Motion of Centre of Mass Questions in English

(A) The system consists of two particles $A$ and $B$ moving under a mutual force of attraction.
Since there is no external force acting on the system $(F_{ext} = 0)$,the acceleration of the center of mass is zero $(a_{cm} = 0)$.
This implies that the velocity of the center of mass $(v_{cm})$ remains constant over time.
Initially,both particles are at rest,so the initial velocity of the center of mass is $v_{cm, initial} = 0$.
Therefore,at any subsequent instant,including when the speeds of $A$ and $B$ are $v$ and $2v$ respectively,the velocity of the center of mass remains $v_{cm} = 0$.

(C) The explosion is an internal force event. According to the law of conservation of momentum,internal forces cannot change the motion of the centre of mass of a system.
Since the external force acting on the shell (gravity) remains unchanged,the acceleration of the centre of mass remains $g$ directed downwards.
Therefore,the centre of mass of the shell fragments continues to follow the same parabolic path that the intact shell would have followed if it had not exploded.

(C) Let the mass of man $A$ be $m_A = 40 \, kg$,mass of man $B$ be $m_B = 60 \, kg$,and mass of the plank be $m_P = 40 \, kg$. The length of the plank is $L = 120 \, cm$.
Since the bottom surface is smooth and there are no external horizontal forces,the position of the center of mass of the system remains fixed.
Let the initial position of the center of the plank be the origin $(0, 0)$.
Initially,man $A$ is at $x_A = -60 \, cm$,man $B$ is at $x_B = 0 \, cm$,and the center of the plank is at $x_P = 0 \, cm$.
The center of mass of the system is $X_{CM} = \frac{m_A x_A + m_B x_B + m_P x_P}{m_A + m_B + m_P} = \frac{40(-60) + 60(0) + 40(0)}{40 + 60 + 40} = \frac{-2400}{140} = -\frac{120}{7} \, cm$.
When $A$ meets $B$,let the plank have shifted by a displacement $d$ to the right. The new position of the plank's center is $x_P' = d$.
Since $B$ remains fixed with respect to the ground,$B$ is at $x_B' = 0$. Since $B$ was at the center of the plank,and the plank shifted by $d$,$B$ is now at the right edge of the plank (relative to the plank's frame). Thus,$A$ must also be at the right edge of the plank to meet $B$.
Using the conservation of center of mass: $m_A x_A' + m_B x_B' + m_P x_P' = m_A x_A + m_B x_B + m_P x_P$.
Since $A$ and $B$ meet at the right end of the plank,$x_A' = x_B' = d + 60$.
$40(d + 60) + 60(0) + 40(d) = -2400$.
$40d + 2400 + 40d = -2400 \Rightarrow 80d = -4800 \Rightarrow d = -60 \, cm$.
The meeting point is $x_B' = 0$. The right end of the plank is at $x_P' + 60 = -60 + 60 = 0$. Thus,they meet at the right end of the plank.

(C) The velocity of the centre of mass $(V_{COM})$ is given by the formula:
$V_{COM} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}}$
Given:
$m_{1} = 10\,kg$,$v_{1} = 14\,m/s$
$m_{2} = 4\,kg$,$v_{2} = 0\,m/s$
Substituting the values:
$V_{COM} = \frac{10 \times 14 + 4 \times 0}{10 + 4}$
$V_{COM} = \frac{140}{14} = 10\,m/s$

(D) The acceleration of the center of mass $(a_{CM})$ of a system is given by the formula $a_{CM} = \frac{F_{ext}}{M_{total}}$.
When both balls are in the air,the only external force acting on each ball is the gravitational force,which is $mg$ acting downwards.
Let the masses of the two balls be $m_1$ and $m_2$. The total external force is $F_{ext} = m_1g + m_2g = (m_1 + m_2)g$.
The total mass of the system is $M_{total} = m_1 + m_2$.
Therefore,$a_{CM} = \frac{(m_1 + m_2)g}{m_1 + m_2} = g$.
Since the acceleration due to gravity $g$ is constant and independent of the mass,velocity,or time of projection,the acceleration of the center of mass remains $g$ downwards.

(B) Let the acceleration of the system be $a$. Since the string is inextensible,both blocks move with the same magnitude of acceleration $a$.
For the block on the $53^{\circ}$ incline,the equation of motion is: $mg \sin 53^{\circ} - T = ma$.
For the block on the $37^{\circ}$ incline,the equation of motion is: $T - mg \sin 37^{\circ} = ma$.
Adding these two equations: $mg(\sin 53^{\circ} - \sin 37^{\circ}) = 2ma$.
$a = \frac{g(\sin 53^{\circ} - \sin 37^{\circ})}{2} = \frac{10(0.8 - 0.6)}{2} = \frac{10(0.2)}{2} = 1 \, m/s^2$.
The acceleration of the centre of mass is given by $\vec{a}_{cm} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$.
The acceleration vectors are $\vec{a}_1 = a(\cos 53^{\circ} \hat{i} + \sin 53^{\circ} \hat{j})$ and $\vec{a}_2 = a(-\cos 37^{\circ} \hat{i} + \sin 37^{\circ} \hat{j})$.
Since $\cos 53^{\circ} = \sin 37^{\circ} = 0.6$ and $\sin 53^{\circ} = \cos 37^{\circ} = 0.8$,we have $\vec{a}_1 = a(0.6 \hat{i} + 0.8 \hat{j})$ and $\vec{a}_2 = a(-0.8 \hat{i} + 0.6 \hat{j})$.
$\vec{a}_{cm} = \frac{a}{2} [(0.6 - 0.8) \hat{i} + (0.8 + 0.6) \hat{j}] = \frac{a}{2} [-0.2 \hat{i} + 1.4 \hat{j}] = a(-0.1 \hat{i} + 0.7 \hat{j})$.
The magnitude is $|\vec{a}_{cm}| = a \sqrt{(-0.1)^2 + (0.7)^2} = 1 \sqrt{0.01 + 0.49} = \sqrt{0.5} = \frac{1}{\sqrt{2}} \, m/s^2$.

(B) Since the surface is frictionless and there are no external horizontal forces,the center of mass of the system remains stationary.
Let the distance moved by the plank be $x$ in the direction opposite to the man's motion.
The man moves a distance $(\ell - x)$ relative to the ground.
Using the principle of conservation of the center of mass: $M(\ell - x) = (3M)x$.
Dividing both sides by $M$: $\ell - x = 3x$.
Solving for $x$: $4x = \ell$,which gives $x = \frac{\ell}{4}$.
The distance moved by the man relative to the ground is $\ell - x = \ell - \frac{\ell}{4} = \frac{3\ell}{4}$.

(B) According to the principle of conservation of linear momentum,if the external force acting on a system is zero,the velocity of the centre of mass remains constant.
In this problem,the shell is in free space,meaning there are no external forces acting on it $(F_{ext} = 0)$.
Therefore,the velocity of the centre of mass after the explosion will be the same as the velocity of the shell before the explosion.
Given initial velocity $v = 60\,m/s$.
Thus,$v_{cm} = 60\,m/s$.

(B) Since the system is on a smooth horizontal plane,there is no external horizontal force acting on the system. Therefore,the centre of mass of the system remains stationary,i.e.,$\Delta X_{cm} = 0$.
Let the plank move by a distance $x$ towards the right when $A$ and $C$ interchange their positions.
Let the initial positions of $A, B, C$ and the plank be $x_A = -2\,m, x_B = 0\,m, x_C = 2\,m$ and $x_P = 0\,m$ respectively.
When $A$ and $C$ interchange positions,their new positions relative to the plank are $x'_A = 2\,m$ and $x'_C = -2\,m$. The plank has shifted by $x$ towards the right.
The displacement of each mass relative to the ground is:
$\Delta x_A = (2 + x) - (-2) = 4 + x$
$\Delta x_B = x - 0 = x$
$\Delta x_C = (-2 + x) - 2 = -4 + x$
$\Delta x_P = x - 0 = x$
Using the formula $m_A \Delta x_A + m_B \Delta x_B + m_C \Delta x_C + m_P \Delta x_P = 0$:
$40(4 + x) + 50(x) + 60(-4 + x) + 90(x) = 0$
$160 + 40x + 50x - 240 + 60x + 90x = 0$
$240x - 80 = 0$
$x = 80/240 = 1/3\,m$.
Since $x$ is positive,the plank (and thus $B$ which is on the plank) shifts $1/3\,m$ towards the right.

(B) Let the acceleration of the system be $a$. Since the blocks are connected by a string,they move with the same magnitude of acceleration $a$.
Applying Newton's second law to the system:
$T - mg \sin 37^{\circ} = ma$
$mg \sin 53^{\circ} - T = ma$
Adding these two equations,we get:
$mg(\sin 53^{\circ} - \sin 37^{\circ}) = 2ma$
$a = \frac{g(\sin 53^{\circ} - \sin 37^{\circ})}{2} = \frac{10(0.8 - 0.6)}{2} = \frac{10(0.2)}{2} = 1 \, m/s^2$.
The acceleration of the first block is $\vec{a}_1 = a \angle 53^{\circ}$ (up the incline) and the second block is $\vec{a}_2 = a \angle 37^{\circ}$ (down the incline).
The acceleration of the centre of mass is $\vec{a}_{cm} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$.
The magnitude is $a_{cm} = \frac{1}{2} \sqrt{a^2 + a^2 + 2a^2 \cos(180^{\circ} - (53^{\circ} + 37^{\circ}))} = \frac{1}{2} \sqrt{2a^2 + 2a^2 \cos(90^{\circ})} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Substituting $a = 1 \, m/s^2$,we get $a_{cm} = \frac{1}{\sqrt{2}} \, m/s^2$.

(B) Given:
Mass $m_1 = 1 \text{ kg}$,velocity $\vec{v}_1 = 2\hat{i} \text{ m/s}$.
Mass $m_2 = 2 \text{ kg}$,velocity $\vec{v}_2 = 2 \cos(30^{\circ})\hat{i} - 2 \sin(30^{\circ})\hat{j} \text{ m/s}$.
The velocity of the centre of mass $\vec{v}_{CM}$ is given by:
$\vec{v}_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$
Substituting the values:
$\vec{v}_{CM} = \frac{1(2\hat{i}) + 2(2 \cos 30^{\circ}\hat{i} - 2 \sin 30^{\circ}\hat{j})}{1 + 2}$
Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\sin 30^{\circ} = \frac{1}{2}$:
$\vec{v}_{CM} = \frac{2\hat{i} + 2(2 \cdot \frac{\sqrt{3}}{2}\hat{i} - 2 \cdot \frac{1}{2}\hat{j})}{3}$
$\vec{v}_{CM} = \frac{2\hat{i} + 2(\sqrt{3}\hat{i} - 1\hat{j})}{3}$
$\vec{v}_{CM} = \frac{2\hat{i} + 2\sqrt{3}\hat{i} - 2\hat{j}}{3}$
$\vec{v}_{CM} = \left( \frac{2 + 2\sqrt{3}}{3} \right)\hat{i} - \frac{2}{3}\hat{j} \text{ m/s}$.

(A) Since there is no external force acting on the system (balloon + monkey) in the vertical direction,the position of the center of mass $(COM)$ of the system remains unchanged.
Let the balloon descend by a distance $y$. As the monkey climbs the rope of length $L$,the monkey moves upward by a distance $(L - y)$ relative to the ground.
Taking the initial position of the $COM$ as the reference point,the condition for the $COM$ to remain stationary is:
$M \cdot y = m \cdot (L - y)$
$My = mL - my$
$My + my = mL$
$y(M + m) = mL$
$y = \frac{mL}{M + m}$
Therefore,the distance by which the balloon descends is $\frac{mL}{M + m}$.

(D) Since there is no external horizontal force acting on the system,the center of mass of the boy-cart system remains stationary relative to the ground.
Let $m_1 = 20\, kg$ be the mass of the boy and $m_2 = 80\, kg$ be the mass of the cart.
Let the boy move $10\, m$ towards the wall relative to the cart. Let the cart move a distance $x$ away from the wall relative to the ground.
The displacement of the boy relative to the ground towards the wall is $\Delta x_1 = 10 - x$.
The displacement of the cart relative to the ground away from the wall is $\Delta x_2 = -x$.
Using the center of mass displacement formula: $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.
$20(10 - x) + 80(-x) = 0$.
$200 - 20x - 80x = 0$.
$100x = 200 \implies x = 2\, m$.
The net displacement of the boy towards the wall relative to the ground is $\Delta x_1 = 10 - 2 = 8\, m$.
The final distance of the boy from the wall is $25 - 8 = 17\, m$.

(C) Let the point of firing be $O$ and the original range be $OQ = 4\, km$. The highest point is $P$,which is at a horizontal distance of $OP = 2\, km$ from the firing point.
Since the explosion is due to internal forces,the center of mass of the system continues to follow the original parabolic path and hits the ground at point $Q$.
Let the mass of the heavier part be $m_1 = 3M/4$ and the lighter part be $m_2 = M/4$.
The heavier part falls vertically from $P$,so its horizontal position is $x_1 = OP = 2\, km$.
Let the horizontal position of the lighter part be $x_2 = OR$.
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 the values:
$4 = \frac{(3M/4) \times 2 + (M/4) \times x_2}{M}$
$4 = \frac{3}{2} + \frac{x_2}{4}$
$4 - 1.5 = \frac{x_2}{4}$
$2.5 = \frac{x_2}{4}$
$x_2 = 10\, km$.
Thus,the horizontal range of the lighter part is $10\, km$.

(C) Let the mass of the wood be $m_1 = 0.03\, kg$ and the mass of the bullet be $m_2 = 0.02\, kg$. The total mass is $M = m_1 + m_2 = 0.05\, kg$.
At the instant of collision,the wood has fallen from the top $(y = 100\, m)$ and the bullet has risen from the ground $(y = 0)$. Since they are dropped/fired at the same time,they collide when the bullet reaches the wood. However,the problem implies the center of mass $(COM)$ approach.
The position of the center of mass from the ground is $Y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \frac{0.03 \times 100 + 0.02 \times 0}{0.05} = \frac{3}{0.05} = 60\, m$.
The velocity of the center of mass is $V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{0.03 \times 0 + 0.02 \times 100}{0.05} = \frac{2}{0.05} = 40\, m/s$.
After the collision,the system moves as a single particle with initial velocity $V_{cm} = 40\, m/s$ at a height of $60\, m$.
The maximum height $H_{max}$ reached by the center of mass from the ground is given by $H_{max} = Y_{cm} + \frac{V_{cm}^2}{2g} = 60 + \frac{40^2}{2 \times 10} = 60 + \frac{1600}{20} = 60 + 80 = 140\, m$.
The height above the top of the building is $140\, m - 100\, m = 40\, m$.

(A) The accelerations of the particles are given by:
$\vec{a}_A = -a\hat{i}$
$\vec{a}_B = a\hat{j}$
$\vec{a}_C = a\hat{i}$
$\vec{a}_D = -a\hat{j}$
The acceleration of the centre of mass is given by:
$\vec{a}_{cm} = \frac{m_A\vec{a}_A + m_B\vec{a}_B + m_C\vec{a}_C + m_D\vec{a}_D}{m_A + m_B + m_C + m_D}$
Substituting the values:
$\vec{a}_{cm} = \frac{m(-a\hat{i}) + 2m(a\hat{j}) + 3m(a\hat{i}) + 4m(-a\hat{j})}{m + 2m + 3m + 4m}$
$\vec{a}_{cm} = \frac{-ma\hat{i} + 2ma\hat{j} + 3ma\hat{i} - 4ma\hat{j}}{10m}$
$\vec{a}_{cm} = \frac{2ma\hat{i} - 2ma\hat{j}}{10m} = \frac{a}{5}\hat{i} - \frac{a}{5}\hat{j} = \frac{a}{5}(\hat{i} - \hat{j})$

(A) The system consists of two blocks $A$ and $B$ connected by a spring. When the system is released,the only forces acting on the blocks are the internal spring forces.
Since there is no external force acting on the system in the horizontal direction,the net external force on the system is zero $(F_{ext} = 0)$.
According to the property of the centre of mass,if the net external force on a system is zero,the acceleration of the centre of mass is zero $(a_{cm} = 0)$.
If the system was initially at rest,the velocity of the centre of mass remains zero,meaning the centre of mass remains stationary.

(B) The system consists of two blocks $A$ and $B$ with masses $m_A = 2\,kg$ and $m_B = 3\,kg$ connected by a spring on a smooth horizontal floor.
Since the floor is smooth,there is no external horizontal force acting on the system.
According to the law of conservation of linear momentum,the velocity of the centre of mass $(v_{cm})$ remains constant if the net external force on the system is zero.
The initial velocity of block $A$ is $v_A = 10\,m/s$ and block $B$ is at rest,$v_B = 0\,m/s$.
The velocity of the centre of mass is given by:
$v_{cm} = \frac{m_A v_A + m_B v_B}{m_A + m_B}$
$v_{cm} = \frac{2 \times 10 + 3 \times 0}{2 + 3} = \frac{20}{5} = 4\,m/s$
Since the net external force is zero,the velocity of the centre of mass will remain $4\,m/s$ at any time $t$,including at $t = 10\,s$.

(C) The velocity of the centre of mass of a system is given by $v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$.
Since the bodies are moving towards each other due to mutual gravitational attraction,there is no external force acting on the system $(F_{ext} = 0)$.
According to the law of conservation of momentum,if the initial velocity of the system is zero,the velocity of the centre of mass remains zero throughout the motion.
Given that the bodies were at rest initially,the initial momentum of the system is $P_i = m_1(0) + m_2(0) = 0$.
Therefore,the velocity of the centre of mass $v_{cm} = 0\,m/s$.

(C) Since there is no external horizontal force acting on the system (man + boat),the center of mass of the system remains stationary.
Let $M = 500\,kg$ be the mass of the boat and $m = 100\,kg$ be the mass of the man.
Let $L = 9\,m$ be the length of the boat.
Let $x$ be the displacement of the boat relative to the water.
When the man moves from one end to the other,his displacement relative to the boat is $L = 9\,m$.
His displacement relative to the water is $(L - x) = (9 - x)$.
Using the principle of conservation of the center of mass: $m(L - x) - Mx = 0$.
Substituting the values: $100(9 - x) - 500x = 0$.
$900 - 100x - 500x = 0$.
$900 = 600x$.
$x = \frac{900}{600} = 1.5\,m$.
Since the boat moves in the direction opposite to the man's motion to keep the center of mass fixed,the displacement is $1.5\,m$ in the direction opposite to the displacement of the man.

(A) For an Atwood machine,the acceleration of the masses is given by $a = \frac{m_1 - m_2}{m_1 + m_2} g$.
Let the downward direction be positive for $m_1$ and the upward direction be positive for $m_2$. Then the acceleration of $m_1$ is $a_1 = a$ (downward) and the acceleration of $m_2$ is $a_2 = -a$ (upward).
The acceleration of the centre of mass $(a_{CM})$ is given by:
$a_{CM} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$
Substituting the values:
$a_{CM} = \frac{m_1(a) + m_2(-a)}{m_1 + m_2} = \frac{(m_1 - m_2)a}{m_1 + m_2}$
Now,substitute $a = \frac{m_1 - m_2}{m_1 + m_2} g$ into the equation:
$a_{CM} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) \left( \frac{m_1 - m_2}{m_1 + m_2} g \right)$
$a_{CM} = \left( \frac{m_1 - m_2}{m_1 + m_2} \right)^2 g$

(D) The position of the centre of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.
For the centre of mass to remain at the same position,the change in the position of the centre of mass must be zero,i.e.,$\Delta R_{cm} = 0$.
Therefore,$m_1 \Delta r_1 + m_2 \Delta r_2 = 0$.
Given that the first particle is moved by a distance $d$ towards the centre of mass,let $\Delta r_1 = d$.
Substituting this into the equation: $m_1 d + m_2 \Delta r_2 = 0$.
Solving for $\Delta r_2$: $\Delta r_2 = -\left( \frac{m_1}{m_2} \right) d$.
The negative sign indicates that the second particle must be moved in the opposite direction to the first particle,also towards the centre of mass,by a distance of $\left( \frac{m_1}{m_2} \right) d$.

(C) Since the only force acting on the system is the mutual gravitational force,there is no external force acting on the system of two masses.
According to the law of conservation of momentum,the center of mass of the system remains at rest if it was initially at rest.
Let $x_1$ be the distance covered by $m_1$ and $x_2$ be the distance covered by $m_2$.
Since the center of mass remains stationary,we have $m_1 x_1 = m_2 x_2$.
Given that $x_1 = x$,we substitute this into the equation:
$m_1 x = m_2 x_2$.
Solving for $x_2$,we get $x_2 = \frac{m_1 x}{m_2}$.

(A) Let the masses be $m_1 = 2m$ and $m_2 = 6m$.
The acceleration of the masses in an Atwood machine is given by $a = \frac{|m_2 - m_1|}{m_1 + m_2} g$.
Substituting the values,$a = \frac{|6m - 2m|}{6m + 2m} g = \frac{4m}{8m} g = \frac{g}{2}$.
The acceleration of the centre of mass is $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.
Here,$a_1 = a$ (upwards) and $a_2 = -a$ (downwards).
So,$a_{cm} = \frac{m_1 a - m_2 a}{m_1 + m_2} = \frac{(2m - 6m) a}{2m + 6m} = \frac{-4m}{8m} a = -\frac{1}{2} a$.
The magnitude of the acceleration of the centre of mass is $|a_{cm}| = \frac{1}{2} a = \frac{1}{2} \times \frac{g}{2} = \frac{g}{4}$.

(C) The system consists of two particles moving under the influence of their internal mutual attraction.
Since there is no external force acting on the system,the net external force $\vec{F}_{ext} = 0$.
According to the properties of the centre of mass,if the net external force on a system is zero,the acceleration of the centre of mass is zero $(\vec{a}_{cm} = 0)$.
This implies that the velocity of the centre of mass remains constant.
Initially,both particles are at rest,so the initial velocity of the centre of mass $\vec{v}_{cm, initial} = 0$.
Therefore,the velocity of the centre of mass at any instant remains $\vec{v}_{cm} = 0$.

(B) The displacement of the centre of mass $(\Delta R_{cm})$ is given by the formula: $\Delta R_{cm} = \frac{m_1 \Delta r_1 + m_2 \Delta r_2}{m_1 + m_2}$.
Given,$m_1 = m$,$\Delta r_1 = x$,$m_2 = m'$,$\Delta r_2 = 0$,and $\Delta R_{cm} = \frac{x}{5}$.
Substituting these values into the equation:
$\frac{x}{5} = \frac{m(x) + m'(0)}{m + m'}$.
$\frac{x}{5} = \frac{mx}{m + m'}$.
Dividing both sides by $x$ (assuming $x \neq 0$):
$\frac{1}{5} = \frac{m}{m + m'}$.
$m + m' = 5m$.
$m' = 4m$.
Therefore,the ratio $\frac{m'}{m} = 4$.

(D) Since there is no external force acting on the system,the center of mass remains stationary. The ratio of the distances moved by the skaters is inversely proportional to their masses (or weights).
Let $m_A$ and $m_B$ be the weights of skaters $A$ and $B$ respectively,where $m_A/m_B = 5/7$.
Let $x_A$ and $x_B$ be the distances moved by $A$ and $B$ respectively. Since they meet,$x_A + x_B = 6\,m$.
From the conservation of momentum,$m_A x_A = m_B x_B$.
Substituting the ratio,$5 x_A = 7 x_B$.
Since $x_B = 6 - x_A$,we have $5 x_A = 7(6 - x_A)$.
$5 x_A = 42 - 7 x_A \implies 12 x_A = 42 \implies x_A = 3.5\,m$.
Then $x_B = 6 - 3.5 = 2.5\,m$.

(B) The particles are moving under mutual attraction,which means the net external force on the system is zero $(F_{ext} = 0)$.
Therefore,the velocity of the centre of mass $(v_{CM})$ remains constant.
Given: $m_1 = 1\,kg$,$x_1 = 2\,m$,$v_1 = 5\,m/s$; $m_2 = 1\,kg$,$x_2 = 8\,m$,$v_2 = -3\,m/s$.
The initial position of the centre of mass is $X_{CM}(0) = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} = \frac{(1)(2) + (1)(8)}{1 + 1} = \frac{10}{2} = 5\,m$.
The velocity of the centre of mass is $v_{CM} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{(1)(5) + (1)(-3)}{1 + 1} = \frac{2}{2} = 1\,m/s$.
The position of the centre of mass at time $t$ is given by $X_{CM}(t) = X_{CM}(0) + v_{CM} \cdot t$.
At $t = 1\,s$,$X_{CM}(1) = 5 + (1)(1) = 6\,m$.

(D) The system consists of two particles $A$ and $B$ moving under a mutual force of attraction.
Since there is no external force acting on the system,the net external force $F_{ext} = 0$.
According to the law of conservation of momentum for the centre of mass,if the net external force is zero,the velocity of the centre of mass remains constant.
Initially,both particles $A$ and $B$ are at rest,so the initial velocity of the centre of mass $v_{cm, initial} = 0$.
Since the velocity of the centre of mass remains constant,the velocity of the centre of mass at any instant,including when the velocities of $A$ and $B$ are $v$ and $2v$ respectively,will be $0$.

(C) The velocity of the center of mass is given by $\overrightarrow{v}_{com} = \frac{m_1 \overrightarrow{v_1} + m_2 \overrightarrow{v_2}}{m_1 + m_2}$.
Since $m_1 = m_2 = m$,we have $\overrightarrow{v}_{com} = \frac{m(\overrightarrow{v_1} + \overrightarrow{v_2})}{2m} = \frac{\overrightarrow{v_1} + \overrightarrow{v_2}}{2} = \frac{2\hat{i} + 2\hat{j}}{2} = (\hat{i} + \hat{j}) \ m/s$.
Similarly,the acceleration of the center of mass is $\overrightarrow{a}_{com} = \frac{m_1 \overrightarrow{a_1} + m_2 \overrightarrow{a_2}}{m_1 + m_2} = \frac{\overrightarrow{a_1} + \overrightarrow{a_2}}{2}$.
Given $\overrightarrow{a_1} = (3\hat{i} + 3\hat{j}) \ m/s^2$ and $\overrightarrow{a_2} = 0$,we get $\overrightarrow{a}_{com} = \frac{3\hat{i} + 3\hat{j}}{2} = 1.5(\hat{i} + \hat{j}) \ m/s^2$.
Since the initial velocity vector $\overrightarrow{v}_{com} = (\hat{i} + \hat{j})$ and the acceleration vector $\overrightarrow{a}_{com} = 1.5(\hat{i} + \hat{j})$ are parallel to each other,the center of mass moves in a straight line.

(C) The system consists of two particles $m_1$ and $m_2$ interacting via an internal force of attraction.
Since there is no external force acting on the system,the net external force $F_{ext} = 0$.
According to the property of the centre of mass,the acceleration of the centre of mass is given by $a_{CM} = F_{ext} / (m_1 + m_2) = 0$.
Since the particles are initially at rest,the initial velocity of the centre of mass $v_{CM} = 0$.
Because $a_{CM} = 0$ and $v_{CM} = 0$,the centre of mass remains at rest throughout the motion.

(V) The speed of the $CM$ remains $V$.
The child is running arbitrarily on a trolley moving with velocity $V$. The running of the child constitutes internal forces within the (trolley + child) system.
According to the law of conservation of momentum,the velocity of the centre of mass $(CM)$ of a system changes only if there is an external force acting on the system.
Since the floor is smooth (frictionless) and there are no other external horizontal forces acting on the system,the net external force on the (trolley + child) system is zero.
Therefore,the velocity of the $CM$ of the system remains unchanged,regardless of the child's motion on the trolley. The speed of the $CM$ remains $V$.

(N/A) rectangular block is shown sliding down an inclined plane $AB$ at an angle $\theta$ without any sidewise movement.
In this motion,the block moves down the plane such that all particles of the body (e.g.,$P_1$ and $P_2$) move together. This means that at any instant of time,every particle of the rigid body has the same velocity vector.
Such motion of a rigid body is defined as pure translational motion.
Therefore,in pure translational motion,at any instant of time,all particles of the body possess the same velocity.

(N/A) $(1)$ To determine the motion of the centre of mass,no knowledge of internal forces of the system of particles is required; only the external forces are needed.
$(2)$ To obtain this equation,we did not need to specify the nature of the system of particles. The system may be a collection of particles with various internal motions,or it may be a rigid body undergoing pure translational motion or a combination of translational and rotational motion.
$(3)$ Regardless of the nature of the system and the motion of its individual particles,the centre of mass always moves according to the equation $M\vec{A} = \vec{F}_{ext}$.

(N/A) As shown in the figure,a projectile following a parabolic path explodes into fragments midway in the air.
The forces leading to the explosion are internal forces,and they contribute nothing to the motion of the centre of mass.
Total external force,which is the force of gravity acting on the body,remains the same before and after the explosion.
Therefore,the centre of mass continues to move under the influence of the external force along the same parabolic trajectory as it would have followed if there were no explosion.

(B) The equation $M\vec{A} = \vec{F}_{ext}$ represents the motion of the center of mass of a system of particles.
Here,$M$ is the total mass of the system,$\vec{A}$ is the acceleration of the center of mass,and $\vec{F}_{ext}$ is the vector sum of all external forces acting on the system.
This is derived from Newton's second law applied to a system of particles,where the internal forces cancel out due to Newton's third law.
Therefore,it is the form of Newton's second law for a system of particles.

Consider a system of $n$ particles. Let $\overrightarrow{r_{1}}, \overrightarrow{r_{2}}, \overrightarrow{r_{3}}, \ldots, \overrightarrow{r_{n}}$ be the position vectors of the particles of masses $m_{1}, m_{2}, m_{3}, \ldots, m_{n}$ respectively with respect to the origin of a coordinate system.
If $\overrightarrow{R}$ is the position vector of the centre of mass,then:
$\overrightarrow{R} = \frac{m_{1} \overrightarrow{r_{1}} + m_{2} \overrightarrow{r_{2}} + \ldots + m_{n} \overrightarrow{r_{n}}}{m_{1} + m_{2} + \ldots + m_{n}}$
Let $M = \sum_{i=1}^{n} m_{i}$ be the total mass of the system. Then:
$M \overrightarrow{R} = m_{1} \overrightarrow{r_{1}} + m_{2} \overrightarrow{r_{2}} + \ldots + m_{n} \overrightarrow{r_{n}} \quad \ldots (1)$
Assuming the mass of the system does not change with time,we differentiate equation $(1)$ with respect to time $t$:
$M \frac{d \overrightarrow{R}}{d t} = m_{1} \frac{d \overrightarrow{r_{1}}}{d t} + m_{2} \frac{d \overrightarrow{r_{2}}}{d t} + \ldots + m_{n} \frac{d \overrightarrow{r_{n}}}{d t}$
Since $\frac{d \overrightarrow{R}}{d t} = \overrightarrow{V}$ (velocity of the centre of mass) and $\frac{d \overrightarrow{r_{i}}}{d t} = \overrightarrow{v_{i}}$ (velocity of the $i$-th particle),we get:
$M \overrightarrow{V} = m_{1} \overrightarrow{v_{1}} + m_{2} \overrightarrow{v_{2}} + \ldots + m_{n} \overrightarrow{v_{n}}$
Therefore,the velocity of the centre of mass is:
$\overrightarrow{V} = \frac{m_{1} \overrightarrow{v_{1}} + m_{2} \overrightarrow{v_{2}} + \ldots + m_{n} \overrightarrow{v_{n}}}{M} = \frac{\sum_{i=1}^{n} m_{i} \overrightarrow{v_{i}}}{\sum_{i=1}^{n} m_{i}}$

(N/A) Let a system consist of $n$ particles with masses $m_1, m_2, \dots, m_n$ moving with velocities $\vec{v}_1, \vec{v}_2, \dots, \vec{v}_n$ respectively.
The total linear momentum $\vec{P}$ of the system is the vector sum of the individual momenta of all particles:
$\vec{P} = \vec{p}_1 + \vec{p}_2 + \dots + \vec{p}_n = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n$ --- $(1)$
The velocity of the centre of mass $\vec{V}$ is defined as:
$\vec{V} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n}{m_1 + m_2 + \dots + m_n}$
Let the total mass of the system be $M = \sum_{i=1}^{n} m_i$. Then:
$\vec{V} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n}{M}$
Rearranging this,we get:
$M \vec{V} = m_1 \vec{v}_1 + m_2 \vec{v}_2 + \dots + m_n \vec{v}_n$ --- $(2)$
Comparing equation $(1)$ and $(2)$,we find:
$\vec{P} = M \vec{V}$
Thus,the total momentum of the system is equal to the product of the total mass of the system and the velocity of the centre of mass.

(N/A) The total linear momentum of a system of particles is given by $\vec{p} = M\vec{v}_{cm}$,where $M$ is the total mass of the system and $\vec{v}_{cm}$ is the velocity of the centre of mass.
Taking the derivative with respect to time $t$ on both sides:
$\frac{d\vec{p}}{dt} = M \frac{d\vec{v}_{cm}}{dt}$
Since the acceleration of the centre of mass is $\vec{a}_{cm} = \frac{d\vec{v}_{cm}}{dt}$,we have:
$\frac{d\vec{p}}{dt} = M\vec{a}_{cm}$
According to the motion of the centre of mass,the net external force acting on the system is $\vec{F}_{ext} = M\vec{a}_{cm}$.
Therefore,substituting this into the equation,we get:
$\frac{d\vec{p}}{dt} = \vec{F}_{ext}$
This is Newton's second law for a system of particles.
Statement: The external force acting on a system of particles is equal to the rate of change of the total linear momentum of the system.

(N/A) If there are no external forces,the centre of mass of a double star system moves like a free particle,as shown in figure $(a)$.
The trajectories of the two stars of equal mass are also shown in figure $(a)$. They appear complex.
If we transform to the centre of mass frame,we find that the two stars are moving in a circular orbit about the centre of mass,which remains at rest. The positions of the stars are diametrically opposite to each other,as shown in figure $(b)$.
In our laboratory frame of reference,the trajectories of the stars are a combination of:
$(i)$ Uniform motion in a straight line of the centre of mass,and
$(ii)$ Circular orbits of the stars about the centre of mass.

(A) The centre of mass $(CM)$ of a system is defined as the point where the total mass of the system is assumed to be concentrated.
In a reference frame attached to the centre of mass,the centre of mass itself is at rest by definition.
Since the velocity of the centre of mass is defined as the rate of change of its position vector,if the position of the $CM$ is constant in this frame,its velocity must be zero.
Therefore,in the centre of mass frame,the velocity of the centre of mass is $0 \ m/s$.

(B) The statement is incorrect.
In a system of particles, the velocity of the centre of mass is defined as the weighted average of the velocities of all individual particles: $V_{cm} = \frac{\sum m_i v_i}{\sum m_i}$.
Individual particles in a system can have different velocities due to internal motion, rotation, or vibration.
For example, in a rotating rigid body, particles at different distances from the axis of rotation have different linear velocities, while the centre of mass may be stationary or moving with a constant velocity.

(B) No,the center of mass will remain stationary.
Since the sphere is stationary and there is no external force acting on the system of gas molecules,the net external force $\vec{F}_{ext} = 0$.
According to the law of conservation of momentum,if the net external force is zero,the velocity of the center of mass remains constant.
Since the sphere is initially at rest,the initial velocity of the center of mass is zero.
Therefore,the center of mass will remain at rest,i.e.,$\vec{v}_{cm} = 0$.

(B) The velocity of the centre of mass $(v_{CM})$ for a system of particles moving in one dimension is given by the formula:
$v_{CM} = \frac{m_1 v_1 + m_2 v_2 + m_3 v_3}{m_1 + m_2 + m_3}$
Given:
$m_1 = 5 \, kg, v_1 = 5 \, m/s$
$m_2 = 4 \, kg, v_2 = 4 \, m/s$
$m_3 = 2 \, kg, v_3 = 2 \, m/s$
Substituting the values into the formula:
$v_{CM} = \frac{(5 \times 5) + (4 \times 4) + (2 \times 2)}{5 + 4 + 2}$
$v_{CM} = \frac{25 + 16 + 4}{11}$
$v_{CM} = \frac{45}{11} \approx 4.09 \, m/s$

(A) The acceleration of the centre of mass of a system depends only on the net external force acting on the system,regardless of the point of application of the force.
According to Newton's second law,the acceleration $a_{cm}$ of the centre of mass is given by $a_{cm} = \frac{F_{net}}{M}$.
Here,the net external force acting on the rod is the applied force $F$.
Therefore,the acceleration of the centre of mass is $a_{cm} = \frac{F}{M}$.
This acceleration is independent of the angle $\theta$ at which the force is applied,as the entire force $F$ contributes to the translational motion of the centre of mass.

(A) The motion of the centre of mass of a system of particles is governed by the net external force acting on the system.
According to Newton's second law for the centre of mass,$F_{ext} = M_{total} \cdot a_{cm}$.
Here,the total mass of the rod is $M$ and the net external force applied is $F$.
Therefore,$F = M \cdot a_{cm}$.
Solving for the acceleration of the centre of mass,we get $a_{cm} = F/M$.

(B) The velocity of the centre of mass $(V_{cm})$ of a system of particles is given by the formula:
$V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$
In this problem,the particles are moving towards each other. Let us define the direction of $m_1$ as positive and $m_2$ as negative (or vice versa) to account for their relative motion.
If $v_1$ is the velocity of $m_1$ in the positive direction,then $v_2$ (moving towards $m_1$) must be in the negative direction,i.e.,$-v_2$.
Substituting these into the formula:
$V_{cm} = \frac{m_1 v_1 + m_2 (-v_2)}{m_1 + m_2} = \frac{m_1 v_1 - m_2 v_2}{m_1 + m_2}$
Thus,the correct option is $B$.

(B) The velocity of the centre of mass $(V_{cm})$ of a system of particles is given by the formula:
$V_{cm} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}$
In this problem,the particles move towards each other. Let the direction of $m_1$ be positive. Then the velocity of $m_1$ is $\vec{v}_1 = v_1$ and the velocity of $m_2$ is $\vec{v}_2 = -v_2$ (since it moves in the opposite direction).
Substituting these values into the formula:
$V_{cm} = \frac{m_1(v_1) + m_2(-v_2)}{m_1 + m_2}$
$V_{cm} = \frac{m_1v_1 - m_2v_2}{m_1 + m_2}$
Thus,the correct option is $B$.

(B) Let the positions of the two particles be $x_1$ and $x_2$ relative to the centre of mass $(CM)$.
By the definition of the centre of mass,$m_1 x_1 = m_2 x_2$.
When the first particle is moved by a distance $x$ towards the $CM$,its new position is $x_1' = x_1 - x$.
To keep the $CM$ at the same position,the second particle must be moved by a distance $x'$ towards the $CM$ such that $m_1(x_1 - x) = m_2(x_2 - x')$.
Expanding this,we get $m_1 x_1 - m_1 x = m_2 x_2 - m_2 x'$.
Since $m_1 x_1 = m_2 x_2$,these terms cancel out,leaving $-m_1 x = -m_2 x'$.
Therefore,$x' = \frac{m_1 x}{m_2}$.

(A) The motion of the centre of mass of a system is governed by the net external force acting on the system,according to Newton's second law for a system of particles: $F_{ext} = M_{total} \cdot a_{cm}$.
Here,the total mass of the rod is $M$ and the net external force applied is $F$.
Therefore,the acceleration of the centre of mass is given by:
$a_{cm} = \frac{F_{ext}}{M_{total}} = \frac{F}{M}$.
This result is independent of the point of application of the force,as the force $F$ acts on the entire mass $M$ to produce translational acceleration of the centre of mass.