On the edge of a wall, we build a brick tower that only holds together because of the bricks' own weight. Our goal is to build a stable tower whose overhang $d$ is greater than the length $\ell$ of a single brick. What is the minimum number of bricks you need? [Hint: Find the center of mass for each brick and add.]

On the edge of a wall, we build a brick tower that only holds together because of the bricks' own weight. Our

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Concept understanding — Center of Mass

What is the Center of Mass?

Imagine you pick up a broom by the handle and try to balance it horizontally on one finger. You instinctively slide your finger along the handle until the broom stays level. That point — the one where the broom doesn't tip — is its center of mass.

Now think about throwing a cricket bat. It spins and wobbles in the air, but there is one point on the bat that follows a smooth, parabolic path, as if all the bat's mass were concentrated there. That point is also the center of mass.

The core idea is simple: the center of mass is the average position of all the mass in an object. It's the point where you could imagine the entire mass of the object being concentrated, and the object would behave the same way under the influence of external forces.

Why does this matter?

When you push an object at its center of mass, it moves in a straight line without rotating. Push it anywhere else, and it will both move and spin. This is why:

A car's stability depends on where its center of mass is (lower = safer).
A tightrope walker holds a long pole — moving the pole shifts their combined center of mass back over the rope.
In projectile motion, the center of mass of a system (like an exploding firework) continues along the original parabolic path, even though the fragments scatter.

The precise definition

For a system of particles, the center of mass is the weighted average of their positions, where the weight is the mass of each particle.

\vec{R}_{\text{CM}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2 + \cdots + m_n \vec{r}_n}{m_1 + m_2 + \cdots + m_n} = \frac{\sum m_i \vec{r}_i}{\sum m_i}

Here:

$\vec{R}_{\text{CM}}$ is the position vector of the center of mass
$m_i$ is the mass of the $i$ -th particle
$\vec{r}_i$ is the position vector of that particle

For a continuous object (like a rod or a sphere), the sum becomes an integral:

\vec{R}_{\text{CM}} = \frac{1}{M} \int \vec{r} \, dm

where $M$ is the total mass and $dm$ is an infinitesimal mass element.

Breaking it down with an example

Take two masses on a light rod: $m_1 = 2 \text{ kg}$ at $x = 0$ , and $m_2 = 3 \text{ kg}$ at $x = 5 \text{ m}$ .

The center of mass is:

x_{\text{CM}} = \frac{(2)(0) + (3)(5)}{2 + 3} = \frac{0 + 15}{5} = 3 \text{ m}

So the center of mass is at $x = 3 \text{ m}$ , closer to the heavier mass. That makes intuitive sense — the heavier mass "pulls" the average toward itself.

Note

The center of mass does not have to be inside the object. A ring or a hollow sphere has its center of mass at the geometric center, which is empty space.

Key properties to remember

External forces only — Internal forces (like collisions between parts of the system) do not affect the motion of the center of mass. Only external forces can change its velocity.
If no external force acts, the center of mass moves with constant velocity (or stays at rest). This is the law of conservation of momentum applied to the whole system.
For symmetric objects with uniform density, the center of mass coincides with the geometric center. For irregular shapes, it shifts toward the region with more mass. …

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Step 1. Set up the stacking rule. Build the tower from the top brick downward. For any group of the topmost $k$ bricks (each of length $\ell$ ), treated as a single rigid unit, the maximum they can overhang the brick immediately below them — without the whole group toppling — is exactly reached when the combined center of mass of the top $k$ bricks sits directly above the very edge of the $(k+1)$ -th brick. Any further overhang and the group topples; this is the greatest stable overhang, since the problem asks for the minimum number of bricks for a given required overhang, using the best (maximum-overhang) stacking at every stage.

Step 2. Find the extra overhang contributed when going from $k$ bricks to $k+1$ bricks. Adding the $(k+1)$ -th brick beneath a stable stack of $k$ bricks (whose combined center of mass is known), the new combined center of mass (of all $k+1$ bricks) shifts by an amount inversely proportional to $k$ ; working through the center-of-mass balance (treating the top $k$ bricks, of total mass $km_0$ for single-brick mass $m_0$ , as concentrated at their own combined center of mass, positioned at the very edge of brick $k+1$ ), the additional overhang gained at this step is

\Delta_k=\frac{\ell}{2k}.

Step 3. Sum the overhangs to get the total overhang after $n$ bricks. Starting from the top (brick 1, contributing nothing extra since it is the very first) and adding each subsequent brick below it, the total overhang beyond the edge of the bottom (table or wall) support, after $n$ bricks, is

d_n=\sum_{k=1}^{n-1}\frac{\ell}{2k}=\frac{\ell}{2}\left(1+\frac12+\frac13+\cdots+\frac{1}{n-1}\right)=\frac{\ell}{2}H_{n-1},

where $H_{n-1}$ is the $(n-1)$ -th harmonic number.

Step 4. Apply the required condition, $d_n>\ell$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 22 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. In the absence of an external force, velocity of the center of mass -$
(a) Remains constant
(b) Decreases
(c) is Zero
(d) Increases
›Reveal solutionSolution

In the absence of an external force, the velocity of the centre of mass of a system remains constant (Newton's first law applied to the system as a whole).

For a system of particles, the equation of motion of the centre of mass is F_ext = M (d v_cm/dt), where M is the total mass and F_ext is the net external force. This follows because all internal forces (between particles of the system) occur in Newton's-third-law pairs and cancel out when summed over the whole system. If F_ext = 0, then d v_cm/dt = 0, meaning the centre of mass m …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. There are two objects of masses 1 kg and 2 kg located at (1, 2) and (-1, 3) respectively. The coordinates of the centre of mass are:$
(a) (2, -1)
(b) (8/3, -1/3)
(c) (-1/3, 8/3)
(d) None of the above
›Reveal solutionSolution

$x_{cm}=\dfrac{m_1x_1+m_2x_2}{m_1+m_2}$ , $y_{cm}=\dfrac{m_1y_1+m_2y_2}{m_1+m_2}$ give $\left(-\tfrac13,\tfrac83\right)$ .

For a system of point masses, the centre of mass is:

$x_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}, \qquad y_{cm}=\frac{m_1y_1+m_2y_2}{m_1+m_2}$

With $m_1=1\text{ kg}$ at $(1,2)$ and $m_2=2\text{ kg}$ at $(-1,3)$ :

$x_{cm}=\frac{1(1)+2(-1)}{1+2}=\frac{1-2}{3}=-\frac13$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If a uniform rod has length l, then its centre of mass will be located at$
(a) l/2
(b) 3l/4
(c) l/3
(d) none of these
›Reveal solutionSolution

A uniform rod's centre of mass is at its midpoint, l/2. Answer (A).

For a uniform rod the linear mass density is constant, so the mass is symmetric about the middle of the rod. By symmetry the centre …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the centre of gravity of a body coincides with its centre of mass, then the gravitational fields acting on different parts of the body are$
(a) zero
(b) equal
(c) different
(d) none of these
›Reveal solutionSolution

Centre of gravity coincides with centre of mass when the gravitational field is uniform (equal on all parts). Answer (B).

The centre of mass depends only on the mass distribution. The centre of gravity is the point where the total weight appears to act, which depends on how g varies over the body.

…

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CBSE 2026Set ANN1 markMCQ
$Q. For which of the following shapes, the centre of mass does not lie on the body?$
(a) Pencil
(b) Dice
(c) Shotput
(d) Bangle
›Reveal solutionSolution
A bangle's centre of mass lies at its geometric centre - empty space - so the centre of mass does not lie on the body.

The centre of mass of a body need not lie within the material of the body; it depends on how the mass is distributed. For symmetric bodies it lies at the geometric centre.

Pencil: centre of mass lies on the pencil (its midpoint).

Dice (cube): centre of mass at the solid centre, inside the body. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Where is the centre of mass of a uniform rod located?$
(a) At one end of the rod
(b) At the midpoint of the rod
(c) At the top of the rod
(d) At the bottom of the rod
›Reveal solutionSolution

For any uniformly (symmetrically) shaped and uniformly dense body like a straight rod, the centre of mass lies at the geometric centre -- the midpoint.

Centre of mass, x_cm = (Integral of x dm) / (Integral of dm), over the length of the rod.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Case study: The 'centre of mass' of a body is a point at which whole of the mass of the body is supposed to be concentrated. The centre of mass of a two body system is- X_cm = (m1 x1 + m2 x2) / (m1 + m2) where symbols have their usual meaning. The velocity of centre of mass is V_cm = dX_cm / dt.$
(1) Which of the following statement is false?
(a) Centre of mass of body may lie outside the body.
(b) Centre of mass of body may lie on the body.
(c) Centre of mass of two bodies of different masses lie at the centre of line joining the masses.
(d) Centre of mass of a ring lies at a point where actually no mass of the body exists.
›Reveal solutionSolution
Statement (c) is the false one — the COM of two unequal masses lies closer to the heavier mass, not at the geometric midpoint.

The centre of mass of a two-particle system lying on a line, with masses $m_1$ and $m_2$ at positions $x_1$ and $x_2$ , is given by:

$X_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

Check each statement:

'Centre of mass of body may lie outside the body' — TRUE. Example: a ring, a horseshoe, or a boomerang — the COM lies in the hollow/empty region, outside the actual material of the body.

'Centre of mass of body may lie on the body' — TRUE. For a uniform, solid, symmetric body like a sphere or cube, the COM lies at the geometric centre, which is a point within the body's material. …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. (2) If two particle of equal mass are placed at a distance 'd' apart, the centre of mass is located$
(a) At one of the particles.
(b) At the midpoint of the distance between them.
(c) Outside the line joining the two particles.
(d) None of above
›Reveal solutionSolution

Two equal masses have their centre of mass exactly at the midpoint of the line joining them.

Let the two particles, each of mass m, be placed at positions $x_1 = 0$ and $x_2 = d$ on a line. Using the centre-of-mass formula:

$X_{cm} = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \dfrac{m(0) + m(d)}{m+m} = \dfrac{md}{2m} = \dfrac{d}{2}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. (3) The motion of centre of mass of a system is determined by$
(a) External forces only
(b) Internal forces only
(c) Both internal and external forces
(d) Number of particles in the system
›Reveal solutionSolution

Only external forces determine the motion of the centre of mass; internal forces always cancel by Newton's third law.

For a system of particles, the total mass M times the acceleration of the centre of mass equals the sum of ALL forces acting on all the particles:

$M a_{cm} = \sum F_{external} + \sum F_{internal}$

By Newton's third law, every internal force (a particle of the system exerting a force on another particle of the same system) comes as an action-reaction pair that is equal in magnitude and opposite in direction. When summed over the whole system, all such internal force pairs cancel exactly:

$\sum F_{internal} = 0$

So the equation of motion for the centre of mass simplifies to:

$M a_{cm} = \sum F_{external}$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. (4) A body moving in a straight line explodes in 20 pieces. The path of centre of mass will be$
(a) Parabolic
(b) Straight line
(c) Circular
(d) Hyperbolic
›Reveal solutionSolution

The COM keeps moving along the original straight-line path, since explosion forces are purely internal.

As derived from Newton's laws applied to a system of particles, the acceleration of the centre of mass of a system is determined ONLY by the net EXTERNAL force acting on it; internal forces (however violent) always cancel in pairs by Newton's third law and cannot change the COM's motion.

When a body moving in a straight line explodes into 20 pieces, the forces of the explosion (which fling the pieces apart) are entirely internal to the system made up of all 20 pieces together. No new external force is introduced by the explosion itself (only gravity, if present, would be external — and the question is framed purely on the explosion's effect).

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Out of two particles of same mass, one is stationary and acceleration of another is a⃗. The acceleration of centre of mass of the system will be (A) zero (B) a⃗/2 (C) a⃗ (D) 2a⃗$
›Reveal solutionSolution

Centre-of-mass acceleration of the two-particle system is $\vec a/2$ .

For a system of particles, $\vec a_{cm}=\dfrac{\sum m_i\vec a_i}{\sum m_i}$ . Here both particles have equal mass $m$ : one is stationary ( $\vec a_1=0$ ) and the other has acceleration $\vec a$ …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. If a stationary firecracker explodes into a number of particles, then the centre of mass will$
(a) move in vertical direction
(b) move in concentric circles
(c) remain stationary
(d) move in parabolic path.
›Reveal solutionSolution

Internal explosive forces cannot move the centre of mass; only an external force can, so it stays at rest.

The forces that blow the firecracker into fragments are internal forces (action-reaction pairs between the fragments) — by Newton's third law they always occur in equal and opposite pairs and cancel out when summed over the whole system. The motion of the centre of mass of a system is governed only by the net external force acting on it: $M\vec{a}_{cm} = \vec{F}_{ext}$ . Since gravity is the only external force here (assume negligible over the short duration, and the cracker was stationary be …

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