Motion of Centre of Mass Questions in English

(A) The position of the centre of mass $(R_{cm})$ is given by $R_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.
For the centre of mass to remain stationary,the change in the position of the centre of mass must be zero: $\Delta R_{cm} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2} = 0$.
This implies $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.
Given that mass $m_1$ is moved towards the centre of mass by distance $d$,we have $\Delta x_1 = d$ (taking the direction towards the centre of mass as positive).
To keep the centre of mass at the same position,mass $m_2$ must move in the opposite direction by distance $d'$,so $\Delta x_2 = -d'$.
Substituting these into the equation: $m_1(d) + m_2(-d') = 0$.
$m_1 d = m_2 d'$.
Therefore,$d' = \frac{m_1}{m_2} d$.

(C) The position of the centre of mass $X_{CM}$ is given by $X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.
To keep the centre of mass unchanged,the change in the position of the centre of mass must be zero,i.e.,$\Delta X_{CM} = 0$.
This implies $\Delta X_{CM} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2} = 0$.
Given $m_1 = 10 \; kg$,$m_2 = 30 \; kg$,and the displacement of the first block $\Delta x_1 = +6 \; cm$ (towards the second block).
Substituting the values: $0 = \frac{10 \times 6 + 30 \times \Delta x_2}{10 + 30}$.
$0 = 60 + 30 \Delta x_2$.
$30 \Delta x_2 = -60$.
$\Delta x_2 = -2 \; cm$.
The negative sign indicates that the $30 \; kg$ block must move in the opposite direction of the $10 \; kg$ block's displacement,which means it must move towards the $10 \; kg$ block by $2 \; cm$.

(D) The initial positions of the masses are $x_1 = 0$ and $x_2 = l$.
The initial position of the centre of mass $(x_{CM})$ at $t=0$ is given by:
$x_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} = \frac{m_1(0) + m_2(l)}{m_1 + m_2} = \frac{m_2 l}{m_1 + m_2}$
The initial velocities of the masses are $v_1 = v_0$ and $v_2 = 0$.
The velocity of the centre of mass $(v_{CM})$ is constant because there is no external force acting on the system:
$v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{m_1 v_0 + m_2(0)}{m_1 + m_2} = \frac{m_1 v_0}{m_1 + m_2}$
At a later time $t$,the position of the centre of mass $(x'_{CM})$ is given by:
$x'_{CM} = x_{CM} + v_{CM} t$
$x'_{CM} = \frac{m_2 l}{m_1 + m_2} + \left( \frac{m_1 v_0}{m_1 + m_2} \right) t$
$x'_{CM} = \frac{m_2 l}{m_1 + m_2} + \frac{m_1 v_0 t}{m_1 + m_2}$
Thus,the correct option is $D$.

(A) The centre of mass of the system is subjected only to the external gravitational force acting downwards.
Internal forces,such as the man throwing the ball or the ball bouncing off the walls,do not affect the motion of the centre of mass of the system.
According to Newton's second law for a system of particles,$F_{ext} = M_{total} \cdot a_{cm}$.
Since the only external force is gravity,the acceleration of the centre of mass is $g$ directed downwards.
Therefore,the centre of mass will move in a straight vertical line downwards,as depicted in Figure $A$.

(D) The explosion force that splits the firecracker is internal to the system,so the path of the centre of mass of the system remains unchanged.
The range of the centre of mass of the system is given by $R = \frac{u^2 \sin 2\theta}{g}$.
Here,$u = 30 \, ms^{-1}$ and the angle with the horizontal is $\theta = 90^{\circ} - 75^{\circ} = 15^{\circ}$.
Therefore,$R = \frac{30 \times 30 \times \sin(2 \times 15^{\circ})}{10} = 90 \times \sin(30^{\circ}) = 90 \times 0.5 = 45 \, m$.
So,the position ($x$-coordinate) of the centre of mass is at $45 \, m$ distance from the origin.
Using the formula for the centre of mass,$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$,where $m_1 = m_2 = m$ and $X_{CM} = 45 \, m$:
$45 = \frac{m(27) + m(x)}{m + m}$
$45 = \frac{27 + x}{2}$
$90 = 27 + x \Rightarrow x = 63 \, m$.
However,the first piece could also fall at $-27 \, m$ (behind the origin) if the explosion is strong enough to reverse its direction:
$45 = \frac{m(-27) + m(x)}{2m}$
$90 = -27 + x \Rightarrow x = 117 \, m$.
Thus,the other piece will fall at $63 \, m$ or $117 \, m$ from the shooting point.

(C) The acceleration of the centre of mass $\vec{a}_{cm}$ for a system of particles is given by the formula:
$\vec{a}_{cm} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2}$
Given that the two particles are identical,let their masses be $m_1 = m_2 = m$.
One particle is at rest,so its acceleration $\vec{a}_1 = 0$.
The other particle has an acceleration $\vec{a}_2 = \vec{f}$.
Substituting these values into the formula:
$\vec{a}_{cm} = \frac{m(0) + m(\vec{f})}{m + m} = \frac{m\vec{f}}{2m} = \frac{\vec{f}}{2}$
Therefore,the acceleration of the centre of mass is $\frac{\vec{f}}{2}$.

(B) The velocity of the centre of mass $(v_{cm})$ of a system is given by the formula: $v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$.
Here,the system is initially at rest,so the initial velocity of the lighter block is $v_2 = 0 \, m/s$.
The heavier block $(m_1 = 5 \, kg)$ is given a velocity $v_1 = 7 \, m/s$.
The total mass of the system is $M = m_1 + m_2 = 5 \, kg + 2 \, kg = 7 \, kg$.
Substituting the values into the formula:
$v_{cm} = \frac{(5 \, kg \times 7 \, m/s) + (2 \, kg \times 0 \, m/s)}{5 \, kg + 2 \, kg}$
$v_{cm} = \frac{35 \, kg \cdot m/s}{7 \, kg}$
$v_{cm} = 5 \, m/s$.

(C) The correct option is $C$.
When a body falls vertically under gravity,the only external force acting on it is the gravitational force,which acts in the vertical downward direction.
According to the law of motion of the centre of mass $(COM)$,the acceleration of the $COM$ is given by $a_{COM} = \frac{F_{ext}}{M}$.
Since there is no external horizontal force acting on the system $(F_{ext, x} = 0)$,the horizontal acceleration of the $COM$ is zero $(a_{COM, x} = 0)$.
Because the initial horizontal velocity of the body is zero,the $COM$ will continue to move only in the vertical direction.
Therefore,the centre of mass of the two parts taken together does not shift horizontally.

(A) Both particles are under the influence of gravity only,so each particle experiences a downward acceleration of $g$.
The acceleration of the centre of mass is given by the formula:
$a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$
Substituting the given values $m_1 = m$,$a_1 = g$ (downwards),$m_2 = 2m$,and $a_2 = g$ (downwards):
$a_{cm} = \frac{m(g) + 2m(g)}{m + 2m}$
$a_{cm} = \frac{3mg}{3m} = g$
Therefore,the acceleration of the centre of mass is $g$ directed downwards.

(C) The system consists of the child and the trolley.
Since the track is smooth,there is no external horizontal force acting on the system (child + trolley).
According to Newton's second law for a system of particles,the acceleration of the centre of mass is given by $a_{com} = \frac{F_{ext}}{M_{total}}$.
Since $F_{ext} = 0$,the acceleration of the centre of mass $a_{com} = 0$.
This implies that the velocity of the centre of mass remains constant.
Initially,the trolley is moving with speed $v$ and the child is standing on it,so the initial velocity of the centre of mass is $v$.
Therefore,even when the child starts running,the centre of mass of the system will continue to move with the same speed $v$.

(B) The system consists of the man and the balloon. There is no net external force acting on the system in the vertical direction (assuming the balloon is in equilibrium in the air).
Initially,the system is at rest,so the velocity of the centre of mass is zero.
Since the net external force is zero,the acceleration of the centre of mass is zero,and the position of the centre of mass remains constant.
Let $y_m$ be the position of the man and $y_b$ be the position of the balloon. The position of the centre of mass $Y_{cm}$ is given by $Y_{cm} = \frac{M y_b + m y_m}{M + m}$.
As the man climbs up,$y_m$ increases. To keep $Y_{cm}$ constant,$y_b$ must decrease.
Therefore,the balloon moves downward.

(A) The acceleration of the centre of mass $(a_{cm})$ for a system of particles is given by the formula:
$a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$
Here,both balls are moving freely under gravity,so the acceleration of each ball is $a_1 = -g$ and $a_2 = -g$ (taking the downward direction as negative).
Substituting the values:
$a_{cm} = \frac{m(-g) + m(-g)}{m + m}$
$a_{cm} = \frac{-2mg}{2m}$
$a_{cm} = -g$
The magnitude of the acceleration of the centre of mass is $g$ directed downwards.

(D) The velocity of the centre of mass $(v_{cm})$ for a system of particles is given by the formula:
$v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$
Given:
Mass of the first block $(m_1)$ = $5 \,kg$
Velocity of the first block $(v_1)$ = $7 \,m/s$
Mass of the second block $(m_2)$ = $2 \,kg$
Velocity of the second block $(v_2)$ = $0 \,m/s$ (initially at rest)
Substituting the values into the formula:
$v_{cm} = \frac{5 \times 7 + 2 \times 0}{5 + 2}$
$v_{cm} = \frac{35 + 0}{7}$
$v_{cm} = \frac{35}{7} = 5 \,m/s$
Therefore,the velocity of the centre of mass is $5 \,m/s$.

(D) Assertion $(A)$ is incorrect because it states that the fragments themselves follow the original path. In reality,only the centre of mass $(C.O.M.)$ of the fragments continues to follow the original parabolic trajectory.
Reason $(R)$ is correct because the explosion is caused by internal chemical forces,and the external force (gravity) remains unchanged during the explosion.
Since the assertion statement is factually incorrect in its phrasing,the correct choice is that $(A)$ is not correct but $(R)$ is correct.

(D) The displacement of the center of mass is given by the formula: $\Delta X_{\text{COM}} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$.
Since the center of mass must remain at its original position,the net displacement of the center of mass is zero,i.e.,$\Delta X_{\text{COM}} = 0$.
Let the displacement of $m_1$ be $\Delta x_1 = +2 \text{ cm}$ (towards the center of mass) and the displacement of $m_2$ be $\Delta x_2 = -x$ (towards the center of mass).
Substituting the values into the equation:
$0 = \frac{3 \times 2 + 2 \times (-x)}{3 + 2}$
$0 = \frac{6 - 2x}{5}$
$6 - 2x = 0$
$2x = 6$
$x = 3 \text{ cm}$.
Therefore,the particle of mass $m_2$ must move by $3 \text{ cm}$ towards the center of mass.

(A) The acceleration of the centre of mass of a system is given by the formula: $\vec{a}_{\text{com}} = \frac{m_1 \vec{a}_1 + m_2 \vec{a}_2}{m_1 + m_2}$.
Since the particles are approaching each other,their acceleration vectors are in opposite directions. Let the direction of the first particle be positive,then $\vec{a}_1 = 2 \ m/s^2$ and $\vec{a}_2 = -4 \ m/s^2$.
Substituting the values: $\vec{a}_{\text{com}} = \frac{(3 \ kg)(2 \ m/s^2) + (6 \ kg)(-4 \ m/s^2)}{3 \ kg + 6 \ kg}$.
$\vec{a}_{\text{com}} = \frac{6 - 24}{9} = \frac{-18}{9} = -2 \ m/s^2$.
The magnitude of the acceleration of the centre of mass is $|\vec{a}_{\text{com}}| = 2 \ m/s^2$.

(C) Let the mass of the man be $m = 100 \ kg$ and the mass of the boat be $M = 500 \ kg$. The length of the boat is $L = 9 \ m$.
Since there is no external horizontal force acting on the system (man + boat),the center of mass of the system remains stationary.
Let $x$ be the displacement of the boat. The man moves a distance $L$ relative to the boat,so his displacement relative to the water is $(L - x)$ in the opposite direction.
Using the center of mass conservation principle: $m(L - x) = Mx$.
$100(9 - x) = 500x$.
$900 - 100x = 500x$.
$600x = 900$.
$x = \frac{900}{600} = 1.5 \ m$.
The boat moves in the direction opposite to the man's motion to keep the center of mass at the same position.

(A) The acceleration of the system is given by $a = \left(\frac{m_2 - m_1}{m_1 + m_2}\right)g = \left(\frac{2 - 1}{2 + 1}\right) \times 10 = \frac{10}{3} \ m/s^2$.
Let the upward direction be positive. The acceleration of the $1 \ kg$ mass is $a_1 = +\frac{10}{3} \ m/s^2$ and the acceleration of the $2 \ kg$ mass is $a_2 = -\frac{10}{3} \ m/s^2$.
The acceleration of the centre of mass is $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.
Substituting the values,$a_{cm} = \frac{1(\frac{10}{3}) + 2(-\frac{10}{3})}{1 + 2} = \frac{\frac{10}{3} - \frac{20}{3}}{3} = \frac{-10/3}{3} = -\frac{10}{9} \ m/s^2$.
The magnitude of acceleration of the centre of mass is $|a_{cm}| = \frac{10}{9} \ m/s^2$.
The distance travelled by the centre of mass in $t = 2 \ s$ starting from rest is $S_{cm} = \frac{1}{2} |a_{cm}| t^2$.
$S_{cm} = \frac{1}{2} \times \frac{10}{9} \times (2)^2 = \frac{1}{2} \times \frac{10}{9} \times 4 = \frac{20}{9} \ m$.

(D) Since there is no external horizontal force on the system (man + plank),the center of mass of the system remains stationary.
Let the plank move a distance $x$ to the left. The man moves a distance $(L - x)$ to the right relative to the ground.
Using the principle of conservation of the center of mass: $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.
Let $m_1 = M$ (man) and $m_2 = 3M$ (plank).
$M(L - x) - 3M(x) = 0$
$L - x - 3x = 0$
$L = 4x$
$x = \frac{L}{4}$ (distance moved by the plank).
The distance moved by the man relative to the ground is $(L - x) = L - \frac{L}{4} = \frac{3L}{4}$.

(C) Let $x_1$ and $x_2$ be the initial distances of particles $m_1$ and $m_2$ from the centre of mass $C$ respectively. For the centre of mass to remain at $C$,the condition is $m_1 x_1 = m_2 x_2$ ... $(i)$
When particle $m_1$ is moved by distance $d$ towards the centre of mass,its new distance from $C$ becomes $(x_1 - d)$.
Let the second particle $m_2$ be moved by a distance $d'$ to keep the centre of mass unchanged. Its new distance from $C$ will be $(x_2 - d')$.
For the centre of mass to remain unchanged,the new condition is $m_1(x_1 - d) = m_2(x_2 - d')$ ... (ii)
Expanding equation (ii): $m_1 x_1 - m_1 d = m_2 x_2 - m_2 d'$
Since $m_1 x_1 = m_2 x_2$ from equation $(i)$,we can substitute this into the expanded equation:
$m_2 x_2 - m_1 d = m_2 x_2 - m_2 d'$
$-m_1 d = -m_2 d'$
$d' = \frac{m_1}{m_2} d$
Since the displacement $d'$ is positive in the direction towards the centre of mass,the second particle must also be moved towards the centre of mass.

(D) Let $x_{1}$ and $x_{2}$ be the distances of masses $m_{1}$ and $m_{2}$ from the centre of mass,respectively.
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 $d$ towards the centre of mass,its new distance becomes $(x_{1} - d)$.
Let the second particle be moved by a distance $D$ to keep the centre of mass unchanged.
For the centre of mass to remain at the same position,the new distances must satisfy: $m_{1}(x_{1} - d) = m_{2}(x_{2} - D)$.
Expanding this,we get $m_{1}x_{1} - m_{1}d = m_{2}x_{2} - m_{2}D$.
Since $m_{1}x_{1} = m_{2}x_{2}$,the equation simplifies to $-m_{1}d = -m_{2}D$.
Thus,$D = \frac{m_{1}}{m_{2}} d$.
Since the first particle moved towards the centre of mass,the second particle must move away from the centre of mass to maintain the balance.

(D) The center of mass velocity is given by $V_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$.
Since the particles are initially at rest,the initial momentum of the system is $P_{initial} = 0$.
There are no external forces acting on the system (only mutual internal attraction),so the net external force $F_{ext} = 0$.
According to the law of conservation of linear momentum,the total momentum of the system remains constant.
Therefore,$P_{final} = P_{initial} = 0$.
Since $P_{final} = M_{total} \times V_{cm}$,and the total mass $M_{total}$ is non-zero,the velocity of the center of mass $V_{cm}$ must be $0$ at all times.

(C) The acceleration of the centre of mass $a_{cm}$ is given by the formula: $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$.
Since both balls are in the air,they are under the influence of gravity alone,so $a_1 = -g$ and $a_2 = -g$.
Substituting these values: $a_{cm} = \frac{m_1(-g) + m_2(-g)}{m_1 + m_2}$.
$a_{cm} = -g \frac{(m_1 + m_2)}{(m_1 + m_2)} = -g$.
Therefore,the magnitude of the acceleration of the centre of mass is equal to $g$,which is the acceleration due to gravity.

(D) For the centre of mass of the system to remain unchanged,the net displacement of the centre of mass must be zero.
Let the masses be $m_A = 10 \ kg$,$m_B = 25 \ kg$,and $m_C = 15 \ kg$.
Let the displacements be $\Delta x_A$,$\Delta x_B$,and $\Delta x_C$.
Taking the direction towards the right as positive:
Block $A$ is moved towards $B$ (right) by $2 \ m$,so $\Delta x_A = +2 \ m$.
Block $C$ is moved towards $B$ (left) by $3 \ m$,so $\Delta x_C = -3 \ m$.
Let block $B$ be moved by $\Delta x_B$.
The condition for the centre of mass to remain unchanged is:
$m_A \Delta x_A + m_B \Delta x_B + m_C \Delta x_C = 0$
Substituting the values:
$(10 \ kg)(2 \ m) + (25 \ kg)(\Delta x_B) + (15 \ kg)(-3 \ m) = 0$
$20 + 25 \Delta x_B - 45 = 0$
$25 \Delta x_B - 25 = 0$
$25 \Delta x_B = 25$
$\Delta x_B = 1 \ m$
Since the result is positive,block $B$ must be moved by $1 \ m$ in the positive direction,which is towards block $C$.

(B) The velocity of the centre of mass is given by the formula: $\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$.
Given: $m_1 = 2 \,kg$,$\vec{v}_1 = 20 \hat{j} \,m \,s^{-1}$ (towards north).
$m_2 = 3 \,kg$,$\vec{v}_2 = 10 \hat{i} \,m \,s^{-1}$ (towards east).
Substituting the values:
$\vec{v}_{cm} = \frac{2(20 \hat{j}) + 3(10 \hat{i})}{2 + 3} = \frac{40 \hat{j} + 30 \hat{i}}{5} = 6 \hat{i} + 8 \hat{j} \,m \,s^{-1}$.
The magnitude of the velocity is $|\vec{v}_{cm}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \,m \,s^{-1}$.

(A) The centre of mass of a system moves according to the net external force acting on the system.
In this problem,the two bodies are moving towards each other due to their mutual gravitational attraction,which is an internal force.
Since there is no external force acting on the system of the two bodies,the net external force $F_{ext} = 0$.
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 $(a_{cm} = 0)$.
Since the bodies are initially at rest,the initial velocity of the centre of mass is $v_{cm, initial} = 0$.
Because the acceleration of the centre of mass is zero,its velocity remains constant over time.
Therefore,the velocity of the centre of mass at any instant,including when the first body attains a velocity $v_0$,will remain $0$.

(B) Let the acceleration of the blocks be $a$. The forces along the incline are $Mg \sin 60^{\circ}$ and $Mg \sin 30^{\circ}$.
The equation of motion is $Mg \sin 60^{\circ} - Mg \sin 30^{\circ} = (M+M)a$.
$a = \frac{g(\sin 60^{\circ} - \sin 30^{\circ})}{2} = \frac{10(\frac{\sqrt{3}}{2} - \frac{1}{2})}{2} = \frac{5(\sqrt{3}-1)}{2} \text{ ms}^{-2}$.
The acceleration of the block on the $60^{\circ}$ incline is $\vec{a}_1 = a(\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j})$ and the block on the $30^{\circ}$ incline is $\vec{a}_2 = a(-\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j})$.
However,using the coordinate system aligned with the inclines as shown in the figure,$\vec{a}_1 = a \hat{i}'$ and $\vec{a}_2 = -a \hat{j}'$ where $\hat{i}'$ and $\hat{j}'$ are unit vectors along the inclines.
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 angle between the two inclines is $90^{\circ}$. Thus,the magnitude is $a_{cm} = \frac{\sqrt{a^2 + a^2}}{2} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
Substituting $a = \frac{5(\sqrt{3}-1)}{2}$,we get $a_{cm} = \frac{5(\sqrt{3}-1)}{2\sqrt{2}}$.

(B) Given: $m_1 = 1 \,g$,$m_2 = 2 \,g$.
Since the particles move towards each other,we assign opposite directions. Let $v_1 = 10 \,ms^{-1}$ and $v_2 = -20 \,ms^{-1}$.
The velocity of the centre of mass $V_{cm}$ is given by the formula:
$V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$
Substituting the values:
$V_{cm} = \frac{(1 \times 10) + (2 \times -20)}{1 + 2}$
$V_{cm} = \frac{10 - 40}{3} = \frac{-30}{3} = -10 \,ms^{-1}$.
The magnitude of the velocity is $10 \,ms^{-1}$.

(A) Let the velocity of the first particle be $\vec{v}_1 = 10 \hat{i} \text{ ms}^{-1}$ and the velocity of the second particle be $\vec{v}_2 = 5 \hat{j} \text{ ms}^{-1}$.
The masses are $m_1 = 10 \text{ g}$ and $m_2 = 15 \text{ g}$.
The velocity of the centre of mass $\vec{v}_{cm}$ is given by:
$\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$
$\vec{v}_{cm} = \frac{10(10 \hat{i}) + 15(5 \hat{j})}{10 + 15} = \frac{100 \hat{i} + 75 \hat{j}}{25} = 4 \hat{i} + 3 \hat{j} \text{ ms}^{-1}$.
The magnitude of the velocity of the centre of mass is:
$|\vec{v}_{cm}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ ms}^{-1}$.

(C) Let the top of the tower be the origin $(0, 25)$ in the $xy$-plane. The masses are $m_1 = 12 \,kg$ and $m_2 = 6 \,kg$.
The initial velocities are $\vec{v}_1 = 15 \hat{j} \,ms^{-1}$ and $\vec{v}_2 = 20 \hat{i} \,ms^{-1}$.
The vertical component of the velocity of the centre of mass is $v_{cm,y} = \frac{m_1 v_{1y} + m_2 v_{2y}}{m_1 + m_2} = \frac{12(15) + 6(0)}{12 + 6} = \frac{180}{18} = 10 \,ms^{-1}$.
The vertical acceleration of the centre of mass is $a_{cm,y} = -g = -10 \,ms^{-2}$.
The height of the centre of mass from the top of the tower is $y_{cm} = v_{cm,y} t - \frac{1}{2} g t^2$.
The maximum height from the top is reached when $v_{cm,y}(t) = 0$,i.e.,$10 - 10t = 0 \implies t = 1 \,s$.
The maximum displacement from the top is $y_{max} = 10(1) - \frac{1}{2}(10)(1)^2 = 10 - 5 = 5 \,m$.
The total height from the ground is $H = 25 + 5 = 30 \,m$.

(D) Consider the system consisting of the mass $m$ and the wedge of mass $M$.
Since the horizontal plane is smooth,there is no external horizontal force acting on the system.
According to the property of the centre of mass,if the net external force on a system in a particular direction is zero,the acceleration of the centre of mass in that direction is zero.
Since the initial velocity of the system is zero,the centre of mass will not move in the horizontal direction.
However,in the vertical direction,the gravitational force acts on the system,which is an external force.
As the mass $m$ slides down the inclined plane,the vertical position of the centre of mass of the system changes (it moves downwards).
Therefore,the centre of mass moves down in the vertical direction and remains unchanged along the horizontal direction.

(D) Let the acceleration of the blocks be $a$. The forces acting along the incline for the block on the $60^{\circ}$ slope is $mg \sin 60^{\circ} - T = ma$ and for the block on the $30^{\circ}$ slope is $T - mg \sin 30^{\circ} = ma$.
Adding these two equations: $mg(\sin 60^{\circ} - \sin 30^{\circ}) = 2ma$.
$a = \frac{g}{2} (\frac{\sqrt{3}}{2} - \frac{1}{2}) = \frac{g(\sqrt{3}-1)}{4}$.
The acceleration vectors for the two blocks are $\vec{a}_1 = a(\cos 60^{\circ} \hat{i} - \sin 60^{\circ} \hat{j})$ and $\vec{a}_2 = a(-\cos 30^{\circ} \hat{i} - \sin 30^{\circ} \hat{j})$.
The acceleration of the centre of mass is $\vec{a}_{CM} = \frac{m\vec{a}_1 + m\vec{a}_2}{2m} = \frac{\vec{a}_1 + \vec{a}_2}{2}$.
$\vec{a}_{CM} = \frac{a}{2} [(\cos 60^{\circ} - \cos 30^{\circ}) \hat{i} - (\sin 60^{\circ} + \sin 30^{\circ}) \hat{j}]$.
Substituting $a = \frac{g(\sqrt{3}-1)}{4}$,$\cos 60^{\circ} = 1/2$,$\cos 30^{\circ} = \sqrt{3}/2$,$\sin 60^{\circ} = \sqrt{3}/2$,$\sin 30^{\circ} = 1/2$:
$|\vec{a}_{CM}| = \frac{a}{2} \sqrt{(\frac{1-\sqrt{3}}{2})^2 + (\frac{\sqrt{3}+1}{2})^2} = \frac{a}{2} \sqrt{\frac{1+3-2\sqrt{3} + 3+1+2\sqrt{3}}{4}} = \frac{a}{2} \sqrt{\frac{8}{4}} = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}$.
$|\vec{a}_{CM}| = \frac{g(\sqrt{3}-1)}{4\sqrt{2}}$.
Thus,option $(d)$ is correct.

(A) Let the masses of the particles be $m_1 = m$ and $m_2 = 2m$. The displacement of the centre of mass $\Delta Y_{cm}$ is given by the formula:
$\Delta Y_{cm} = \frac{m_1 \Delta y_1 + m_2 \Delta y_2}{m_1 + m_2}$
Given that $\Delta Y_{cm} = 2 \ cm$,$\Delta y_1 = 9 \ cm$,$m_1 = m$,and $m_2 = 2m$,we substitute these values into the equation:
$2 = \frac{m(9) + 2m(\Delta y_2)}{m + 2m}$
$2 = \frac{9m + 2m(\Delta y_2)}{3m}$
$2 = \frac{9 + 2\Delta y_2}{3}$
$6 = 9 + 2\Delta y_2$
$2\Delta y_2 = 6 - 9$
$2\Delta y_2 = -3$
$\Delta y_2 = -1.5 \ cm$
The negative sign indicates that the heavier particle must be moved $1.5 \ cm$ downward.

(A) The velocity of the centre of mass $(CM)$ is given by the formula $V_{CM} = \frac{m_A v_A + m_B v_B}{m_A + m_B}$.
Since the particles are initially at rest,the initial velocity of the centre of mass is $V_{CM, initial} = 0$.
According to the law of conservation of momentum,if the net external force on a system is zero,the velocity of the centre of mass remains constant.
In this system,the particles move under mutual forces of attraction,which are internal forces.
Therefore,the net external force on the system is zero.
Since the initial velocity of the centre of mass was zero,it must remain zero at all times,regardless of the individual speeds of the particles.

(A) The velocity of the centre of mass $(v_{CM})$ is given by:
$v_{CM} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2} = \frac{\overrightarrow{v}_1 + \overrightarrow{v}_2}{2}$ (since $m_1 = m_2 = m$).
$v_{CM} = \frac{4 \hat{i} + 4 \hat{j}}{2} = (2 \hat{i} + 2 \hat{j}) \text{ ms}^{-1}$.
The acceleration of the centre of mass $(a_{CM})$ is given by:
$a_{CM} = \frac{m_1 \overrightarrow{a}_1 + m_2 \overrightarrow{a}_2}{m_1 + m_2} = \frac{\overrightarrow{a}_1 + 0}{2} = \frac{5 \hat{i} + 5 \hat{j}}{2} = (2.5 \hat{i} + 2.5 \hat{j}) \text{ ms}^{-2}$.
Since the acceleration vector $\overrightarrow{a}_{CM}$ is constant,the centre of mass moves with constant acceleration.
Because the initial velocity $\overrightarrow{v}_{CM}$ and the acceleration $\overrightarrow{a}_{CM}$ are parallel (both are in the direction of $\hat{i} + \hat{j}$),the path of the centre of mass is a straight line.

(A) The acceleration $a$ of the system is given by:
$a = \left( \frac{m_2 - m_1}{m_1 + m_2} \right) g = \left( \frac{2m - m}{2m + m} \right) \times 10 = \frac{10}{3} \ ms^{-2}$
After time $t = 5.4 \ s$,the speed of each block is:
$v = at = \frac{10}{3} \times 5.4 = 18 \ ms^{-1}$
The block of mass $m$ moves upwards with velocity $v_1 = 18 \ ms^{-1}$ (let this be positive direction),and the block of mass $2m$ moves downwards with velocity $v_2 = -18 \ ms^{-1}$.
The velocity of the centre of mass is:
$v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{m(18) + 2m(-18)}{m + 2m} = \frac{18m - 36m}{3m} = \frac{-18m}{3m} = -6 \ ms^{-1}$
The speed is the magnitude of the velocity,which is $6 \ ms^{-1}$.

(D) Let $M = 90 \ kg$ be the mass of the plank and $m = 20 \ kg$ be the mass of the girl.
The length of the plank is $l = 3.3 \ m$.
Since there is no external horizontal force acting on the system (plank + girl),the center of mass of the system remains stationary.
Let the displacement of the plank be $\Delta x$ in the direction opposite to the girl's motion.
The displacement of the girl relative to the water is $(l - \Delta x)$.
Using the center of mass principle: $M \Delta x = m(l - \Delta x)$.
$90 \Delta x = 20(3.3 - \Delta x)$.
$90 \Delta x = 66 - 20 \Delta x$.
$110 \Delta x = 66$.
$\Delta x = \frac{66}{110} = 0.6 \ m$.
Converting to centimeters: $0.6 \ m = 60 \ cm$.

(B) Given masses are $m_1 = m$, $m_2 = 2m$, and $m_3 = 3m$.
The velocities are directed as follows:
Particle $A$ moves north: $\vec{v}_1 = 6 \hat{j} \,ms^{-1}$
Particle $B$ moves south: $\vec{v}_2 = -12 \hat{j} \,ms^{-1}$
Particle $C$ moves east: $\vec{v}_3 = 8 \hat{i} \,ms^{-1}$
The velocity of the centre of mass $\vec{v}_{cm}$ is given by:
$\vec{v}_{cm} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2 + m_3 \vec{v}_3}{m_1 + m_2 + m_3}$
Substituting the values:
$\vec{v}_{cm} = \frac{m(6 \hat{j}) + 2m(-12 \hat{j}) + 3m(8 \hat{i})}{m + 2m + 3m}$
$\vec{v}_{cm} = \frac{6m \hat{j} - 24m \hat{j} + 24m \hat{i}}{6m}$
$\vec{v}_{cm} = \frac{24m \hat{i} - 18m \hat{j}}{6m} = 4 \hat{i} - 3 \hat{j} \,ms^{-1}$
The magnitude of the velocity of the centre of mass is:
$v_{cm} = \sqrt{(4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \,ms^{-1}$

(D) Let $x_1$ and $x_2$ be the initial distances of $m_1$ and $m_2$ from the centre of mass,respectively.
By the definition of the centre of mass,$m_1 x_1 = m_2 x_2$.
If $m_1$ is moved towards the centre of mass by a distance $d$,its new distance from the centre of mass becomes $(x_1 - d)$.
Let the second particle $m_2$ be moved by a distance $s$ towards the centre of mass to keep the centre of mass at the same position.
Its new distance from the centre of mass becomes $(x_2 - s)$.
For the centre of mass to remain at the same position,the new condition must satisfy $m_1(x_1 - d) = m_2(x_2 - s)$.
Expanding this,we get $m_1 x_1 - m_1 d = m_2 x_2 - m_2 s$.
Since $m_1 x_1 = m_2 x_2$,we can subtract these terms from both sides to get $-m_1 d = -m_2 s$.
Solving for $s$,we get $s = \frac{m_1}{m_2} d$.

(C) 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,$m_1 \Delta r_1 + m_2 \Delta r_2 = 0$.
Here,the particle of mass $m_1$ is moved by a distance $d$ towards the centre of mass,so $\Delta r_1 = -d$ (assuming the direction towards the centre of mass is negative).
Substituting this into the equation: $m_1 (-d) + m_2 \Delta r_2 = 0$.
$m_2 \Delta r_2 = m_1 d$.
$\Delta r_2 = \left(\frac{m_1}{m_2}\right) d$.
Thus,the particle of mass $m_2$ must be moved by a distance of $\left(\frac{m_1}{m_2}\right) d$ to keep the centre of mass at the original position.

(C) The velocity of the centre of mass $(v_{CM})$ is given by the formula:
$v_{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$
Substituting the given values:
$v_{CM} = \frac{4 \times 5 \hat{i} + 2 \times 10 \hat{i}}{4 + 2}$
$v_{CM} = \frac{20 \hat{i} + 20 \hat{i}}{6} = \frac{40 \hat{i}}{6} = \frac{20}{3} \hat{i} \text{ m/s}$
The kinetic energy of the centre of mass is calculated as:
$K_{CM} = \frac{1}{2} (m_1 + m_2) v_{CM}^2$
$K_{CM} = \frac{1}{2} \times (4 + 2) \times \left( \frac{20}{3} \right)^2$
$K_{CM} = \frac{1}{2} \times 6 \times \frac{400}{9}$
$K_{CM} = 3 \times \frac{400}{9} = \frac{400}{3} \text{ J}$

(B) Let the masses be $m_1 = 2 \,kg$ and $m_2 = 1 \,kg$. The acceleration of the system is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(2 - 1)g}{2 + 1} = \frac{g}{3}$.
Taking the downward direction as positive for $m_1$ and upward for $m_2$,the accelerations are $a_1 = \frac{g}{3}$ (downward) and $a_2 = -\frac{g}{3}$ (upward).
The acceleration of the centre of mass is $a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \frac{2(g/3) + 1(-g/3)}{2 + 1} = \frac{g/3}{3} = \frac{g}{9}$.
The distance traversed by the centre of mass in $t = 2 \,s$ starting from rest is $S = \frac{1}{2} a_{cm} t^2$.
Substituting the values: $S = \frac{1}{2} \times \frac{10}{9} \times (2)^2 = \frac{1}{2} \times \frac{10}{9} \times 4 = \frac{20}{9} \approx 2.22 \,m$.

(B) Mass of the boat,$M = 750 \ kg$. Mass of the man,$m = 80 \ kg$. Length of the boat,$L = 10 \ m$. Since there is no external horizontal force acting on the system (boat + man),the centre of mass of the system remains stationary. Let the boat displace by a distance $x$ in the direction opposite to the man's motion. The man moves a distance $L$ relative to the boat,so his displacement relative to the water is $(L - x)$. The displacement of the boat's centre of mass relative to the water is $x$ in the opposite direction. Using the principle of conservation of the centre of mass: $m \Delta x_m + M \Delta x_B = 0$. Here,$\Delta x_m = (L - x)$ and $\Delta x_B = -x$. Substituting the values: $80(10 - x) + 750(-x) = 0$. $800 - 80x - 750x = 0$. $830x = 800$. $x = \frac{800}{830} \approx 0.96 \ m$. Thus,the boat moves $0.96 \ m$ in the direction opposite to the man's displacement.

(C) In the absence of any external horizontal force, the position of the Centre of Mass $(COM)$ of the system remains unchanged.
Let the shore be the origin $(x = 0)$.
The mass of the plank is $M = 10 \,kg$ and its length is $L = 10 \,m$. Its $COM$ is at $x_p = 5 \,m$.
The mass of the boy is $m = 30 \,kg$. Initially, he is at the far edge, so his position is $x_b = 10 \,m$.
The initial position of the system's $COM$ is:
$X_{COM} = \frac{M x_p + m x_b}{M + m} = \frac{10 \times 5 + 30 \times 10}{10 + 30} = \frac{50 + 300}{40} = \frac{350}{40} = 8.75 \,m$.
When the boy walks to the near edge, the plank moves by a distance $d$ away from the shore. The new position of the plank's $COM$ is $x_p' = 5 + d$, and the new position of the boy is $x_b' = d$.
Since the $COM$ remains at the same position:
$X_{COM} = \frac{M x_p' + m x_b'}{M + m}$
$8.75 = \frac{10(5 + d) + 30(d)}{40}$
$8.75 \times 40 = 50 + 10d + 30d$
$350 = 50 + 40d$
$300 = 40d$
$d = \frac{300}{40} = 7.5 \,m$.

(A) The velocity of the centre of mass is given by the formula:
$v_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}$
Given: $m_1 = 4 \,kg$,$m_2 = 5 \,kg$. The velocity $\vec{v}_1$ is along the east (x-axis) and $\vec{v}_2$ is along the north (y-axis).
So,$\vec{v}_1 = 5 \hat{i} \,m/s$ and $\vec{v}_2 = 3 \hat{j} \,m/s$.
Substituting the values:
$\vec{v}_{CM} = \frac{4(5 \hat{i}) + 5(3 \hat{j})}{4 + 5}$
$\vec{v}_{CM} = \frac{20 \hat{i} + 15 \hat{j}}{9} = \frac{20}{9} \hat{i} + \frac{15}{9} \hat{j} \,m/s$
The magnitude of the velocity is:
$|v_{CM}| = \sqrt{(\frac{20}{9})^2 + (\frac{15}{9})^2}$
$|v_{CM}| = \sqrt{\frac{400 + 225}{81}} = \sqrt{\frac{625}{81}}$
$|v_{CM}| = \frac{25}{9} \,m/s$

(C) Given,$m_1 = 6 \,kg, m_2 = 4 \,kg$ and $\overrightarrow{v}_1 = 5 \hat{i}-2 \hat{j}+10 \hat{k}, \overrightarrow{v}_2 = 10 \hat{i}-2 \hat{j}+5 \hat{k}$.
The velocity of the centre of mass $\overrightarrow{v}_{cm}$ is given by the formula:
$\overrightarrow{v}_{cm} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2}$
Substituting the values:
$\overrightarrow{v}_{cm} = \frac{6(5 \hat{i}-2 \hat{j}+10 \hat{k}) + 4(10 \hat{i}-2 \hat{j}+5 \hat{k})}{6 + 4}$
$\overrightarrow{v}_{cm} = \frac{(30 \hat{i}-12 \hat{j}+60 \hat{k}) + (40 \hat{i}-8 \hat{j}+20 \hat{k})}{10}$
$\overrightarrow{v}_{cm} = \frac{70 \hat{i}-20 \hat{j}+80 \hat{k}}{10}$
$\overrightarrow{v}_{cm} = 7 \hat{i}-2 \hat{j}+8 \hat{k}$