Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact) Questions in English

(A) When the $6^{th}$ ball just leaves the table,$6$ balls are hanging vertically and $4$ balls are on the horizontal table.
The total mass of the system is $M = 10m$.
The driving force is the weight of the $6$ hanging balls,which is $F = 6mg$.
The acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{6mg}{10m} = \frac{3g}{5}$.
To find the tension $T$ between the $7^{th}$ and $8^{th}$ ball,we consider the system of the $3$ balls $(8^{th}, 9^{th}, 10^{th})$ remaining on the table.
The only horizontal force acting on this system is the tension $T$.
Applying Newton's second law to these $3$ balls: $T = (3m)a$.
Substituting the value of $a$: $T = 3m \times \frac{3g}{5} = \frac{9mg}{5}$.
Given $m = 2 \; kg$ and taking $g = 10 \; m/s^2$:
$T = \frac{9 \times 2 \times 10}{5} = \frac{180}{5} = 36 \; N$.

(A) Let the mass of each particle be $m$,the angular speed be $\omega$,and the length of each string be $l$.
For the outer particle (mass $m$ at distance $2l$ from the center),the tension $T_2$ provides the necessary centripetal force:
$T_2 = m(2l)\omega^2 = 2ml\omega^2 \quad \dots (1)$
For the inner particle (mass $m$ at distance $l$ from the center),the net force is the difference between the tensions $T_1$ and $T_2$,which provides the centripetal force:
$T_1 - T_2 = m(l)\omega^2 = ml\omega^2 \quad \dots (2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{T_1 - T_2}{T_2} = \frac{ml\omega^2}{2ml\omega^2} = \frac{1}{2}$
$2(T_1 - T_2) = T_2$
$2T_1 - 2T_2 = T_2$
$2T_1 = 3T_2$
$\frac{T_1}{T_2} = \frac{3}{2}$

(C) The maximum tension the string can bear is $T_{\text{max}} = 50 \,kg \times g = 50 \times 9.8 \,N = 490 \,N$.
Let the mass of the man be $M = 98 \,kg$ and his downward acceleration be $a$.
The forces acting on the man are his weight $Mg$ downwards and the tension $T$ upwards.
According to Newton's second law,$Mg - T = Ma$.
To find the minimum acceleration $a$ for the man to descend without breaking the string,we set $T = T_{\text{max}} = 50 \,g$.
Substituting the values: $98 \,g - 50 \,g = 98 \,a$.
$48 \,g = 98 \,a$.
$a = \frac{48 \times 9.8}{98} = \frac{48}{10} = 4.8 \,m/s^2$.

(C) The total mass of the system is $M_{\text{total}} = 3\; kg + 5\; kg + 7\; kg = 15\; kg$.
The external force applied is $F = 120\; N$.
The acceleration of the system is $a = \frac{F}{M_{\text{total}}} = \frac{120}{15} = 8\; m/s^2$.
For the $7\; kg$ mass,the only horizontal force acting is the tension $T_2$. Thus,$T_2 = m_3 \cdot a = 7 \times 8 = 56\; N$.
For the $5\; kg$ mass,the forces acting are $T_1$ (pulling forward) and $T_2$ (pulling backward). Thus,$T_1 - T_2 = m_2 \cdot a$.
$T_1 - 56 = 5 \times 8 = 40$.
$T_1 = 40 + 56 = 96\; N$.
Therefore,$T_1 = 96\; N$ and $T_2 = 56\; N$.

(A) Let the mass of the man be $m$ and his weight be $w = mg$.
The breaking strength of the rope is given as $T_{\max} = \eta w = \eta mg$.
When the man slides down with an acceleration $a$,the equation of motion is given by Newton's Second Law:
$mg - T = ma$
To prevent the rope from breaking,the tension $T$ must be less than or equal to $T_{\max}$. For the limiting case where the rope is just about to break,we set $T = T_{\max} = \eta mg$.
Substituting this into the equation of motion:
$mg - \eta mg = ma$
Dividing both sides by $m$:
$g(1 - \eta) = a$
Thus,the maximum acceleration the man can have without breaking the rope is $a = g(1 - \eta)$.

(D) The total mass of the rod is $M = 3.0 \,kg$ and total length is $L = 30 \,cm$.
The linear mass density is $\lambda = \frac{M}{L} = \frac{3.0 \,kg}{30 \,cm} = 0.1 \,kg/cm$.
Mass of the $10 \,cm$ part is $m_1 = 0.1 \times 10 = 1.0 \,kg$.
Mass of the $20 \,cm$ part is $m_2 = 0.1 \times 20 = 2.0 \,kg$.
The net force on the system is $F_{\text{net}} = 32 \,N - 20 \,N = 12 \,N$.
The acceleration of the system is $a = \frac{F_{\text{net}}}{M} = \frac{12 \,N}{3.0 \,kg} = 4 \,m/s^2$.
Let $T$ be the force exerted by the $20 \,cm$ part on the $10 \,cm$ part. Considering the $10 \,cm$ part,the forces acting on it are $20 \,N$ (left) and $T$ (right).
Applying Newton's second law: $T - 20 = m_1 a$.
$T - 20 = 1.0 \times 4$.
$T = 20 + 4 = 24 \,N$.

(C) The total mass of the system is $M = 1 \, kg + 2 \, kg = 3 \, kg$.
The external force applied is $F = 6 \, N$.
The acceleration of the system is given by $a = \frac{F}{M} = \frac{6 \, N}{3 \, kg} = 2 \, m/s^2$.
For the $2 \, kg$ block,the only horizontal force acting on it is the normal contact force $N$ exerted by the $1 \, kg$ block. Thus,the net force on the $2 \, kg$ block is $F_{\text{net}, 2kg} = m_2 \cdot a = 2 \, kg \times 2 \, m/s^2 = 4 \, N$.
For the $1 \, kg$ block,the forces acting on it are the applied force $F = 6 \, N$ in the forward direction and the reaction force $N = 4 \, N$ from the $2 \, kg$ block in the backward direction. Thus,the net force on the $1 \, kg$ block is $F_{\text{net}, 1kg} = F - N = 6 \, N - 4 \, N = 2 \, N$.
Therefore,the net forces acting on the $1 \, kg$ and $2 \, kg$ blocks are $2 \, N$ and $4 \, N$,respectively.

(C) The reading of the dynamometer is equal to the tension $T$ in the spring.
First,calculate the acceleration $a$ of the system by applying Newton's second law to the combined system of two blocks:
$F_{\text{net}} = M_{\text{total}} a$
$50 \, N - 30 \, N = (6 \, kg + 4 \, kg) a$
$20 \, N = 10 \, kg \times a$
$a = 2 \, m/s^2$ (directed towards the $6 \, kg$ block).
Now,consider the $4 \, kg$ block. The forces acting on it are the tension $T$ (pulling towards the $6 \, kg$ block) and the $30 \, N$ force (pulling to the right). Since the system accelerates to the left:
$T - 30 \, N = 4 \, kg \times 2 \, m/s^2$
$T - 30 \, N = 8 \, N$
$T = 38 \, N$.
Alternatively,considering the $6 \, kg$ block:
$50 \, N - T = 6 \, kg \times 2 \, m/s^2$
$50 \, N - T = 12 \, N$
$T = 38 \, N$.
Thus,the reading of the dynamometer is $38 \, N$.

(B) The system consists of two blocks of masses $m_1 = 2 \,kg$ and $m_2 = 3 \,kg$ connected by a string on a smooth horizontal surface.
First,calculate the acceleration of the system. The horizontal component of the applied force $F = 10 \,N$ is $F_x = F \cos 60^{\circ} = 10 \times 0.5 = 5 \,N$.
The total mass of the system is $M = m_1 + m_2 = 2 \,kg + 3 \,kg = 5 \,kg$.
Using Newton's second law,$F_x = M a$,we get $5 = 5 \times a$,which gives $a = 1 \,m/s^2$.
Now,consider the free body diagram of the $2 \,kg$ block. The only horizontal force acting on it is the tension $T$ in the string.
Applying Newton's second law to the $2 \,kg$ block: $T = m_1 a = 2 \,kg \times 1 \,m/s^2 = 2 \,N$.
Thus,the tension in the string is $2 \,N$.

(A) First,calculate the acceleration of the system. The total mass is $M = 3 + 2 + 1 = 6\,kg$.
The forces acting along the incline are $F_1 = 60\,N$ (upwards) and $F_2 = 18\,N$ (downwards).
The component of gravity acting downwards for all blocks is $Mg \sin 30^{\circ} = 6 \times 10 \times 0.5 = 30\,N$.
The net force is $F_{\text{net}} = 60 - 18 - 30 = 12\,N$.
The acceleration is $a = \frac{F_{\text{net}}}{M} = \frac{12}{6} = 2\,ms^{-2}$ (upwards).
Now,consider the $1\,kg$ block. Let $N$ be the normal reaction force between the $2\,kg$ and $1\,kg$ blocks.
The forces on the $1\,kg$ block are $N$ (upwards),$F_2 = 18\,N$ (downwards),and $mg \sin 30^{\circ} = 1 \times 10 \times 0.5 = 5\,N$ (downwards).
Applying Newton's second law: $N - 18 - 5 = m \times a \implies N - 23 = 1 \times 2 \implies N = 25\,N$.

(A) The total mass of the system is $M = m_A + m_B + m_C = 5 \text{ kg} + 3 \text{ kg} + 2 \text{ kg} = 10 \text{ kg}$.
Since the surface is smooth,the acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{80 \text{ N}}{10 \text{ kg}} = 8 \text{ m/s}^2$.
For block $A$ $(5 \text{ kg})$,the only horizontal force acting is the tension $T_1$. Therefore,$T_1 = m_A \times a = 5 \text{ kg} \times 8 \text{ m/s}^2 = 40 \text{ N}$.
For block $B$ $(3 \text{ kg})$,the forces acting are $T_2$ (pulling forward) and $T_1$ (pulling backward). Therefore,$T_2 - T_1 = m_B \times a$.
Substituting the values,$T_2 - 40 \text{ N} = 3 \text{ kg} \times 8 \text{ m/s}^2 = 24 \text{ N}$.
Thus,$T_2 = 40 \text{ N} + 24 \text{ N} = 64 \text{ N}$.
Therefore,the tensions $T_1$ and $T_2$ are $40 \text{ N}$ and $64 \text{ N}$ respectively.

(B) Given: Force $F = 10 \,N$, mass of block $A$ $(m_A) = 2 \,kg$, mass of block $B$ $(m_B) = 3 \,kg$.
Since the blocks are in contact and moving together, the acceleration $(a)$ of the system is given by Newton's second law:
$F = (m_A + m_B) a$
$10 = (2 + 3) a$
$10 = 5a$
$a = 2 \,m/s^2$
Now, consider the free body diagram of block $B$. The only horizontal force acting on block $B$ is the force exerted by block $A$ $(F_{AB})$, which causes it to accelerate at $2 \,m/s^2$:
$F_{AB} = m_B \times a$
$F_{AB} = 3 \,kg \times 2 \,m/s^2 = 6 \,N$
Therefore, the force exerted by block $A$ on block $B$ is $6 \,N$.

(C) Let the acceleration of the system be $a$.
For the block on the incline,the equation of motion is: $Mg \sin \theta - T = Ma$
For the block on the horizontal surface,the equation of motion is: $T = Ma$
Adding both equations: $Mg \sin \theta = 2Ma$
Solving for acceleration: $a = \frac{g \sin \theta}{2}$
Substituting $a$ into the second equation: $T = M \left( \frac{g \sin \theta}{2} \right) = \frac{Mg \sin \theta}{2}$

(D) Since all surfaces are smooth,there is no friction between the two blocks $m_1$ and $m_2$.
When a horizontal force $F$ is applied to the lower block $m_1$,it will move with an acceleration $a = \frac{F}{m_1}$.
Because there is no friction between the surfaces,no horizontal force is transferred to the upper block $m_2$.
Therefore,the net horizontal force acting on $m_2$ is zero,and its horizontal acceleration is zero.

(C) Let $M = 20 \ kg$ be the mass of the trolley and $m = 4 \ kg$ be the mass of the hanging block.
The driving force is the weight of the hanging block,$F_{drive} = mg = 4 \times 10 = 40 \ N$.
The normal force on the trolley is $N = Mg = 20 \times 10 = 200 \ N$.
The kinetic friction force acting on the trolley is $f_k = \mu_k N = 0.02 \times 200 = 4 \ N$.
The net force on the system is $F_{net} = F_{drive} - f_k = 40 - 4 = 36 \ N$.
The total mass of the system is $M_{total} = M + m = 20 + 4 = 24 \ kg$.
Using Newton's second law,$a = \frac{F_{net}}{M_{total}} = \frac{36}{24} = 1.5 \ ms^{-2}$.

(B) The forces acting on the lift are the tension $T$ in the cable (upward) and the weight $mg$ (downward).
According to Newton's second law of motion,the net force is $F_{net} = ma$.
For an upward acceleration $a$,the equation of motion is $T - mg = ma$.
Rearranging for tension: $T = m(g + a)$.
Substituting the given values: $m = 200 \ kg$,$g = 10 \ m/s^2$,and $a = 4 \ m/s^2$.
$T = 200 \times (10 + 4)$.
$T = 200 \times 14$.
$T = 2800 \ N$.

(B) Let the mass of the fireman be $m$ and his acceleration be $a$. The weight of the fireman is $mg$.
The breaking strength of the rope is given as $\frac{2}{3}mg$.
When the fireman slides down with acceleration $a$,the tension $T$ in the rope is given by $T = m(g - a)$.
For the rope not to break,the tension $T$ must be less than or equal to the breaking strength.
To find the minimum acceleration $a$,we set the tension equal to the breaking strength:
$m(g - a) = \frac{2}{3}mg$
Dividing both sides by $m$:
$g - a = \frac{2}{3}g$
$a = g - \frac{2}{3}g = \frac{g}{3}$
Therefore,the minimum acceleration is $\frac{g}{3}$.

(C) Given: $m_1 = 6 \ kg$,$m_2 = 4 \ kg$,and $F = 5 \ N$.
Since the blocks are in contact and moving together,they share the same acceleration $a$.
The total mass of the system is $M = m_1 + m_2 = 6 \ kg + 4 \ kg = 10 \ kg$.
Using Newton's second law for the whole system: $F = M \times a$.
$5 \ N = 10 \ kg \times a \implies a = \frac{5}{10} = 0.5 \ m/s^2$.
The force exerted on the lighter block $(m_2)$ is the contact force $F_{12}$ exerted by the $6 \ kg$ block on the $4 \ kg$ block.
Applying Newton's second law to the $4 \ kg$ block: $F_{12} = m_2 \times a$.
$F_{12} = 4 \ kg \times 0.5 \ m/s^2 = 2 \ N$.

(B) The force equation for the elevator moving upward with acceleration $a$ is given by $T_{actual} = m(g+a)$.
For a safe journey,the maximum tension $T$ that the rope can bear must be at least equal to the actual tension,so $T = m(g+a)$.
The stress in the rope is defined as $\sigma = \frac{T}{A}$,where $A$ is the cross-sectional area of the rope.
Assuming the rope is cylindrical with diameter $d$,the area $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$.
Substituting this into the tension equation: $T = \frac{m(g+a)}{\pi d^2 / 4} = \frac{4 m(g+a)}{\pi d^2}$.
Rearranging for $d^2$,we get $d^2 = \frac{4 m(g+a)}{\pi T}$.
Taking the square root of both sides,the minimum diameter is $d = [\frac{4 m(g+a)}{\pi T}]^{1/2}$.

(D) Given: Mass $m = 4 \times 10^{7} \,kg$, Force $F = 5 \times 10^{4} \,N$, Initial velocity $u = 0$, Distance $s = 4 \,m$.
Using Newton's second law, $F = ma$:
$5 \times 10^{4} = 4 \times 10^{7} \times a$
$a = \frac{5 \times 10^{4}}{4 \times 10^{7}} = 1.25 \times 10^{-3} \,ms^{-2}$.
Using the kinematic equation $v^{2} = u^{2} + 2as$:
$v^{2} = 0^{2} + 2 \times (1.25 \times 10^{-3}) \times 4$
$v^{2} = 2 \times 1.25 \times 4 \times 10^{-3} = 10 \times 10^{-3} = 10^{-2} \,m^{2}s^{-2}$.
$v = \sqrt{10^{-2}} = 0.1 \,ms^{-1}$.

(B) When a lift of mass $m$ is ascending with an upward acceleration $a$, the forces acting on the lift are:
$1$. The tension $T$ in the cable acting upwards.
$2$. The weight $mg$ of the lift acting downwards.
According to Newton's second law of motion, the net force $F_{\text{net}}$ is equal to the product of mass and acceleration $(F_{\text{net}} = ma)$.
Since the lift is moving upwards, the tension $T$ must be greater than the weight $mg$.
Therefore, $T - mg = ma$.
Rearranging the equation, we get $T = mg + ma = m(g+a)$.

(B) The system consists of a block of mass $M$ and a rope of mass $m$ being pulled together as a single unit.
Since the surface is smooth,there is no friction.
The total mass of the system is $(M + m)$.
According to Newton's second law of motion,the force $F$ applied to the system is equal to the product of the total mass and the acceleration $a$.
$F = (M + m) a$
Therefore,the acceleration $a$ is given by:
$a = \frac{F}{M + m}$

(D) Let $m_1 = 5 \,kg$ and $m_2 = 3 \,kg$. The system is accelerating upwards with $a = 2 \,m/s^2$.
For the lower mass $m_2$ $(3 \,kg)$:
The forces acting are tension $T_2$ upwards and weight $m_2 g$ downwards.
The equation of motion is: $T_2 - m_2 g = m_2 a$
$T_2 - 3 \times 9.8 = 3 \times 2$
$T_2 - 29.4 = 6$
$T_2 = 35.4 \,N$
For the upper mass $m_1$ $(5 \,kg)$:
The forces acting are tension $T_1$ upwards, and tension $T_2$ and weight $m_1 g$ downwards.
The equation of motion is: $T_1 - T_2 - m_1 g = m_1 a$
$T_1 - 35.4 - 5 \times 9.8 = 5 \times 2$
$T_1 - 35.4 - 49 = 10$
$T_1 - 84.4 = 10$
$T_1 = 94.4 \,N$

(A) Let the mass of the body be $m$. The initial velocity $u = 2 \,m/s$.
When the body is falling freely, it has velocity $v = 2 \,m/s$.
Now, an upward air resistance force $F_{air} = mg$ acts on the body.
The net force on the body is $F_{net} = F_{air} - mg = mg - mg = 0$.
However, if the air resistance is constant and opposes the motion, we must consider the deceleration. If the air resistance is $F = mg$, the net force is $F_{net} = mg - mg = 0$ (terminal velocity).
Assuming the question implies the air resistance is $2mg$ (net upward force $mg$), the deceleration $a = F_{net}/m = (2mg - mg)/m = g = 10 \,m/s^2$.
Using $v^2 = u^2 + 2as$, where $v = 0$, $u = 2 \,m/s$, and $a = -10 \,m/s^2$:
$0 = (2)^2 + 2(-10)s$
$20s = 4$
$s = 4/20 = 0.2 \,m$.

(B) According to the question, the system is in equilibrium. Let $T$ be the tension in the string connected to mass $m$, and $T_1$ be the tension in the string connected to the $1 \,kg$ mass.
For the $1 \,kg$ mass, the tension $T_1 = 1g$.
For the $2 \,kg$ mass on the inclined plane, the forces acting along the incline are the component of its weight $2g \sin 30^{\circ}$ and the tension $T_1$ acting upwards. The net force along the incline is $(2g - T_1) \sin 30^{\circ}$ (assuming the $2 \,kg$ mass is pulled by the tension $T$ up the incline).
For the system to be in equilibrium, the tension $T$ must balance the net force along the incline:
$T = (2g - T_1) \sin 30^{\circ}$
Substituting $T_1 = 1g$:
$T = (2g - 1g) \sin 30^{\circ} = g \sin 30^{\circ} = g \times 0.5 = 0.5g$
For the mass $m$, the tension $T$ must balance its weight:
$T = mg$
Equating the two expressions for $T$:
$mg = 0.5g$
$m = 0.5 \,kg$

(C) Let the acceleration of the system be $a$. The net force acting on the system is $F_{net} = F_2 - F_1 = 7.5t - 4.2t = 3.3t \text{ N}$.
The total mass of the system is $M = m + 2m = 3m$.
Using Newton's second law,$F_{net} = Ma$,we get $3.3t = (3m)a$,so $a = \frac{1.1t}{m}$.
Now,consider the block of mass $2m$. The forces acting on it are $F_2$ in the forward direction and tension $T$ in the backward direction.
Applying Newton's second law to this block: $F_2 - T = (2m)a$.
Substituting the values: $7.5t - T = 2m \left( \frac{1.1t}{m} \right) = 2.2t$.
Thus,$T = 7.5t - 2.2t = 5.3t$.
Given that the tension $T = 10.6 \text{ N}$,we have $5.3t = 10.6$.
Solving for $t$,we get $t = \frac{10.6}{5.3} = 2 \text{ s}$.

(C) Let $m$ be the mass of the person and $T$ be the tension in the rope.
The breaking strength of the rope is given as $T_{max} = \frac{4}{3} mg$.
When the person climbs up with an acceleration $a$,the equation of motion is $T - mg = ma$.
For safe climbing,the tension $T$ must not exceed the breaking strength $T_{max}$.
So,$T_{max} - mg = ma_{max}$.
Substituting $T_{max} = \frac{4}{3} mg$ into the equation:
$\frac{4}{3} mg - mg = ma_{max}$.
$\frac{1}{3} mg = ma_{max}$.
Therefore,$a_{max} = \frac{g}{3}$.

(A) Given:
Mass of block $M = 18.5 \ kg$
Force $F = 40 \ N$
Length of rope $L = 3 \ m$
Linear density of rope $\mu = 0.5 \ kg \ m^{-1}$
Distance $s = 9 \ m$
Mass of the rope $m_r = \mu \times L = 0.5 \times 3 = 1.5 \ kg$
Total mass of the system $M_{total} = M + m_r = 18.5 + 1.5 = 20 \ kg$
Acceleration of the system $a = \frac{F}{M_{total}} = \frac{40}{20} = 2 \ m \ s^{-2}$
Using the equation of motion $s = ut + \frac{1}{2}at^2$,where initial velocity $u = 0$:
$9 = 0 + \frac{1}{2} \times 2 \times t^2$
$9 = t^2$
$t = 3 \ s$

(B) The total mass of the system is $M = 40 \ kg + 60 \ kg = 100 \ kg$.
Applying Newton's second law to the entire system,the acceleration $a$ is given by:
$a = \frac{F}{M} = \frac{1000 \ N}{100 \ kg} = 10 \ m/s^2$.
Now,consider the $40 \ kg$ block. The only horizontal force acting on it is the tension $T$ in the string.
Applying Newton's second law to the $40 \ kg$ block:
$T = m_1 \times a$
$T = 40 \ kg \times 10 \ m/s^2 = 400 \ N$.

(A) The component of the force responsible for horizontal motion is $F \cos \theta$.
So,the acceleration of the bar is $a = \frac{F \cos \theta}{m}$.
Since acceleration $a = v \frac{d v}{d x}$,we have:
$v \frac{d v}{d x} = \frac{F \cos \theta}{m}$
Given $\theta = k x$,we have $d \theta = k d x$,or $d x = \frac{d \theta}{k}$.
Substituting these into the equation:
$v d v = \frac{F \cos \theta}{m} \cdot \frac{d \theta}{k}$
Integrating both sides with initial conditions $v = 0$ at $\theta = 0$:
$\int_{0}^{v} v d v = \frac{F}{m k} \int_{0}^{\theta} \cos \theta d \theta$
$\frac{v^2}{2} = \frac{F}{m k} [\sin \theta]_{0}^{\theta}$
$\frac{v^2}{2} = \frac{F \sin \theta}{m k}$
$v^2 = \frac{2 F \sin \theta}{m k}$
$v = \sqrt{\frac{2 F \sin \theta}{m k}}$

(B) Let $F$ be the upward buoyant force acting on the balloon.
In the first case,the balloon of mass $M$ descends with acceleration $a$. According to Newton's second law:
$M g - F = M a \Rightarrow F = M g - M a$ ...$(i)$
In the second case,a mass $M^{\prime}$ is removed,so the new mass is $(M - M^{\prime})$. The balloon now ascends with acceleration $a$. According to Newton's second law:
$F - (M - M^{\prime}) g = (M - M^{\prime}) a$
Substituting $F$ from Eq. $(i)$:
$(M g - M a) - (M - M^{\prime}) g = (M - M^{\prime}) a$
$M g - M a - M g + M^{\prime} g = M a - M^{\prime} a$
$M^{\prime} g + M^{\prime} a = M a + M a$
$M^{\prime} (g + a) = 2 M a$
$M^{\prime} = \frac{2 M a}{g + a}$

(B) Let $a$ be the common acceleration of the system of blocks when a force $F$ is applied.
According to Newton's second law for the whole system:
$F = (M_P + M_Q) a$
$\Rightarrow a = \frac{F}{M_P + M_Q}$
Now,consider the free body diagram of block $Q$. The only horizontal force acting on block $Q$ is the contact force $R$ exerted by block $P$.
Applying Newton's second law to block $Q$:
$R = M_Q a$
Substituting the value of $a$:
$R = M_Q \left( \frac{F}{M_P + M_Q} \right)$
$R = \frac{M_Q F}{M_P + M_Q}$

(A) Given:
Mass of block $(M) = 48 \ kg$
Linear density of string $(\lambda) = 0.5 \ kg \ m^{-1}$
Length of string $(l) = 4 \ m$
Force applied $(F) = 25 \ N$
Mass of the string $(m_s) = \lambda \times l = 0.5 \times 4 = 2 \ kg$
Total mass of the system $(m_{total}) = M + m_s = 48 + 2 = 50 \ kg$
Net acceleration of the system $(a_{sys}) = \frac{F}{m_{total}} = \frac{25}{50} = 0.5 \ m \ s^{-2}$
Let the tension at the point connecting the end of the string with the block be $T$.
From the Free Body Diagram $(FBD)$ of the block:
$T = M \times a_{sys} = 48 \times 0.5 = 24 \ N$
Hence,the force acting on the block is $24 \ N$.

(B) Let the maximum accelerations of the two boys be $a_1$ and $a_2$ respectively.
The total tension $T$ in the rope must not exceed $60 \,kg$-wt.
The equation of motion for the two boys climbing up is:
$T = m_1(g + a_1) + m_2(g + a_2)$
Given $T = 60 \,kg$-wt,$m_1 = 20 \,kg$,and $m_2 = 30 \,kg$:
$60g = 20(g + a_1) + 30(g + a_2)$
$60g = 20g + 20a_1 + 30g + 30a_2$
$60g = 50g + 20a_1 + 30a_2$
$10g = 20a_1 + 30a_2$
Dividing by $10$:
$g = 2a_1 + 3a_2$
To find the ratio of maximum accelerations,we consider the individual limits for each boy. If only one boy were climbing,the maximum acceleration would be $a = (T/m) - g$. For boy $1$: $a_{1,max} = (60/20)g - g = 2g$. For boy $2$: $a_{2,max} = (60/30)g - g = g$.
However,the question asks for the ratio of accelerations when they climb together. Assuming they climb with the same acceleration $a$ to find the limit,$g = 2a + 3a = 5a \Rightarrow a = g/5$. If we look for the ratio of their individual maximum possible accelerations,it is $2g : g = 2 : 1$. Given the options,the intended ratio is $a_1 : a_2 = 2 : 1$.

(B) Let $m = 200 \text{ g} = 0.2 \text{ kg}$ be the mass of each sphere.
Let $L = 1 \text{ m}$ be the total length of the string.
Since $P$ is the mid-point,the distance $OP = 0.5 \text{ m}$ and $PQ = 0.5 \text{ m}$.
Let $\omega$ be the constant angular speed.
$1$. Tension $T_1$ in the string between $P$ and $Q$ provides the centripetal force for sphere $Q$:
$T_1 = m \cdot r_Q \cdot \omega^2 = m \cdot L \cdot \omega^2 = 0.2 \cdot 1 \cdot \omega^2 = 0.2 \omega^2$.
$2$. Tension $T_2$ in the string between $O$ and $P$ provides the centripetal force for both spheres $P$ and $Q$:
$T_2 = m \cdot r_P \cdot \omega^2 + T_1 = m \cdot (0.5) \cdot \omega^2 + 0.2 \omega^2 = 0.2 \cdot 0.5 \cdot \omega^2 + 0.2 \omega^2 = 0.1 \omega^2 + 0.2 \omega^2 = 0.3 \omega^2$.
$3$. The ratio of the tension between $P$ and $Q$ to that between $P$ and $O$ is:
$\frac{T_1}{T_2} = \frac{0.2 \omega^2}{0.3 \omega^2} = \frac{2}{3}$.

(C) Let $R$ be the upward buoyant force acting on the balloon.
Case $1$: The balloon is moving down with acceleration $a_0$.
The equation of motion is: $Mg - R = Ma_0$ ....$(i)$
Case $2$: After removing mass $m$,the balloon moves up with acceleration $2a_0$.
The equation of motion is: $R - (M - m)g = (M - m)(2a_0)$ ....(ii)
Adding equations $(i)$ and (ii):
$(Mg - R) + (R - (M - m)g) = Ma_0 + (M - m)(2a_0)$
$Mg - Mg + mg = Ma_0 + 2Ma_0 - 2ma_0$
$mg = 3Ma_0 - 2ma_0$
$mg + 2ma_0 = 3Ma_0$
$m(g + 2a_0) = 3Ma_0$
$m = \frac{3Ma_0}{g + 2a_0}$

(A) First,consider the system as a whole. The total mass of the system is $(m_1 + m_2)$.
Since the surface is smooth,the acceleration $(a)$ of the system is given by Newton's second law: $a = \frac{F}{m_1 + m_2}$.
Now,consider the free body diagram of mass $m_1$. The only horizontal force acting on $m_1$ is the tension $(T)$ in the string.
Applying Newton's second law to mass $m_1$: $T = m_1 \times a$.
Substituting the value of acceleration $(a)$: $T = m_1 \times \left(\frac{F}{m_1 + m_2}\right)$.
Therefore,the tension in the string is $T = \left(\frac{m_1}{m_1 + m_2}\right) F$.

(A) First,calculate the mass of block $A$ using Newton's second law: $F = m_A a \Rightarrow 10 = m_A \times 20 \Rightarrow m_A = 0.5 \,kg$.
When block $A$ is placed against block $B$ and a force $F^{\prime} = 20 \,N$ is applied,both blocks move together with a common acceleration $a$.
The equation of motion for block $A$ is: $F^{\prime} - N = m_A a \Rightarrow 20 - N = 0.5 a$ $(i)$.
The equation of motion for block $B$ is: $N = m_B a \Rightarrow N = 1.5 a$ $(ii)$.
Adding equations $(i)$ and $(ii)$: $20 = 2 a \Rightarrow a = 10 \,m/s^2$.
Substituting $a$ into equation $(ii)$: $N = 1.5 \times 10 = 15 \,N$.
Thus,the force on block $B$ is $15 \,N$.

(C) For the given system,let $a$ be the acceleration of the blocks and $T$ be the tension in the string.
For mass $m_1$ moving horizontally: $T = m_1 a$
For mass $m_2$ moving vertically downwards: $m_2 g - T = m_2 a$
Adding these two equations: $m_2 g = (m_1 + m_2) a$
Therefore,the acceleration of the system is $a = \frac{m_2 g}{m_1 + m_2}$.
Using the kinematic equation $v^2 = u^2 + 2ah$,where initial velocity $u = 0$ and displacement is $h$:
$v^2 = 0 + 2 \left( \frac{m_2 g}{m_1 + m_2} \right) h$
$v = \sqrt{\left( \frac{m_2}{m_1 + m_2} \right) 2 g h}$.

(C) Mass of the elevator,$m = 500 \,kg$.
Acceleration of the elevator,$a = 2 \,m/s^2$ (upward).
When the elevator is moving upward,the tension $T$ in the cable is given by the equation of motion: $T - mg = ma$.
Therefore,$T = m(g + a) = 500(10 + 2) = 500(12) = 6000 \,N$.
The work done by the tension force $W$ is given by $W = T \times s$,where $s = 12 \,m$ is the displacement.
$W = 6000 \,N \times 12 \,m = 72000 \,J$.
Converting to kilojoules,$W = 72 \,kJ$.

(A) Given: Mass $m = 1 \ kg$,acceleration $a = 4.9 \ ms^{-2}$,acceleration due to gravity $g \approx 9.8 \ ms^{-2}$.
$(i)$ When the body is lifted up with acceleration $a$,the equation of motion is:
$T_1 - mg = ma$
$T_1 = m(g + a) = 1 \times (9.8 + 4.9) = 14.7 \ N$
(ii) When the body is lowered with acceleration $a$,the equation of motion is:
$mg - T_2 = ma$
$T_2 = m(g - a) = 1 \times (9.8 - 4.9) = 4.9 \ N$
The ratio of tension in the first and second case is:
$\frac{T_1}{T_2} = \frac{14.7}{4.9} = \frac{3}{1} = 3:1$

(A) The total mass of the system is $M = m + 2m + 3m = 6m$.
The acceleration of the system is $a = \frac{F}{6m}$.
To find $N_{1}$,consider the motion of the blocks $2m$ and $3m$ together. The force $N_{1}$ pushes these two blocks:
$N_{1} = (2m + 3m)a = 5m \times \frac{F}{6m} = \frac{5}{6}F$.
To find $N_{2}$,consider the motion of the block $3m$ alone. The force $N_{2}$ pushes this block:
$N_{2} = (3m)a = 3m \times \frac{F}{6m} = \frac{3}{6}F = \frac{1}{2}F$.
Comparing the values,we have $F = \frac{6}{6}F$,$N_{1} = \frac{5}{6}F$,and $N_{2} = \frac{3}{6}F$.
Therefore,$F > N_{1} > N_{2}$.

(B) The total mass of the system is $M = 4 \ kg + 2 \ kg + 1 \ kg = 7 \ kg$.
Since the surface is frictionless,the acceleration $a$ of the system is given by $a = \frac{F}{M} = \frac{14 \ N}{7 \ kg} = 2 \ m/s^2$.
To find the contact force $N$ between the $4 \ kg$ and $2 \ kg$ blocks,we consider the motion of the combined $2 \ kg$ and $1 \ kg$ blocks (total mass $m' = 3 \ kg$).
The force $N$ is the only horizontal force acting on this combined system of $3 \ kg$ that causes it to accelerate at $2 \ m/s^2$.
Therefore,$N = m' \times a = 3 \ kg \times 2 \ m/s^2 = 6 \ N$.

(B) Step $1$: Calculate the common acceleration of the system.
The total mass of the system is $M = 2 \,kg + 1 \,kg = 3 \,kg$.
The applied force is $F = 3 \,N$.
Using Newton's second law, $F = Ma$, we get:
$a = \frac{F}{M} = \frac{3 \,N}{3 \,kg} = 1 \,m/s^2$.
Step $2$: Calculate the contact force between the blocks.
Consider the $1 \,kg$ block. The only horizontal force acting on it is the contact force $N_1$ exerted by the $2 \,kg$ block.
Applying Newton's second law to the $1 \,kg$ block:
$N_1 = m_2 \times a = 1 \,kg \times 1 \,m/s^2 = 1 \,N$.
Therefore, the contact force between the two blocks is $1 \,N$.

(D) Given: Mass $m = 100 \text{ g} = 0.1 \text{ kg}$,Force $\vec{F} = (5\hat{i} + 10\hat{j}) \text{ N}$,Time $t = 2 \text{ s}$,Initial velocity $\vec{u} = 0$.
Acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{5\hat{i} + 10\hat{j}}{0.1} = (50\hat{i} + 100\hat{j}) \text{ m/s}^2$.
Using the equation of motion $\vec{r} = \vec{u}t + \frac{1}{2}\vec{a}t^2$:
$\vec{r} = 0 + \frac{1}{2}(50\hat{i} + 100\hat{j})(2)^2 = \frac{1}{2}(50\hat{i} + 100\hat{j})(4) = 2(50\hat{i} + 100\hat{j}) = (100\hat{i} + 200\hat{j}) \text{ m}$.
Comparing this with the given position $(2x\hat{i} + 5y\hat{j}) \text{ m}$:
$2x = 100 \Rightarrow x = 50$.
$5y = 200 \Rightarrow y = 40$.
Therefore,the ratio $x : y = 50 : 40 = 5 : 4$.