Concept Card: Centre of Mass of a Two-Particle System
1. Physical Framework & Core Meaning:
The centre of mass (CM) of a system of particles is the unique spatial point that moves as though all the mass of the system were concentrated at that point and all external forces were applied directly to it.
- Internal vs. External Forces: Mutual internal forces between particles ($\vec{F}_{12}$ and $\vec{F}_{21}$) cancel out in pairs in accordance with Newton’s Third Law ($\vec{F}_{12} = -\vec{F}_{21}$). Consequently, internal forces cannot alter the net momentum or the acceleration of the centre of mass:
$$\Sigma \vec{F}_{\text{net}} = \Sigma \vec{F}_{\text{ext}} = M \vec{a}_{\text{cm}}$$ - If $\Sigma \vec{F}_{\text{ext}} = \vec{0}$, the centre of mass moves with constant linear velocity $\vec{v}_{\text{cm}} = \text{constant}$. If initially at rest, it remains permanently at rest.
2. Mathematical Formulations for Two Particles
Consider two particles of masses $m_1$ and $m_2$ located at position vectors $\vec{r}_1$ and $\vec{r}_2$ with respect to an arbitrary origin $O$:
- Vector Position of CM:
$$\mathbf{\vec{r}_{\text{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2}}$$ - Cartesian Coordinates:
$$x_{\text{cm}} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}, \quad y_{\text{cm}} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}, \quad z_{\text{cm}} = \frac{m_1 z_1 + m_2 z_2}{m_1 + m_2}$$ - 1D Separation & The Inverse Mass Ratio Law:
Let two particles be separated by a distance $d$. If the origin is placed at $m_1$ ($x_1 = 0, x_2 = d$):
– Distance of CM from $m_1$: $r_1 = \left(\frac{m_2}{m_1 + m_2}\right)d$
– Distance of CM from $m_2$: $r_2 = \left(\frac{m_1}{m_1 + m_2}\right)d$
– Inverse Ratio Rule:
$$\mathbf{\frac{r_1}{r_2} = \frac{m_2}{m_1} \iff m_1 r_1 = m_2 r_2}$$
The centre of mass lies on the line segment joining the two masses and divides it internally in the inverse ratio of their masses. It is always located closer to the heavier mass.
3. The Centre of Mass Reference Frame (Zero-Momentum Frame)
- Position vectors of particles relative to the CM: $\vec{r}’_1 = \vec{r}_1 – \vec{r}_{\text{cm}}$ and $\vec{r}’_2 = \vec{r}_2 – \vec{r}_{\text{cm}}$.
- Fundamental Invariant: The first moment of mass about the centre of mass is strictly zero:
$$\mathbf{m_1 \vec{r}’_1 + m_2 \vec{r}’_2 = \vec{0}}$$ - In the CM frame, the total linear momentum is identically zero at all times: $\vec{p}_{\text{cm}} = \vec{0}$.
4. Shift & Displacement Theorems
When the particles undergo small displacements $\Delta \vec{r}_1$ and $\Delta \vec{r}_2$:
$$\Delta \vec{r}_{\text{cm}} = \frac{m_1 \Delta \vec{r}_1 + m_2 \Delta \vec{r}_2}{m_1 + m_2}$$
- Stationary CM Condition ($\Delta \vec{r}_{\text{cm}} = \vec{0}$):
If $\Sigma \vec{F}_{\text{ext}} = \vec{0}$ and the system is released from rest, the centre of mass cannot shift:
$$\mathbf{m_1 \Delta \vec{r}_1 + m_2 \Delta \vec{r}_2 = \vec{0} \implies m_1 \Delta x_1 + m_2 \Delta x_2 = 0 \implies \Delta x_2 = -\frac{m_1}{m_2}\Delta x_1}$$ - Classic applications: A person walking on a free boat or plank, mutual gravitational attraction, expanding/contracting spring-mass systems on smooth floors.
5. Centre of Mass (CM) vs. Centre of Gravity (CG)
- Centre of Mass: Point determined solely by the mass distribution ($\vec{r}_{\text{cm}} = \frac{\Sigma m_i \vec{r}_i}{\Sigma m_i}$).
- Centre of Gravity: Point where the resultant gravitational force acts and about which the net gravitational torque is zero ($\vec{r}_{\text{cg}} = \frac{\Sigma m_i g_i \vec{r}_i}{\Sigma m_i g_i}$).
- CM and CG coincide only in a uniform gravitational field ($\vec{g} = \text{constant}$). For massive or vertically extensive bodies in a non-uniform field, CG lies slightly below CM.
6. Common JEE Pitfalls & Traps
- Trap 1 (Material Requirement): The centre of mass does not need to lie within physical material. In hollow spheres, rings, or separated particle pairs, the CM is located in empty space.
- Trap 2 (Vector Displacement Signs): Displacements are vectors. If a person walks forward ($+x$), the plank shifts backward ($-x$).
- Trap 3 (Relative vs. Ground Displacement): When a person of mass $m$ walks distance $L$ relative to a floating boat of mass $M$, the boat shifts backward by $x_b$ relative to the water, so the person’s true ground displacement is $L – x_b$, giving $x_b = \frac{m}{m + M} L$.
Solved Examples
Example 1 (Direct Conceptual Application – 3D Vector Coordinates & Separation):
Two particles of masses $m_1 = 3.0\text{ kg}$ and $m_2 = 1.0\text{ kg}$ are located in space at position vectors:
$$\vec{r}_1 = (2.0\hat{i} + 4.0\hat{j} – \hat{k})\text{ m}, \qquad \vec{r}_2 = (-2.0\hat{i} + 8.0\hat{j} + 7.0\hat{k})\text{ m}$$
(a) Determine the position vector $\vec{r}_{\text{cm}}$ of the centre of mass.
(b) Find the distance of the centre of mass from particle 1 ($r_1$) and from particle 2 ($r_2$).
(c) Verify the inverse mass ratio relation $m_1 r_1 = m_2 r_2$.
Solution:
(a) Position vector of the centre of mass:
$$\vec{r}_{\text{cm}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} = \frac{3.0(2.0\hat{i} + 4.0\hat{j} – \hat{k}) + 1.0(-2.0\hat{i} + 8.0\hat{j} + 7.0\hat{k})}{3.0 + 1.0}$$
Numerator:
$$m_1 \vec{r}_1 + m_2 \vec{r}_2 = (6.0\hat{i} + 12.0\hat{j} – 3.0\hat{k}) + (-2.0\hat{i} + 8.0\hat{j} + 7.0\hat{k}) = (4.0\hat{i} + 20.0\hat{j} + 4.0\hat{k})\text{ kg}\cdot\text{m}$$
Dividing by total mass $M = 4.0\text{ kg}$:
$$\vec{r}_{\text{cm}} = \frac{4.0\hat{i} + 20.0\hat{j} + 4.0\hat{k}}{4.0} = (\hat{i} + 5.0\hat{j} + \hat{k})\text{ m}$$
(b) Displacement vector from particle 1 to the centre of mass:
$$\vec{r}_{\text{cm}} – \vec{r}_1 = (1.0 – 2.0)\hat{i} + (5.0 – 4.0)\hat{j} + (1.0 – (-1.0))\hat{k} = -\hat{i} + \hat{j} + 2.0\hat{k}\text{ m}$$
Distance:
$$r_1 = |\vec{r}_{\text{cm}} – \vec{r}_1| = \sqrt{(-1.0)^2 + 1.0^2 + 2.0^2} = \sqrt{1 + 1 + 4} = \sqrt{6}\text{ m} \approx 2.449\text{ m}$$
Displacement vector from particle 2 to the centre of mass:
$$\vec{r}_{\text{cm}} – \vec{r}_2 = (1.0 – (-2.0))\hat{i} + (5.0 – 8.0)\hat{j} + (1.0 – 7.0)\hat{k} = 3.0\hat{i} – 3.0\hat{j} – 6.0\hat{k}\text{ m}$$
Distance:
$$r_2 = |\vec{r}_{\text{cm}} – \vec{r}_2| = \sqrt{3.0^2 + (-3.0)^2 + (-6.0)^2} = \sqrt{9 + 9 + 36} = \sqrt{54} = 3\sqrt{6}\text{ m} \approx 7.348\text{ m}$$
(c) Checking the inverse ratio:
$$m_1 r_1 = 3.0 \times \sqrt{6} = 3\sqrt{6}\text{ kg}\cdot\text{m}$$
$$m_2 r_2 = 1.0 \times 3\sqrt{6} = 3\sqrt{6}\text{ kg}\cdot\text{m}$$
$$m_1 r_1 = m_2 r_2 \implies \frac{r_1}{r_2} = \frac{1}{3} = \frac{m_2}{m_1}$$
The centre of mass divides the line segment internally in the exact inverse ratio of the masses.
Example 2 (Mathematical Manipulation – Diatomic Molecule & Reduced Mass):
In a carbon monoxide ($\text{CO}$) molecule, the internuclear bond length between the carbon atom ($m_{\text{C}} = 12.0\text{ u}$) and the oxygen atom ($m_{\text{O}} = 16.0\text{ u}$) is $d = 1.13\text{ \AA} = 1.13 \times 10^{-10}\text{ m}$.
(a) Locate the centre of mass of the molecule relative to the carbon nucleus.
(b) Prove that the moment of inertia $I$ of the molecule about an axis passing through its centre of mass and perpendicular to the internuclear axis is $I = \mu d^2$, where $\mu = \frac{m_{\text{C}} m_{\text{O}}}{m_{\text{C}} + m_{\text{O}}}$ is the reduced mass.
(c) Calculate the numerical value of $I$ in $\text{kg}\cdot\text{m}^2$ ($1\text{ u} \approx 1.66 \times 10^{-27}\text{ kg}$).
Solution:
(a) Taking the carbon atom as the origin ($x_{\text{C}} = 0$):
$$x_{\text{cm}} = \frac{m_{\text{O}}}{m_{\text{C}} + m_{\text{O}}} d = \frac{16.0}{12.0 + 16.0}(1.13\text{ \AA}) = \frac{16.0}{28.0}(1.13\text{ \AA}) = \frac{4}{7}(1.13\text{ \AA}) \approx 0.646\text{ \AA}$$
The centre of mass is located $0.646\text{ \AA}$ from the carbon nucleus (and $1.13 – 0.646 = 0.484\text{ \AA}$ from the oxygen nucleus).
(b) The distances of the atoms from the centre of mass are:
$$r_{\text{C}} = \frac{m_{\text{O}}}{m_{\text{C}} + m_{\text{O}}} d, \qquad r_{\text{O}} = \frac{m_{\text{C}}}{m_{\text{C}} + m_{\text{O}}} d$$
The moment of inertia about the CM axis:
$$I = m_{\text{C}} r_{\text{C}}^2 + m_{\text{O}} r_{\text{O}}^2 = m_{\text{C}}\left(\frac{m_{\text{O}} d}{m_{\text{C}} + m_{\text{O}}}\right)^2 + m_{\text{O}}\left(\frac{m_{\text{C}} d}{m_{\text{C}} + m_{\text{O}}}\right)^2 = \frac{m_{\text{C}} m_{\text{O}}^2 + m_{\text{O}} m_{\text{C}}^2}{(m_{\text{C}} + m_{\text{O}})^2} d^2$$
Factoring out $m_{\text{C}} m_{\text{O}}$:
$$I = \frac{m_{\text{C}} m_{\text{O}}(m_{\text{C}} + m_{\text{O}})}{(m_{\text{C}} + m_{\text{O}})^2} d^2 = \left(\frac{m_{\text{C}} m_{\text{O}}}{m_{\text{C}} + m_{\text{O}}}\right) d^2 = \mu d^2$$
(c) Numerical calculation:
$$\mu = \frac{12.0 \times 16.0}{28.0}\text{ u} = \frac{48.0}{7.0}\text{ u} \approx 6.857\text{ u} = 6.857 \times 1.66 \times 10^{-27}\text{ kg} \approx 1.138 \times 10^{-26}\text{ kg}$$
$$I = \mu d^2 = (1.138 \times 10^{-26}\text{ kg})(1.13 \times 10^{-10}\text{ m})^2 = (1.138 \times 10^{-26})(1.2769 \times 10^{-20}) \approx 1.45 \times 10^{-46}\text{ kg}\cdot\text{m}^2$$
Example 3 (Standard JEE Advanced Scenario – Person Walking on a Floating Boat):
A person of mass $m = 60.0\text{ kg}$ stands at one end of a boat of mass $M = 140.0\text{ kg}$ and length $L = 5.0\text{ m}$ floating stationary on calm water. Friction between the boat and water is negligible. The person walks from one end of the boat to the other end.
(a) Explain why the centre of mass of the (person + boat) system does not move.
(b) Calculate the displacement of the boat relative to the water / shore.
(c) Calculate the net displacement of the person relative to the water / shore.
Solution:
(a) The water is frictionless, so no external horizontal forces act on the (person + boat) system ($\Sigma F_{x,\text{ext}} = 0$). The force exerted between the person’s feet and the boat deck is purely internal. Since the system was initially at rest, the velocity of the centre of mass remains zero, and its position along the horizontal direction remains strictly fixed: $\Delta x_{\text{cm}} = 0$.
(b) Choosing the direction of the person’s movement as $+x$:
Let the boat shift backward by distance $x_b$ relative to the water (displacement $\Delta x_{\text{boat}} = -x_b$).
The person walks a distance $L$ relative to the boat. Therefore, the person’s displacement relative to the water is:
$$\Delta x_{\text{person}} = L – x_b$$
Applying the stationary centre of mass condition:
$$m \Delta x_{\text{person}} + M \Delta x_{\text{boat}} = 0 \implies m(L – x_b) + M(-x_b) = 0$$
$$m L – (m + M)x_b = 0 \implies x_b = \left(\frac{m}{m + M}\right)L$$
Substituting the given values:
$$x_b = \left(\frac{60.0}{60.0 + 140.0}\right)(5.0) = \left(\frac{60.0}{200.0}\right)(5.0) = 0.30 \times 5.0 = 1.50\text{ meters}$$
The boat shifts backward by $1.50\text{ m}$ relative to the water.
(c) Net displacement of the person relative to the water:
$$\Delta x_{\text{person}} = L – x_b = 5.0 – 1.50 = +3.50\text{ meters}$$
Example 4 (Edge Case – Two Connected Blocks with Spring on a Smooth Surface):
Two blocks of masses $m_1 = 2.0\text{ kg}$ and $m_2 = 4.0\text{ kg}$ are connected by a light horizontal spring of force constant $k = 1200\text{ N/m}$ on a frictionless horizontal floor. Initially, the blocks are at rest and the spring is in its relaxed state. A sudden horizontal impulse imparts an initial velocity $u_1 = 6.0\text{ m/s}$ to block 1 directed directly away from block 2.
(a) Determine the velocity $v_{\text{cm}}$ of the centre of mass of the system.
(b) Calculate the maximum elongation $x_{\text{max}}$ of the spring during the subsequent motion.
(c) Find the velocities of both blocks at the instant of maximum elongation.
Solution:
(a) Since no external horizontal forces act on the system after the impulse:
$$v_{\text{cm}} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{2.0(6.0) + 4.0(0)}{2.0 + 4.0} = \frac{12.0}{6.0} = 2.0\text{ m/s}$$
The centre of mass moves with a constant speed of $2.0\text{ m/s}$ along $+x$.
(b) At maximum elongation of the spring, the relative speed between the two blocks is zero. Both blocks move with the identical velocity $v_{\text{cm}} = 2.0\text{ m/s}$.
Initial mechanical energy of the system:
$$E = K_i = \frac{1}{2}m_1 u_1^2 = \frac{1}{2}(2.0)(6.0)^2 = 36.0\text{ Joules}$$
Kinetic energy at maximum extension:
$$K_{\text{cm}} = \frac{1}{2}(m_1 + m_2)v_{\text{cm}}^2 = \frac{1}{2}(6.0)(2.0)^2 = 12.0\text{ Joules}$$
By conservation of mechanical energy, the deficit in kinetic energy is stored as spring elastic potential energy:
$$\frac{1}{2}k x_{\text{max}}^2 = K_i – K_{\text{cm}} = 36.0 – 12.0 = 24.0\text{ Joules}$$
$$\frac{1}{2}(1200) x_{\text{max}}^2 = 24.0 \implies 600 x_{\text{max}}^2 = 24.0 \implies x_{\text{max}}^2 = \frac{24.0}{600} = 0.04\text{ m}^2$$
$$x_{\text{max}} = \sqrt{0.04} = 0.20\text{ m} = 20.0\text{ cm}$$
(c) At maximum extension, relative velocity is zero, so both block 1 and block 2 have velocities equal to $v_{\text{cm}} = 2.0\text{ m/s}$ in the $+x$-direction.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
The centre of mass of a system of particles:
(A) Always coincides with the geometric centroid
(B) Must necessarily lie within the physical boundary of the system
(C) Depends only on the masses of the particles and their relative positions
(D) Changes its location when the coordinate axes are translated
Problem 2 (JEE Main – Single Correct):
Two particles of masses $1.0\text{ kg}$ and $3.0\text{ kg}$ are separated by a distance of $8.0\text{ m}$. The distance of the centre of mass from the $1.0\text{ kg}$ mass is:
(A) $2.0\text{ m}$
(B) $4.0\text{ m}$
(C) $6.0\text{ m}$
(D) $8.0\text{ m}$
Problem 3 (JEE Main – Single Correct):
In the absence of external forces, two particles initially at rest move toward each other under mutual gravitational attraction. Their centre of mass:
(A) Accelerates toward the heavier mass
(B) Accelerates toward the lighter mass
(C) Remains permanently at rest
(D) Oscillates about the midpoint
Problem 4 (JEE Main – Single Correct):
A boy of mass $40.0\text{ kg}$ stands at one end of a uniform wooden plank of mass $60.0\text{ kg}$ and length $10.0\text{ m}$ resting on frictionless ice. If the boy walks from one end to the other end of the plank, the distance moved by the plank relative to the ice is:
(A) $2.0\text{ m}$
(B) $4.0\text{ m}$
(C) $6.0\text{ m}$
(D) $10.0\text{ m}$
Problem 5 (JEE Main – Single Correct):
Two particles $A$ of mass $m$ and $B$ of mass $2m$ are placed on the $x$-axis at $x = 2.0\text{ m}$ and $x = 8.0\text{ m}$ respectively. Where must a third particle $C$ of mass $3m$ be placed on the $x$-axis so that the centre of mass of the combined three-particle system is located at $x = 5.0\text{ m}$?
(A) $x = 3.0\text{ m}$
(B) $x = 4.0\text{ m}$
(C) $x = 5.0\text{ m}$
(D) $x = 6.0\text{ m}$
Problem 6 (JEE Advanced – One or More Correct):
For any isolated two-particle system with masses $m_1$ and $m_2$ separated by distance $r$:
(A) The position vector of the centre of mass always lies on the line segment connecting the two particles.
(B) The sum of the first mass moments about the centre of mass is identically zero: $m_1 \vec{r}’_1 + m_2 \vec{r}’_2 = \vec{0}$.
(C) The centre of mass is always located closer to the heavier mass.
(D) The total kinetic energy in the laboratory frame satisfies $K_{\text{lab}} = K_{\text{cm}} + K_{\text{rel}}$, where $K_{\text{rel}} = \frac{1}{2}\mu v_{\text{rel}}^2$ and $\mu$ is the reduced mass.
Problem 7 (JEE Advanced – One or More Correct):
Two blocks of masses $m_1$ and $m_2$ are placed on a smooth horizontal surface connected by an ideal compressed spring. The system is released from rest:
(A) The centre of mass remains stationary throughout the motion.
(B) At any instant, the ratio of their speeds is $\frac{v_1}{v_2} = \frac{m_2}{m_1}$.
(C) At any instant, the ratio of their accelerations is $\frac{a_1}{a_2} = \frac{m_2}{m_1}$.
(D) At any instant, the ratio of their kinetic energies is $\frac{K_1}{K_2} = \frac{m_2}{m_1}$.
Problem 8 (JEE Advanced – One or More Correct):
A projectile of mass $2m$ is launched from the ground with velocity $u$ at an angle $\theta$ to the horizontal. At the apex (highest point) of its parabolic trajectory, an internal explosion splits it into two equal fragments of mass $m$ each. Immediately after the explosion, one fragment falls vertically downward from rest:
(A) The other fragment is projected horizontally forward with speed $2u\cos\theta$.
(B) The centre of mass of the system continues to move along the original parabolic trajectory until the first fragment hits the ground.
(C) The centre of mass of the two fragments strikes the ground at horizontal range $R = \frac{u^2 \sin 2\theta}{g}$.
(D) The second fragment hits the ground at a horizontal distance of $1.5 R$ from the point of launch.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
Two bodies of masses $m_1 = 4.0\text{ kg}$ and $m_2 = 6.0\text{ kg}$ are located on the $x$-axis at $x_1 = 0\text{ m}$ and $x_2 = 10.0\text{ m}$. If mass $m_1$ is shifted to the right by $3.0\text{ m}$, by what distance (in meters) must mass $m_2$ be shifted so that the location of the centre of mass remains completely unchanged?
Problem 10 (JEE Main / Advanced – Numerical Value Type):
In a hydrogen chloride ($\text{HCl}$) molecule, the separation between the hydrogen nucleus (mass $1.0\text{ u}$) and the chlorine nucleus (mass $35.0\text{ u}$) is $1.44\text{ \AA}$. Calculate the distance of the centre of mass of the molecule from the hydrogen nucleus in $\text{\AA}$.
Solutions & Explanations
Answer Key Summary:
1. (C) | 2. (C) | 3. (C) | 4. (B) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 2 | 10. 1.4
Solution 1:
The centre of mass is defined as $\vec{r}_{\text{cm}} = \frac{\Sigma m_i \vec{r}_i}{\Sigma m_i}$. It depends solely on the distribution and magnitude of masses. It does not depend on the choice of coordinate axes, need not coincide with the geometric center (unless mass distribution is uniform and symmetric), and can lie in empty space.
Correct Option: (C)
Solution 2:
Let $m_1 = 1.0\text{ kg}, m_2 = 3.0\text{ kg}$, and $d = 8.0\text{ m}$.
Distance from $m_1$:
$$r_1 = \left(\frac{m_2}{m_1 + m_2}\right)d = \left(\frac{3.0}{1.0 + 3.0}\right)(8.0) = \frac{3}{4} \times 8.0 = 6.0\text{ m}$$
Correct Option: (C)
Solution 3:
Mutual gravitational attraction is an internal force pair ($\vec{F}_{12} = -\vec{F}_{21}$). Since $\Sigma \vec{F}_{\text{ext}} = \vec{0}$ and the particles start from rest ($\vec{v}_{\text{cm}} = \vec{0}$), the centre of mass remains permanently at rest at its initial location.
Correct Option: (C)
Solution 4:
Since the ice is frictionless, $\Sigma F_{x,\text{ext}} = 0$, so $\Delta x_{\text{cm}} = 0$.
$$m \Delta x_{\text{boy}} + M \Delta x_{\text{plank}} = 0$$
If the plank shifts backward by distance $x$, the boy’s ground displacement is $L – x$:
$$m(L – x) – M x = 0 \implies x = \left(\frac{m}{m + M}\right)L$$
$$x = \left(\frac{40.0}{40.0 + 60.0}\right)(10.0) = \left(\frac{40.0}{100.0}\right)(10.0) = 4.0\text{ m}$$
Correct Option: (B)
Solution 5:
Formula for centre of mass of three particles:
$$x_{\text{cm}} = \frac{m_A x_A + m_B x_B + m_C x_C}{m_A + m_B + m_C}$$
Given $x_{\text{cm}} = 5.0\text{ m}$:
$$5.0 = \frac{m(2.0) + 2m(8.0) + 3m(x_C)}{m + 2m + 3m} = \frac{2m + 16m + 3m x_C}{6m} = \frac{18 + 3x_C}{6}$$
$$30.0 = 18.0 + 3x_C \implies 3x_C = 12.0 \implies x_C = 4.0\text{ m}$$
Correct Option: (B)
Solution 6:
– (A) True: $\vec{r}_{\text{cm}} = \vec{r}_1 + \frac{m_2}{m_1+m_2}(\vec{r}_2 – \vec{r}_1)$, which defines points along the line connecting 1 and 2.
– (B) True: Definition of CM reference frame ensures $\Sigma m_i \vec{r}’_i = \vec{0}$.
– (C) True: $r_1/r_2 = m_2/m_1 < 1 \implies r_1 < r_2$.
– (D) True: König’s theorem states $K_{\text{lab}} = \frac{1}{2}M v_{\text{cm}}^2 + \frac{1}{2}\mu v_{\text{rel}}^2$.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: $\Sigma \vec{F}_{\text{ext}} = \vec{0}$ and system starts from rest, so CM is stationary.
– (B) True: $m_1 v_1 = m_2 v_2 \implies v_1/v_2 = m_2/m_1$.
– (C) True: Spring force magnitude $F_s$ is identical on both: $a_1 = F_s/m_1, a_2 = F_s/m_2 \implies a_1/a_2 = m_2/m_1$.
– (D) True: $K_1/K_2 = \frac{p_1^2 / (2m_1)}{p_2^2 / (2m_2)} = \frac{m_2}{m_1}$ since $p_1 = p_2$.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: At the apex, projectile velocity is horizontal: $\vec{v}_{\text{apex}} = (u\cos\theta)\hat{i}$. Momentum before explosion: $(2m)(u\cos\theta)\hat{i}$. First fragment has zero horizontal velocity, so: $(2m)(u\cos\theta)\hat{i} = m(0) + m \vec{v}_2 \implies \vec{v}_2 = (2u\cos\theta)\hat{i}$.
– (B) True: The explosion forces are internal; external force is only gravity, so the CM follows the identical original parabola.
– (C) True: Since CM follows the original path, it lands at original range $R$.
– (D) True: First fragment lands at $x_1 = R/2$ (falls vertically from apex). By CM formula: $x_{\text{cm}} = \frac{m x_1 + m x_2}{2m} = R \implies \frac{R/2 + x_2}{2} = R \implies R/2 + x_2 = 2R \implies x_2 = 1.5 R$.
Correct Options: (A, B, C, D)
Solution 9:
For the centre of mass to remain stationary ($\Delta x_{\text{cm}} = 0$):
$$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$$
Given $m_1 = 4.0\text{ kg}, \Delta x_1 = +3.0\text{ m}, m_2 = 6.0\text{ kg}$:
$$4.0(+3.0) + 6.0(\Delta x_2) = 0 \implies 12.0 + 6.0\Delta x_2 = 0 \implies \Delta x_2 = -2.0\text{ m}$$
Mass $m_2$ must be shifted by a distance of $2.0\text{
Mass $m_2$ must be shifted by a distance of $2.0\text{ m}$ to the left.
Correct Answer: 2
Solution 10:
Given $m_{\text{H}} = 1.0\text{ u}, m_{\text{Cl}} = 35.0\text{ u}$, separation $d = 1.44\text{ \AA}$.
Distance from hydrogen nucleus:
$$r_{\text{H}} = \left(\frac{m_{\text{Cl}}}{m_{\text{H}} + m_{\text{Cl}}}\right)d = \left(\frac{35.0}{1.0 + 35.0}\right)(1.44\text{ \AA}) = \left(\frac{35}{36}\right)(1.44) = 35 \times 0.04 = 1.40\text{ \AA}$$
Correct Answer: 1.4