Q.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 a 1-D elastic collision of $P$ (mass $m_P=1.5\,\mathrm{kg}$, speed $u=10\,\mathrm{ms}^{-1}$) with $Q$ at rest:

Given $\dfrac{v_P'}{v_Q'}=\dfrac{1}{3}$:

So $m_Q=\dfrac{1.5}{3}=0.5\,\mathrm{kg}$.

The centre-of-mass velocity is unchanged by the collision:

Concept & Intuition

The centre of mass velocity depends only on total momentum and total mass. When two bodies move along the same line, their relative direction changes the total momentum dramatically. By setting up expressions for $v_\text{cm}$ in both cases and using the given factor of 9, we can eliminate the unknown masses and find the velocity ratio.

1. Define variables Let $v_A$ and $v_B$ be the velocities of A (2 kg) and B (3 kg) respectively. We are told that the linear momentum of A is greater than that of B:

2. Centre of mass velocity – same direction When both move in the same direction (say to the right),

3. Centre of mass velocity – opposite directions When they move in opposite directions, we need to assign signs. Let A move to the right ($+v_A$) and B to the left ($-v_B$). Then

(The problem states “moving in opposite directions” without specifying which is positive; the magnitude is what matters, and the sign will be handled by the ratio.)

4. Apply the given condition The problem says: “The velocity of centre of mass when moving in the same direction is 9 times the velocity of centre of mass when moving in opposite directions.” This means

Substituting:

The factor $1/5$ cancels, giving

5. Solve for the ratio Expand and simplify:

Bring terms together:

Hence

So the ratio $v_A : v_B = 15 : 8$.

Concept and Intuition

A wire bent into a circular ring has all its mass at the same distance from the center. The moment of inertia of a thin ring about a diameter is a standard result: $I = \frac{1}{2} M R^2$. Here, the mass $M$ is given (100 g = 0.1 kg), and the moment of inertia about a diameter is provided. We can solve for the radius $R$, and then the length of the wire is simply the circumference $L = 2\pi R$. The trick is to keep units consistent (kg, m, then convert to cm).

Step-by-step solution

1. Write the known quantities in SI units

   Mass: $M = 100\ \text{g} = 0.1\ \text{kg}$

   Moment of inertia about a diameter: $I = 98 \times 10^{-5}\ \text{kg m}^2$

2. Recall the formula for moment of inertia of a thin ring about a diameter

   For a thin circular ring of mass $M$ and radius $R$, the moment of inertia about any diameter is

This is because the ring’s mass is distributed at a constant distance $R$ from the center, and the perpendicular axis theorem gives $I_{\text{diameter}} = \frac{1}{2} I_{\text{axis through center}}$.

3. Substitute the known values and solve for $R^2$

Multiply both sides by 2:

Divide by 0.1:

So $R = \sqrt{1.96 \times 10^{-2}} = 0.14\ \text{m}$ (since $\sqrt{1.96}=1.4$ and $\sqrt{10^{-2}}=0.1$).

4. Find the length of the wire (circumference)

Simplify: $2 \times 0.14 = 0.28$, so

Convert to cm: $0.88\ \text{m} = 88\ \text{cm}$.

The problem gives you angular momentum $L$, mass $m$, and angular speed $\omega$ (in rev/min), and asks for the radius. The direct link between these quantities is the moment of inertia. For a uniform circular disc rotating about its central perpendicular axis, the moment of inertia is $I = \frac12 m R^2$. Angular momentum is $L = I\omega$, so you can solve for $R$.

The only trap is the units: angular speed is in rev/min, but angular momentum is in Js (which is kg m²/s). You must convert $\omega$ to rad/s before plugging in.

1. Convert angular speed to rad/s. $150$ rev/min means $150$ revolutions per minute. One revolution is $2\pi$ radians, and one minute is $60$ seconds.

2. Write the expression for angular momentum. For the disc, $I = \frac12 m R^2$. So

3. Substitute the known values. $m = \frac{40}{\pi}$ kg, $\omega = 5\pi$ rad/s, $L = 6.25$ Js.

4. Simplify. The $\pi$ cancels:

5. Solve for $R$.

Convert to cm: $0.25$ m = $25$ cm.

The centre of mass of a system moves as if all the mass were concentrated there and all external forces acted there. For velocity, we use:

That is, the total momentum divided by total mass. So we just need to find the vector sum of the momenta.

1. Assign directions

   Let north be $+\hat{j}$, south be $-\hat{j}$, east be $+\hat{i}$.

2. Write each particle’s momentum

   • Particle A: mass $m$, velocity $6\ \mathrm{m/s}$ north → $\vec{p}_A = m \cdot 6\hat{j} = 6m\,\hat{j}$
   • Particle B: mass $2m$, velocity $12\ \mathrm{m/s}$ south → $\vec{p}_B = 2m \cdot (-12\hat{j}) = -24m\,\hat{j}$
   • Particle C: mass $3m$, velocity $8\ \mathrm{m/s}$ east → $\vec{p}_C = 3m \cdot 8\hat{i} = 24m\,\hat{i}$

3. Sum the momenta

   North-south components: $6m - 24m = -18m\,\hat{j}$ (i.e., $18m$ south)

   East-west components: $24m\,\hat{i}$ (east)

   So total momentum $\vec{P} = 24m\,\hat{i} - 18m\,\hat{j}$

4. Magnitude of total momentum

5. Velocity of centre of mass Total mass $M = m + 2m + 3m = 6m$

Direction: $\theta = \tan^{-1}\left(\frac{18}{24}\right)$ south of east, but the question only asks for speed.

Concept & Intuition

When an element exists as a mixture of isotopes, its average (atomic) mass is not a simple arithmetic mean — it’s a weighted average. Each isotope contributes to the average in proportion to how much of it is present. Think of it like a class grade: if homework is worth 20%, quizzes 30%, and exams 50%, your final grade is $0.2 \times \text{homework} + 0.3 \times \text{quizzes} + 0.5 \times \text{exams}$. Here, the “weights” are the fractional abundances of each isotope. The key is to convert the given ratio into fractions that sum to 1.

Step-by-step solution

1. Understand the ratio

   The relative abundances of isotopes A, B, and C are given as $2 : 3 : 5$. This means that for every $2 + 3 + 5 = 10$ atoms of the element, 2 are of type A, 3 of type B, and 5 of type C.

2. Convert the ratio to fractional abundances

   The fraction of isotope A is $\frac{2}{10} = 0.2$, of B is $\frac{3}{10} = 0.3$, and of C is $\frac{5}{10} = 0.5$.

   These fractions add up to $0.2 + 0.3 + 0.5 = 1$, as they must.

3. Apply the weighted average formula

   The average mass $\bar{m}$ is the sum of each isotope’s mass multiplied by its fractional abundance:

4. Match with the options This expression is exactly option (A). Option (B) uses the raw ratio numbers (2, 3, 5) without dividing by the total, which would give a sum larger than any individual mass — clearly wrong. Option (C) uses 0.4, 0.6, and 1, which don’t sum to 1. Option (D) is just 2 times option (B), also incorrect.

Taking the new balance point at $35$ cm as the pivot, two opposing torques act.

Coins (total mass $= 2 \times 9 = 18$ g) at the 10 cm mark:

Weight of the scale acting at its centre of mass (50 cm mark):

Balancing the torques about the pivot (let $M$ be the mass of the scale):

The centre of mass velocity is simply the total momentum divided by the total mass. Since momentum is a vector, we treat each component separately — this is the cleanest way to avoid sign errors.

1. Write the velocities clearly

   $\vec{v}_A = \hat{i} + 7\hat{j}$

   $\vec{v}_B = -6\hat{i} + \hat{j}$ (because $\hat{j} - 6\hat{i} = -6\hat{i} + \hat{j}$)

2. Find total mass

   $m_A = 100\ \text{g},\quad m_B = 300\ \text{g}$

   $M = 100 + 300 = 400\ \text{g}$

3. Compute total momentum (mass × velocity) for each component

   • For $\hat{i}$-component: $p_{i} = (100)(1) + (300)(-6) = 100 - 1800 = -1700\ \text{g·m/s}$
   • For $\hat{j}$-component: $p_{j} = (100)(7) + (300)(1) = 700 + 300 = 1000\ \text{g·m/s}$

4. Velocity of centre of mass

   $\vec{v}_{cm} = \frac{\vec{p}}{M} = \frac{-1700}{400}\hat{i} + \frac{1000}{400}\hat{j}$

   Simplify:

   $\vec{v}_{cm} = -\frac{17}{4}\hat{i} + \frac{5}{2}\hat{j}\ \text{m/s}$

Concept and Intuition

The center of mass of a symmetric object lies on its axis of symmetry. For a semicircular plate positioned with its diameter along the x-axis and extending into the positive y-region, the center of mass must lie on the y-axis (by symmetry, $x_{cm} = 0$).

The key is finding how far up the y-axis the center of mass sits. We need to integrate over the area, using the fact that for a uniform plate, the center of mass is the geometric centroid.

Finding the y-coordinate of the Center of Mass

1. Set up the coordinate system

The semicircle has radius $R = 9$ cm, with its diameter along the x-axis from $(-9, 0)$ to $(9, 0)$, and the curved part in the upper half-plane ($y \geq 0$).

2. Use the centroid formula

For a uniform lamina (2D plate), the y-coordinate of the center of mass is:

where the denominator is just the total area.

3. Calculate using polar coordinates

It's easier to use polar coordinates centered at the origin. For the semicircle:

• $r$ ranges from $0$ to $R = 9$
• $\theta$ ranges from $0$ to $\pi$

In polar coordinates: $y = r\sin\theta$ and $dA = r \, dr \, d\theta$

4. Evaluate the denominator (total area)

This confirms the semicircle area formula.

5. Evaluate the numerator

6. Compute the center of mass

7. Substitute $R = 9$ cm

8. Compare with options

The exact answer is $\frac{12}{\pi} \approx 3.82$ cm, but this doesn't match any option exactly. Option (B) gives 4.5 cm, which is the closest approximation among the choices (possibly using $\pi \approx \frac{8}{3}$ or a rounded value).