracting \vec{v}_{\text{cm}} from every velocity removes the net drift. In the CM frame, the system as a whole goes nowhere — all motion is relative, and the total momentum is exactly zero by construction.

This is the frame where collisions look simplest. In an elastic collision viewed from the CM frame, each body simply reverses its velocity — because both total momentum (zero) and total kinetic energy must be conserved, the only solution is to flip the sign of every velocity. All the complexity of the lab-frame formulas — the mass ratios, the (m_1 - m_2)/(m_1 + m_2) fractions — is just the coordinate transformation into and out of this frame.

For an explosion in the CM frame, fragments fly out symmetrically: whatever momentum one fragment gains, another fragment gains an equal and opposite momentum. The total stays zero.

Momentum as a restatement of Newton's second law

Equation (5) can be rewritten using momentum. Since \vec{v}_{\text{cm}} = \vec{p}_{\text{total}}/M and M is constant:

\vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} = M\frac{d\vec{v}_{\text{cm}}}{dt} = \frac{d(M\vec{v}_{\text{cm}})}{dt} = \frac{d\vec{p}_{\text{total}}}{dt}
\boxed{\vec{F}_{\text{ext}} = \frac{d\vec{p}_{\text{total}}}{dt}}

Why: this is the most general form of Newton's second law. The total external force equals the rate of change of total momentum. When \vec{F}_{\text{ext}} = 0, total momentum is constant — which is conservation of momentum. The CM theorem and momentum conservation are two faces of the same coin.

Variable mass — the rocket equation preview

The derivation above assumed M is constant. But what about a rocket that burns fuel? As the rocket ejects exhaust, its mass decreases. The CM of the total system (rocket plus expelled fuel) still obeys \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}}, because no mass leaves the system — it just changes form from fuel to exhaust.

But if you want an equation for the rocket alone, you need the thrust term:

\vec{F}_{\text{ext}} + \vec{v}_{\text{rel}}\frac{dm}{dt} = m\vec{a}_{\text{rocket}}

Here \vec{v}_{\text{rel}} is the exhaust velocity relative to the rocket and dm/dt is the rate of mass ejection. The product \vec{v}_{\text{rel}} \cdot dm/dt is the thrust. It looks like a force on the rocket, but from the total-system perspective it is an internal force — the rocket pushes the exhaust backward, the exhaust pushes the rocket forward.

ISRO's PSLV burns about 12 tonnes of solid propellant in its first stage, ejecting exhaust at roughly 2.6 km/s. The thrust is what lifts the 320-tonne vehicle off the launchpad. But from the CM perspective, the entire PSLV-plus-exhaust system still obeys \vec{F}_{\text{ext}} = M\vec{a}_{\text{cm}} — the thrust is internal, and only gravity and atmospheric drag (external forces) determine where the CM goes.

Where this leads next