In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = $10^{-10}$ m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10⁻¹⁰ m). Find

Exercises · 6.2

10^{-10}

Sikkim CbseNCERTSubjective· 3mImportance★★★★★est

25% · 14/57 Questions

✓ Free question

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.
In a uniform gravitational field, the center of mass and the center of gravity are the same point. (They differ only if gravity varies significantly across the object — not something you'll see in school problems.)

A final intuition

Think of the center of mass as the balance point of an object. If you could place a tiny, invisible support exactly at that point, the object would be perfectly balanced in any orientation. Every piece of mass on one side is exactly counterbalanced by the pieces on the other side.

That's why, when you jump off a boat, the boat moves backward — your center of mass and the boat's center of mass shift relative to each other, but the center of mass of the whole system (you + boat) stays put (if no external horizontal force acts). This is the heart of why the center of mass concept is so powerful: it lets you treat a complicated, spinning, wobbling object as a single point for many problems.

Looking up "Center of Mass: definition, formula & real-world examples" is a good habit before an exam, and it is worth knowing that Center of Mass is a core, NCERT-aligned topic from the System of Particles and Rotational Motion portion of the Class 11 Physics curriculum, and it is tested regularly in CBSE board exams as well as in JEE Main and NEET. Cross-checking this explanation against the relevant NCERT Physics chapter and solving a few past-year questions will round out your preparation.

The center of mass of the HCl molecule is found by treating it as a two-point mass system, with the heavier chlorine atom pulling the CM closer to itself. The CM is located approximately 0.035 Å from the chlorine nucleus.

The center of mass (CM) of a system is a unique point where the entire mass of the system can be considered to be concentrated for translational motion. It's essentially the "average" position of all the mass, weighted by how much mass is at each location. For a molecule like HCl, which consists of two atoms, we can approximate each atom as a point mass located at its nucleus, as the problem states that nearly all the mass is concentrated there.

Intuitively, the center of mass will always be closer to the heavier part of the system. Imagine trying to balance a seesaw with a child on one end and an adult on the other; the pivot point (CM) must be much closer to the adult to achieve balance. In the HCl molecule, the chlorine atom is significantly more massive than the hydrogen atom, so we expect the CM to be located much closer to the chlorine nucleus.

We will use the formula for the center of mass of a two-particle system, which is a weighted average of their positions.

Set up a coordinate system and define positions.

Let's place the hydrogen nucleus at the origin of our one-dimensional coordinate system. This simplifies the calculation as one position will be zero.
- Position of Hydrogen nucleus ( $x_H$ ): $0 \text{ Å}$
- Position of Chlorine nucleus ( $x_{Cl}$ ): $1.27 \text{ Å}$ (given separation)
Define the masses of the atoms.

Let $m_H$ be the mass of the hydrogen atom. We are given that a chlorine atom is about 35.5 times as massive as a hydrogen atom.
- Mass of Hydrogen atom ( $m_H$ ): $m$ (we can use a generic mass unit $m$ )
- Mass of Chlorine atom ( $m_{Cl}$ ): $35.5 m$
Apply the formula for the center of mass.

For a system of two point masses $m_1$ and $m_2$ located at positions $x_1$ and $x_2$ respectively, the position of the center of mass ( $X_{CM}$ ) is given by:
$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$
Substituting our specific values for HCl:

X_{CM} = \frac{m_H x_H + m_{Cl} x_{Cl}}{m_H + m_{Cl}}

Substitute values and calculate. Now, we plug in the values for masses and positions:

X_{CM} = \frac{(m)(0 \text{ Å}) + (35.5 m)(1.27 \text{ Å})}{m + 35.5 m}

The term $(m)(0 \text{ Å})$ becomes $0$. The mass unit $m$ will cancel out from the numerator and denominator, which is a common feature when dealing with mass ratios.

X_{CM} = \frac{35.5 m \times 1.27 \text{ Å}}{36.5 m}

X_{CM} = \frac{35.5 \times 1.27 \text{ Å}}{36.5}

X_{CM} = \frac{45.085 \text{ Å}}{36.5}

X_{CM} \approx 1.235 \text{ Å}

Interpret the result.

The calculated value $X_{CM} = 1.235 \text{ Å}$ represents the position of the center of mass from our chosen origin, which is the hydrogen nucleus.
- So, the CM is $1.235 \text{ Å}$ away from the hydrogen nucleus.
- To find its distance from the chlorine nucleus, we subtract this value from the total separation: Distance from Cl nucleus $= 1.27 \text{ Å} - 1.235 \text{ Å} = 0.035 \text{ Å}$ .
This result confirms our intuition: the center of mass is very close to the much heavier chlorine atom, only $0.035 \text{ Å}$ away from it, while being $1.235 \text{ Å}$ away from the lighter hydrogen atom.

Watch out

When stating the location of the center of mass, always specify the reference point. A numerical value alone is insufficient.

✓Final answer

The approximate location of the center of mass of the HCl molecule is 0.035 Å from the chlorine nucleus .

←Previous question6.1 — Give the location of the centre of mass of a (i) sphere, (ii) cylinder…→Next question6.3 — A child sits stationary at one end of a long trolley moving uniformly …

🎓Unlock everything free for 14 days

Unlock everything free for 14 days