Velocity and acceleration of the CM
The position of the centre of mass of \(n\) particles of total mass \(M = m_1 + m_2 + \dots + m_n\) is the mass-weighted average position \(M\vec R = m_1\vec r_1 + m_2\vec r_2 + \dots + m_n\vec r_n\). Treat each mass as a constant and differentiate once with respect to time. The result is the velocity of the centre of mass:
$$ M\vec V = m_1\vec v_1 + m_2\vec v_2 + \dots + m_n\vec v_n $$
So \(\vec V_{\text{cm}}\) is just the mass-weighted average of the individual particle velocities. Differentiate once more and the same averaging carries over to acceleration:
$$ M\vec A = m_1\vec a_1 + m_2\vec a_2 + \dots + m_n\vec a_n $$
Here \(\vec A = \vec A_{\text{cm}}\) is the acceleration of the centre of mass. Nothing has been assumed yet about why the particles accelerate — these two relations are pure kinematics, the direct time-derivatives of the definition of the CM.
The master equation M·a_cm = F_ext
Now bring in dynamics. By Newton's second law the net force on particle \(i\) is \(\vec F_i = m_i\vec a_i\). Substituting into the acceleration relation gives
$$ M\vec A = \vec F_1 + \vec F_2 + \dots + \vec F_n $$
NCERT states the meaning of this sum precisely: "the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles." But each \(\vec F_i\) is itself a sum of an external part (from bodies outside the system) and internal parts (from the other particles). The internal parts cancel — established in the next section — leaving only the external forces. The equation collapses to its final form:
$$ M\vec a_{\text{cm}} = \vec F_{\text{ext}} $$
"The centre of mass of a system of particles moves as if all the mass of the system was concentrated at the centre of mass and all the external forces were applied at that point." — NCERT §6.3
The practical payoff is enormous. To find the translational motion of any extended body — a wobbling hammer, a spinning wheel, a cloud of fragments — you do not track millions of particles. You track one point, give it the total mass, and apply only the external forces. This is exactly the silent assumption made in earlier chapters when bodies were treated as point particles; §6.3 is its formal justification.
Why internal forces cancel
The cancellation is a direct consequence of Newton's third law. Sort the force on each particle into the force from outside the system (external) and the forces from the other particles inside it (internal). When you add up the forces over all particles, the internal forces "occur in equal and opposite pairs and in the sum of forces their contribution is zero," in NCERT's words. The NIOS lesson states the same for a rigid body: "the sum of the internal forces is zero because they cancel each other in pairs."
This is why the result is so general. NCERT stresses that the derivation "did not need to specify the nature of the system of particles": it may have all kinds of internal motion, or be a rigid body in pure translation, or a rigid body that is also rotating. Whatever the internal story, the centre of mass obeys \(M\vec a_{\text{cm}} = \vec F_{\text{ext}}\).
The exploding-projectile parabola
NCERT calls this "a good illustration of Eq. (6.11)." A projectile follows the usual parabolic trajectory and then explodes into fragments midway through its flight. The forces that drive the explosion are internal forces; they contribute nothing to the motion of the centre of mass. The one external force — gravity — is the same before and after the blast.
If a question states that one fragment lands at a particular spot, you can locate the others by insisting the CM remain on the original parabola. If the explosion happens at the very top of the flight, the CM continues to the same range and the same landing point as the intact shell — the fragments simply distribute themselves around that point.
Worked example 1 — exploding shell into three fragments
A shell of mass \(m\) is at rest initially. It explodes into three fragments having masses in the ratio \(2:2:1\). The two equal-mass fragments fly off along mutually perpendicular directions, each with speed \(v\). Find the speed of the third (lighter) fragment.
Set the CM equation to work. The shell is at rest and the explosion forces are internal, so the total external force in the horizontal plane is zero. The centre of mass therefore stays at rest: total momentum before equals total momentum after, both zero.
Assign masses. Let the fragments be \(2m', 2m', m'\). The two heavy fragments move along perpendicular directions: momentum \(2m'v\,\hat i\) and \(2m'v\,\hat j\).
Combine the two perpendicular momenta. Their resultant has magnitude \(\sqrt{(2m'v)^2 + (2m'v)^2} = 2\sqrt{2}\,m'v\), pointing into the first quadrant.
The third fragment must cancel it. For zero net momentum, \(m'v' = 2\sqrt{2}\,m'v\), so \(\boxed{v' = 2\sqrt{2}\,v}\) directed opposite to the resultant of the other two. The CM stays put exactly because no external force acts in the plane of the blast.
Worked example 2 — two-block system on a frictionless surface
Two blocks of mass \(2\,\text{kg}\) and \(3\,\text{kg}\) rest on a frictionless horizontal table connected by a compressed spring held by a thread. The thread is burnt; the spring pushes the blocks apart. (a) What is the acceleration of the centre of mass while the spring acts? (b) The \(2\,\text{kg}\) block leaves with speed \(3\,\text{m s}^{-1}\). Find the speed of the \(3\,\text{kg}\) block.
(a) The CM acceleration. Horizontally, the only forces are the spring forces between the two blocks — these are internal to the system. The external horizontal force is zero, so \(M\vec a_{\text{cm}} = \vec F_{\text{ext}} = 0\). The centre of mass does not accelerate; it stays exactly where it was. (Vertically, weight and normal cancel.)
(b) Speed of the second block. Since the CM stays at rest, \(\vec V_{\text{cm}} = 0\) throughout. With \(M\vec V_{\text{cm}} = m_1\vec v_1 + m_2\vec v_2 = 0\) we get \(2 \times 3 = 3 \times v_2\), so \(v_2 = 2\,\text{m s}^{-1}\), moving opposite to the \(2\,\text{kg}\) block.
Takeaway. The spring force, however large, never moves the centre of mass — it only redistributes momentum so the two parts recoil in inverse ratio of their masses.
The CM-at-rest argument is really momentum conservation in disguise — see linear momentum of a system of particles for the full treatment.
Worked example 3 — man walking on a boat
A man of mass \(60\,\text{kg}\) stands at one end of a stationary boat of mass \(120\,\text{kg}\) floating on still water. He walks \(3\,\text{m}\) toward the other end. Neglecting water resistance, how far does the boat move, and in which direction?
Identify the external horizontal force. The water exerts no horizontal force (resistance neglected), so the horizontal external force on the man-plus-boat system is zero. Therefore the horizontal position of the centre of mass cannot change — it stays fixed in the ground frame.
Set up displacements. Let the boat slide back by a distance \(x\). The man's displacement relative to the water is \((3 - x)\) forward (he advances \(3\,\text{m}\) relative to the boat, but the boat itself retreats \(x\)). For the CM to stay fixed: \(m_{\text{man}}(3 - x) = m_{\text{boat}}\,x\).
Solve. \(60(3 - x) = 120\,x \Rightarrow 180 - 60x = 120x \Rightarrow 180 = 180x \Rightarrow x = 1\,\text{m}\). The boat slides \(1\,\text{m}\) backward (opposite to the man's walk); the man advances \(2\,\text{m}\) relative to the water.
Check. Man moves \(+2\,\text{m}\), boat \(-1\,\text{m}\): \(60(2) + 120(-1) = 120 - 120 = 0\). The mass-weighted displacement of the CM is zero, as required.
Applications at a glance
The single idea — when \(\vec F_{\text{ext}} = 0\) the centre of mass keeps its state of motion — drives a whole family of NEET situations. The table collects them.
| Situation | External horizontal force | What the CM does |
|---|---|---|
| Shell explodes in flight | Gravity only (vertical) | CM continues on the original parabola |
| Shell at rest explodes | Zero in the blast plane | CM stays at rest; fragment momenta sum to zero |
| Spring pushes two blocks apart | Zero (frictionless) | CM stays at rest; blocks recoil in inverse mass ratio |
| Man walks on a floating boat | Zero (water resistance neglected) | CM position fixed; boat slides opposite to the man |
| Gun recoils on firing | Zero (horizontal) | CM of gun + bullet stays at rest; gun recoils |
In every row the internal force can be enormous — the blast, the spring, the muscles, the propellant — yet none of it shifts the centre of mass. Only an external agent can do that. Reading a problem, your first move is to ask which forces are external; everything internal is irrelevant to \(\vec a_{\text{cm}}\).
Motion of the centre of mass in one breath
- \(\vec V_{\text{cm}} = \dfrac{\sum m_i\vec v_i}{M}\) and \(\vec A_{\text{cm}} = \dfrac{\sum m_i\vec a_i}{M}\) — both are mass-weighted averages.
- \(M\vec a_{\text{cm}} = \vec F_{\text{ext}}\): the CM moves as if all mass were concentrated there and all external forces acted on it.
- Internal forces cancel in equal-and-opposite third-law pairs, so they never affect \(\vec a_{\text{cm}}\).
- An exploding projectile's CM stays on the original parabola — gravity is unchanged by the blast.
- When \(\vec F_{\text{ext}} = 0\), the CM keeps its velocity (or stays at rest): the basis of recoil, boat–man, and two-block problems.
- The result holds for rigid bodies, rotating bodies, and loose particle clouds alike.