Variable Mass System and Rocket Problem Questions in English

(C) The thrust force $F$ exerted on the rocket is given by $F = v_r \left( \frac{dm}{dt} \right)$,where $v_r$ is the relative velocity of the ejected gas and $\frac{dm}{dt}$ is the rate of mass ejection.
Given that the rocket ejects $1/60$ of its mass $m$ in $1$ second,we have $\frac{dm}{dt} = \frac{m}{60} \times \frac{1}{1} = \frac{m}{60}$.
The thrust force is $F = 2400 \times \frac{m}{60} = 40m$.
According to Newton's second law,$F = ma$,so $ma = 40m$.
Therefore,the acceleration $a = 40\,m/s^2$.

(C) During the first phase $(t_1 = 60\,s)$:
Acceleration $a = 10\,m/s^2$.
Height $h_1 = \frac{1}{2}at_1^2 = \frac{1}{2} \times 10 \times (60)^2 = 18000\,m = 18\,km$.
Velocity $v_1 = at_1 = 10 \times 60 = 600\,m/s$.
During the second phase (free fall):
Maximum additional height $h_2 = \frac{v_1^2}{2g} = \frac{600^2}{2 \times 10} = \frac{360000}{20} = 18000\,m = 18\,km$.
Total maximum height $H_{\max} = h_1 + h_2 = 18\,km + 18\,km = 36\,km$.
For total time of flight:
Using $s = ut + \frac{1}{2}at^2$ for the whole journey,where $s = 0$ (displacement from ground):
$0 = v_1 t_2 - \frac{1}{2}g t_2^2$ (where $t_2$ is time after fuel finishes).
$-18000 = 600 t_2 - 5 t_2^2 \implies t_2^2 - 120 t_2 - 3600 = 0$.
Solving for $t_2$: $t_2 = \frac{120 + \sqrt{14400 - 4(1)(-3600)}}{2} = \frac{120 + \sqrt{28800}}{2} = 60 + 60\sqrt{2}\,s$.
Total time $T = t_1 + t_2 = 60 + 60 + 60\sqrt{2} = (120 + 60\sqrt{2})\,s$.
Thus,both $(a)$ and $(b)$ are correct.

(D) The $Assertion$ is incorrect because a rocket does not require surrounding air to move forward. In fact,rockets operate efficiently in the vacuum of space.
The rocket moves forward by ejecting its own fuel combustion products (exhaust gases) at high velocity backwards. According to Newton's third law of motion,the rocket exerts a force on the gases,and the gases exert an equal and opposite reaction force on the rocket,providing the necessary thrust.
Since the $Assertion$ is false and the $Reason$ is a scientifically correct statement regarding the principle of rocket propulsion (Newton's third law),the correct choice is that the $Assertion$ is incorrect but the $Reason$ is correct. However,based on the provided options,if we consider the $Assertion$ false,the most appropriate classification is that the $Assertion$ is incorrect.

(A) Mass of the rocket,$m = 20,000\; kg$.
Initial acceleration,$a = 5.0\; m s^{-2}$.
Acceleration due to gravity,$g = 10\; m s^{-2}$.
Using Newton's second law of motion,the net force acting on the rocket is the difference between the thrust $(F)$ and the weight $(mg)$.
The equation of motion is: $F - mg = ma$.
Rearranging for thrust: $F = m(g + a)$.
Substituting the values: $F = 20,000 \times (10 + 5) = 20,000 \times 15 = 3 \times 10^{5}\; N$.

(N/A) No,the flight of a rocket cannot be considered projectile motion.
Projectile motion is defined as the motion of an object under the sole influence of gravity,where the object is given an initial velocity and then allowed to move freely.
$A$ rocket,however,is propelled by the continuous thrust generated by the combustion of its fuel.
Since the rocket is subject to an external force (thrust) throughout its flight,it does not satisfy the condition of moving solely under the influence of gravity.

(N/A) No,the formula $\vec{F} = m\vec{a}$ is not valid in all circumstances.
Newton's Second Law is fundamentally defined as $\vec{F} = \frac{d\vec{p}}{dt}$,where $\vec{p} = m\vec{v}$ is the linear momentum.
Expanding this,we get $\vec{F} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}$.
If the mass $m$ is constant,then $\frac{dm}{dt} = 0$,which simplifies the equation to $\vec{F} = m\vec{a}$.
However,in cases where mass changes (e.g.,rocket propulsion) or at relativistic speeds where mass depends on velocity,the formula $\vec{F} = m\vec{a}$ does not hold.

(A) Let $M$ be the mass of the rocket at time $t$ and $v$ be its velocity.
In time interval $\Delta t$,the rocket ejects gas of mass $\Delta m$ with a relative speed $u$.
The velocity of the ejected gas with respect to the ground is $(v - u)$.
The velocity of the rocket after time $\Delta t$ becomes $(v + \Delta v)$.
The kinetic energy of the system at time $t$ is $(KE)_t = \frac{1}{2} M v^2$.
At time $t + \Delta t$,the kinetic energy of the system is $(KE)_{t+\Delta t} = \frac{1}{2}(M - \Delta m)(v + \Delta v)^2 + \frac{1}{2} \Delta m(v - u)^2$.
Expanding this,we get $(KE)_{t+\Delta t} \approx \frac{1}{2} M v^2 + M v \Delta v - \Delta m v u + \frac{1}{2} \Delta m u^2$.
From the conservation of momentum,the thrust force $F = M \frac{dv}{dt} = u \frac{dm}{dt}$,so $M \Delta v = \Delta m u$.
Substituting $M \Delta v = \Delta m u$ into the expression for the change in kinetic energy $\Delta K = (KE)_{t+\Delta t} - (KE)_t$:
$\Delta K = (M \Delta v - \Delta m u) v + \frac{1}{2} \Delta m u^2$.
Since $M \Delta v = \Delta m u$,the first term becomes zero.
Thus,$\Delta K = \frac{1}{2} \Delta m u^2$.
By the work-energy theorem,the work done by the internal mechanism is equal to the change in kinetic energy,so $\Delta W = \frac{1}{2} \Delta m u^2$.

(D) The rate of change of mass is given by $\frac{dM(t)}{dt} = bv^2$.
According to Newton's second law for a variable mass system,the thrust force exerted on the spaceship due to the accretion of dust is $F_{\text{thrust}} = -v \frac{dM(t)}{dt}$.
The negative sign indicates that the force acts in the direction opposite to the velocity,causing deceleration.
Using $F = M(t)a$,we have $M(t)a = -v \left( bv^2 \right)$.
Therefore,the instantaneous acceleration is $a = -\frac{bv^3}{M(t)}$.

(C) The force required to hold the gun is equal to the rate of change of momentum of the bullets fired.
The formula for the force is given by $F = n \cdot m \cdot v$,where:
$n$ is the number of bullets fired per second $(4 \, s^{-1})$,
$m$ is the mass of each bullet $(20 \, g = 0.02 \, kg)$,
$v$ is the velocity of the bullet $(300 \, m/s)$.
Substituting the values:
$F = 4 \times 0.02 \, kg \times 300 \, m/s$
$F = 4 \times 6 = 24 \, N$.
Therefore,the force required to hold the gun is $24 \, N$.

(D) The thrust force exerted on the rocket is given by $F_{\text{thrust}} = \left(\frac{dm}{dt}\right) V_{\text{rel}}$.
The net force acting on the rocket is $F_{\text{net}} = F_{\text{thrust}} - mg = ma$.
Substituting the given values: $m = 1000 \, \text{kg}$,$a = 20 \, \text{m s}^{-2}$,$V_{\text{rel}} = 500 \, \text{m s}^{-1}$,and $g = 10 \, \text{m s}^{-2}$.
$\left(\frac{dm}{dt}\right) \times 500 - (1000 \times 10) = 1000 \times 20$
$\left(\frac{dm}{dt}\right) \times 500 - 10000 = 20000$
$\left(\frac{dm}{dt}\right) \times 500 = 30000$
$\frac{dm}{dt} = \frac{30000}{500} = 60 \, \text{kg s}^{-1}$.

(C) The force exerted by the water jet on the block is given by the rate of change of momentum of the water striking the block.
$F = \frac{dp}{dt} = v \frac{dm}{dt}$
Given,the speed of water $v = 10 \,m \,s^{-1}$ and the rate of mass flow $\frac{dm}{dt} = 1 \,kg \,s^{-1}$.
Substituting these values,we get:
$F = 10 \times 1 = 10 \,N$
According to Newton's second law,$F = Ma$,where $M$ is the mass of the block and $a$ is its acceleration.
$10 = 2 \times a$
$a = \frac{10}{2} = 5 \,m \,s^{-2}$
Therefore,the initial acceleration of the block is $5 \,m \,s^{-2}$.

(D) Given: Rate of mass flow $\frac{dm}{dt} = 0.5 \, kg s^{-1}$,Velocity of belt $v = 5 \, m s^{-1}$.
When sand falls on the belt,the belt must exert a force to accelerate the sand to the velocity of the belt.
The force required to maintain the constant velocity is equal to the thrust force: $F = \frac{dm}{dt} \times v$.
$F = 0.5 \, kg s^{-1} \times 5 \, m s^{-1} = 2.5 \, N$.
The power required to keep the belt moving at a constant velocity is $P = F \times v$.
$P = 2.5 \, N \times 5 \, m s^{-1} = 12.5 \, W$.

(B) The average force $F$ exerted on the balloon due to the escaping air is given by the rate of change of momentum,which is $F = v \cdot \frac{dm}{dt}$.
Here,the total mass $m = 10\,g$ escapes in time $t = 5\,s$.
Therefore,the rate of mass loss is $\frac{dm}{dt} = \frac{10\,g}{5\,s} = 2\,g/s$.
The velocity of the escaping air is $v = 4.5\,cm/s$.
Substituting these values into the formula:
$F = 2\,g/s \times 4.5\,cm/s = 9\,g \cdot cm/s^2$.
Since $1\,dyne = 1\,g \cdot cm/s^2$,the average force is $9\,dyne$.

(C) To resist the downward acceleration,the upward force exerted by the bullets (thrust force) must balance the weight of the soldier and the gun.
Let $M$ be the total mass of the soldier and the gun,and $m$ be the mass of one bullet.
The force exerted by the bullets is equal to the rate of change of momentum of the bullets: $F = \frac{N}{\Delta t} \times m \times v$,where $\frac{N}{\Delta t}$ is the number of bullets fired per second.
Equating this force to the weight of the soldier: $Mg = \frac{N}{\Delta t} \times m \times v$.
Given: $\frac{N}{\Delta t} = 40 \,s^{-1}$,$m = 49 \,g = 0.049 \,kg$,$v = 500 \,m/s$,and $g = 9.8 \,m/s^2$.
Substituting the values: $M \times 9.8 = 40 \times 0.049 \times 500$.
$M \times 9.8 = 40 \times 24.5 = 980$.
$M = \frac{980}{9.8} = 100 \,kg$.

(A) According to the principle of conservation of linear momentum,if no external horizontal force acts on the system (cart + sand),the horizontal velocity of the sand remains unchanged as it falls vertically.
Let the mass of the cart be $M$ and the mass of the sand be $m$. Initially,both are moving with velocity $v$.
When the sand falls vertically,its horizontal component of velocity remains $v$ because there is no horizontal force acting on the sand during its fall.
Therefore,the sand continues to move in the same direction with the same horizontal speed $v$ as the cart.
Thus,the sand is moving with the cart.
Hence,option $A$ is correct.

(B) The force exerted on a rocket due to the ejection of fuel is given by the formula for a variable mass system: $F = v \cdot \frac{dm}{dt}$.
Here,the rate of fuel consumption is $\frac{dm}{dt} = 2 \, kg/s$.
The velocity of the ejected fuel is $v = 80 \, km/s = 80 \times 10^3 \, m/s$.
Substituting these values into the formula:
$F = (80 \times 10^3 \, m/s) \times (2 \, kg/s) = 1,60,000 \, N$.
Therefore,the correct option is $B$.

(B) The force exerted on the rocket due to the ejection of gases is given by the thrust formula for a variable mass system:
$F = v \cdot \frac{dm}{dt}$
Given:
Rate of mass ejection $\frac{dm}{dt} = 0.05 \,kg/s$
Velocity of gases $v = 400 \,m/s$
Substituting the values:
$F = 400 \,m/s \times 0.05 \,kg/s = 20 \,N$
Therefore,the accelerating force on the rocket is $20 \,N$.

(B) The thrust force on the rocket is given by $F_{thrust} = v_{rel} \left( \frac{dm}{dt} \right)$.
Given: Initial mass $M_0 = 5700 \,kg$,rate of mass ejection $\frac{dm}{dt} = 15 \,kg/s$,relative speed $v = 12 \,km/s = 12000 \,m/s$,and $g = 10 \,m/s^2$.
After $t = 60 \,s$ ($1$ minute),the mass of the rocket is $m = M_0 - \left( \frac{dm}{dt} \right) t = 5700 - (15 \times 60) = 5700 - 900 = 4800 \,kg$.
The net force acting on the rocket is $F_{net} = F_{thrust} - mg = v \left( \frac{dm}{dt} \right) - mg$.
The acceleration $a$ is given by $a = \frac{F_{net}}{m} = \frac{v \left( \frac{dm}{dt} \right)}{m} - g$.
Substituting the values: $a = \frac{12000 \times 15}{4800} - 10$.
$a = \frac{180000}{4800} - 10 = 37.5 - 10 = 27.5 \,m/s^2$.

(B) The average force $F$ exerted on the balloon due to the escaping air is given by the rate of change of momentum,which is $F = v \frac{dm}{dt}$.
Here,the velocity of the air $v = 4 \,m/s$.
The total mass of air $m = 2 \,g = 0.002 \,kg$.
The time taken for the air to escape $dt = 2.5 \,s$.
Substituting these values into the formula:
$F = 4 \times \left( \frac{0.002 \,kg}{2.5 \,s} \right)$
$F = 4 \times 0.0008 \,N$
$F = 0.0032 \,N$.
Therefore,the average force acting on the balloon is $0.0032 \,N$.

(D) The thrust force $F$ produced by the rocket engine is given by the formula $F = v_{e} \left( \frac{dm}{dt} \right)$,where $v_{e}$ is the exhaust speed and $\frac{dm}{dt}$ is the rate of mass ejection.
To overcome the weight of the rocket,the thrust must be equal to the gravitational force acting on the rocket,so $F = mg$.
Given: Mass of the rocket $m = 6000 \,kg$,acceleration due to gravity $g \approx 10 \,m/s^2$,and exhaust speed $v_{e} = 1000 \,m/s$.
Equating the forces: $v_{e} \left( \frac{dm}{dt} \right) = mg$.
Substituting the values: $1000 \times \left( \frac{dm}{dt} \right) = 6000 \times 10$.
$1000 \times \left( \frac{dm}{dt} \right) = 60000$.
$\frac{dm}{dt} = \frac{60000}{1000} = 60 \,kg/s$.
Therefore,the amount of gas that must be ejected each second is $60 \,kg$.

(C) The force required to maintain a constant velocity when mass is being added to a system is given by the thrust force formula: $F = v \frac{dm}{dt}$.
Here,the velocity of the train is $v = 10 \,m/s$.
The rate at which mass is added is $\frac{dm}{dt} = 5 \,kg/s$.
Since the rain falls vertically (zero horizontal velocity),the engine must provide an additional force to accelerate this added mass to the velocity of the train.
Therefore,$F = 10 \,m/s \times 5 \,kg/s = 50 \,N$.

(C) Let $F$ be the upward buoyant force acting on the balloon.
For the initial case,the equation of motion is:
$F - Mg = Ma$
$F = M(a + g)$
Let $x$ be the mass released from the balloon. The new mass of the balloon is $(M - x)$.
For the second case,the equation of motion is:
$F - (M - x)g = (M - x)(3a)$
Substitute the value of $F$ from the first equation into the second:
$M(a + g) - (M - x)g = (M - x)(3a)$
$Ma + Mg - Mg + xg = 3Ma - 3xa$
$Ma + xg = 3Ma - 3xa$
$xg + 3xa = 3Ma - Ma$
$x(g + 3a) = 2Ma$
$x = \frac{2Ma}{g + 3a}$

(D) The power $P$ required to maintain a constant speed $v$ for a conveyor belt onto which mass is being dropped is given by $P = F \cdot v$.
Since the belt moves at a constant speed,the force $F$ required to overcome the momentum change of the added sand is $F = \frac{dp}{dt} = v \frac{dm}{dt}$.
Given that the rate of mass deposition is $\frac{dm}{dt} = k\sqrt{v}$,where $k$ is a constant.
Substituting this into the force equation: $F = v(k\sqrt{v}) = kv^{3/2}$.
Now,calculating the power: $P = F \cdot v = (kv^{3/2}) \cdot v = kv^{5/2}$.
Squaring both sides of the equation,we get $P^2 = k^2 v^5$.
Therefore,$P^2 \propto v^5$.

(C) The thrust force $F$ exerted on the rocket is given by $F = v_{rel} \cdot \frac{dm}{dt}$.
Given: $v_{rel} = 5 \ km/s = 5000 \ m/s$ and $\frac{dm}{dt} = 10 \ kg/s$.
So,$F = 5000 \times 10 = 50000 \ N$.
The mass of the rocket at time $t$ is $M(t) = M_0 - (\frac{dm}{dt})t$.
At $t = 50 \ s$,$M(50) = 1500 - (10 \times 50) = 1500 - 500 = 1000 \ kg$.
The acceleration $a$ is given by $a = \frac{F}{M(t)} = \frac{50000}{1000} = 50 \ m/s^2$.

(D) The motion of a rocket is based on the principle of conservation of linear momentum. As the fuel burns,the rocket ejects hot gases at high speed in the downward direction. According to Newton's third law of motion,the gases exert an equal and opposite force on the rocket,providing the necessary thrust. Since there is no external force acting on the system (rocket + fuel),the total linear momentum of the system remains conserved.

(D) The force required to maintain a constant velocity is given by the rate of change of momentum of the system.
Since the sand is dropped vertically onto the belt,it has no initial horizontal velocity.
As the sand lands on the belt,it must be accelerated from a horizontal velocity of $0$ to $V$ to match the belt's speed.
The rate at which mass is added to the belt is $dm/dt = M$.
The force $F$ required is given by $F = v \cdot (dm/dt)$.
Substituting the given values,$F = V \cdot M = MV \text{ N}$.

(C) The force exerted by the gas on the cracker is equal to the rate of change of momentum of the gas.
Given:
Mass rate of gas,$\frac{dm}{dt} = 25 \text{ g/s} = 25 \times 10^{-3} \text{ kg/s}$.
Velocity of gas,$v = 400 \text{ m/s}$.
The force $F$ is given by the formula:
$F = v \times \frac{dm}{dt}$
$F = 400 \text{ m/s} \times 25 \times 10^{-3} \text{ kg/s}$
$F = 400 \times 0.025 \text{ N}$
$F = 10 \text{ N}$.

(D) The force exerted by the bullets on the rifle is equal to the rate of change of momentum of the bullets.
Given:
Mass of rifle $M = 20 \,kg$
Number of bullets per second $n = 4$
Mass of each bullet $m = 35 \times 10^{-3} \,kg$
Velocity of each bullet $v = 400 \,ms^{-1}$
The force $F$ required to keep the rifle stationary is equal to the recoil force exerted by the bullets.
The recoil force is given by the rate of change of momentum of the bullets:
$F = n \times (m \times v)$
$F = 4 \times (35 \times 10^{-3} \,kg) \times (400 \,ms^{-1})$
$F = 4 \times 35 \times 0.4$
$F = 140 \times 0.4 = 56 \,N$
To prevent the rifle from moving backwards,an external force of $56 \,N$ must be applied in the direction of the bullet's motion.

(D) The force exerted on a rocket due to the ejection of fuel is given by the thrust formula $F = v \frac{dm}{dt}$.
Given, the rate of fuel consumption is $\frac{dm}{dt} = 1 \,kg/s$.
The velocity of the ejected fuel is $v = 60 \,km/s = 60000 \,m/s$.
Substituting these values into the formula:
$F = 60000 \,m/s \times 1 \,kg/s = 60000 \,N$.
Therefore, the force exerted on the rocket is $60000 \,N$.

(A) The equation of motion for a rocket with variable mass is given by $F_{\text{ext}} + v_{\text{rel}} \frac{dm}{dt} = m \frac{dv}{dt}$.
Given that the rocket moves with constant acceleration $a$,we have $\frac{dv}{dt} = a$.
Assuming no external forces $(F_{\text{ext}} = 0)$,the equation becomes $v_{\text{rel}} \frac{dm}{dt} = m a$.
Here,$v_{\text{rel}} = v$ (the exhaust velocity relative to the rocket).
Rearranging the terms,we get $\frac{dm}{m} = \frac{a}{v} dt$.
Integrating both sides from initial mass $m_0$ at $t=0$ to mass $m$ at time $t$:
$\int_{m_0}^{m} \frac{dm}{m} = \int_{0}^{t} \frac{a}{v} dt$.
$\ln \left( \frac{m}{m_0} \right) = \frac{a t}{v}$.
Taking the exponential of both sides,we get $m = m_0 e^{-\frac{at}{v}}$.

(A) Let $F_B$ be the upward buoyant force acting on the balloon.
When the balloon of mass $m$ is descending with acceleration $a$,the equation of motion is: $mg - F_B = ma$,which implies $F_B = m(g - a)$.
Let $m'$ be the mass to be removed,so the new mass of the balloon is $(m - m')$.
When the balloon moves vertically up with acceleration $a$,the equation of motion is: $F_B - (m - m')g = (m - m')a$.
Substituting $F_B = m(g - a)$ into the equation: $m(g - a) - (m - m')g = (m - m')a$.
$mg - ma - mg + m'g = ma - m'a$.
$m'g + m'a = 2ma$.
$m'(g + a) = 2ma$.
Therefore,$m' = \frac{2ma}{g+a}$.

(B) Given:
Mass of the block,$M = 2 \,kg$
Rate of mass flow of water,$\frac{dm}{dt} = 1 \,kg \,s^{-1}$
Velocity of water jet,$v = 5 \,ms^{-1}$
The force exerted by the water jet on the block is given by the rate of change of momentum:
$F = \frac{dp}{dt} = \frac{d(mv)}{dt} = v \cdot \frac{dm}{dt}$
Substituting the values:
$F = 5 \,ms^{-1} \times 1 \,kg \,s^{-1} = 5 \,N$
According to Newton's second law,the acceleration $a$ of the block is:
$a = \frac{F}{M} = \frac{5 \,N}{2 \,kg} = 2.5 \,ms^{-2}$

(D) Given: Mass of the block $M = 4 \ kg$,velocity of water $v = 10 \ m \ s^{-1}$,and rate of mass flow $\frac{dm}{dt} = 2 \ kg \ s^{-1}$.
According to Newton's second law,the force exerted by the water jet on the block is given by the rate of change of momentum:
$F = v \cdot \left(\frac{dm}{dt}\right)$
Substituting the given values:
$F = 10 \times 2 = 20 \ N$
Now,using Newton's second law for the block $(F = Ma)$:
$a = \frac{F}{M} = \frac{20 \ N}{4 \ kg} = 5 \ m \ s^{-2}$
Therefore,the acceleration of the block is $5 \ m \ s^{-2}$.

(A) Given, resultant vertical acceleration $a = 10 \,m/s^2$, time $t = 1 \,min = 60 \,s$, and initial velocity $u = 0$.
Height reached during the powered phase $(h_1)$:
$h_1 = ut + \frac{1}{2}at^2 = 0 \times 60 + \frac{1}{2} \times 10 \times (60)^2 = 18000 \,m = 18 \,km$.
Velocity at the end of the powered phase $(v)$:
$v = u + at = 0 + 10 \times 60 = 600 \,m/s$.
After the fuel is exhausted, the rocket moves under gravity $(a = -g = -9.8 \,m/s^2)$ until its velocity becomes zero.
Height reached during the unpowered phase $(h_2)$:
Using $v_f^2 - v^2 = 2ah_2$, where $v_f = 0$:
$0 - (600)^2 = 2(-9.8)h_2 \Rightarrow h_2 = \frac{360000}{19.6} \approx 18367.3 \,m \approx 18.4 \,km$.
Total maximum height $H = h_1 + h_2 = 18 \,km + 18.4 \,km = 36.4 \,km$.

(C) The velocity of a rocket at any time $t$ is given by the Tsiolkovsky rocket equation: $v = u \ln\left(\frac{m_0}{m}\right) - gt$.
Given that gravitational forces are ignored,$g = 0$,so the equation simplifies to $v = u \ln\left(\frac{m_0}{m}\right)$.
Here,$u = 5 \ km/s$ is the exhaust speed of the gases relative to the rocket.
The initial mass is $m_0$ and the final mass is $m = \frac{1}{20}m_0$.
Therefore,the ratio $\frac{m_0}{m} = 20$.
Substituting these values into the equation:
$v = 5 \ln(20) \ km/s$.
Thus,the correct option is $C$.

(C) Step $1$: Calculate the velocity and height at the end of the powered flight $(t = 5 \,s)$.
The initial velocity $u = 0$, acceleration $a = 20 \,m/s^2$, and time $t = 5 \,s$.
The velocity at $t = 5 \,s$ is $V_{max} = u + at = 0 + 20 \times 5 = 100 \,m/s$.
The height reached at $t = 5 \,s$ is $S = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 20 \times 5^2 = 250 \,m$.
Step $2$: Calculate the final speed when it hits the ground.
After $t = 5 \,s$, the rocket is in free fall under gravity $(g = 10 \,m/s^2)$.
Using the work-energy theorem or kinematic equations, the total energy at the peak height is conserved, or we can use $V^2 = u^2 + 2aS$.
Here, the rocket starts from height $S = 250 \,m$ with an initial upward velocity $V_{max} = 100 \,m/s$.
When it hits the ground, the final velocity $V$ is given by $V^2 = V_{max}^2 + 2gS$.
$V^2 = (100)^2 + 2 \times 10 \times 250$.
$V^2 = 10000 + 5000 = 15000$.
$V = \sqrt{15000} = \sqrt{2500 \times 6} = 50 \sqrt{6} \,m/s$.

(D) $1$. Calculate the height of the rocket at $t = 20 \ s$: $h = \frac{1}{2} a t^2 = \frac{1}{2} \times 1 \times (20)^2 = 200 \ m$.
$2$. Calculate the velocity of the rocket at $t = 20 \ s$: $v = a t = 1 \times 20 = 20 \ m \ s^{-1}$.
$3$. The piece breaks off and acts as a projectile with initial velocity $u = 20 \ m \ s^{-1}$ upwards from a height $h = 200 \ m$.
$4$. Using the equation of motion $s = ut + \frac{1}{2} a t^2$ for the piece,where $s = -200 \ m$ (downward displacement),$u = 20 \ m \ s^{-1}$,and $a = -g = -10 \ m \ s^{-2}$:
$-200 = 20 t - \frac{1}{2} \times 10 \times t^2$
$-200 = 20 t - 5 t^2$
$5 t^2 - 20 t - 200 = 0$
$t^2 - 4 t - 40 = 0$.
$5$. Solving for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$t = \frac{4 \pm \sqrt{16 - 4(1)(-40)}}{2} = \frac{4 \pm \sqrt{176}}{2} = 2 \pm \sqrt{44} \approx 2 \pm 6.63$.
$6$. Since time must be positive,$t \approx 8.63 \ s$.

(B) Let the initial mass be $m$ and initial velocity be $v$. The initial kinetic energy is $(K.E.)_i = \frac{1}{2}mv^2$ and initial momentum is $p_i = mv$.
After the fuel is burned,the final mass is $m' = \frac{m}{2}$ and the final kinetic energy is $(K.E.)_f = 16 \times (K.E.)_i = 16 \times \frac{1}{2}mv^2 = 8mv^2$.
Let the final velocity be $v'$. Then $(K.E.)_f = \frac{1}{2}m'v'^2 = \frac{1}{2}(\frac{m}{2})v'^2 = \frac{1}{4}mv'^2$.
Equating the two expressions for $(K.E.)_f$: $\frac{1}{4}mv'^2 = 8mv^2 \Rightarrow v'^2 = 32v^2 \Rightarrow v' = \sqrt{32}v = 4\sqrt{2}v$.
The final momentum is $p_f = m'v' = (\frac{m}{2})(4\sqrt{2}v) = 2\sqrt{2}mv$.
The factor by which momentum increases is $\frac{p_f}{p_i} = \frac{2\sqrt{2}mv}{mv} = 2\sqrt{2}$.