Conservation of Linear Momentum Questions in English

(C) Let $m = 80 \, kg$ be the mass of the man and $M = 320 \, kg$ be the mass of the trolley.
Since the system is on frictionless rails,there is no external horizontal force acting on the system. Therefore,the momentum of the system remains conserved.
Initially,the system is at rest,so the total initial momentum is $0$.
Let $v_m$ be the velocity of the man relative to the ground and $v_t$ be the velocity of the trolley relative to the ground.
Given that the velocity of the man relative to the trolley is $v_{rel} = 1 \, m/s$,we have $v_m - v_t = 1$,which implies $v_m = 1 + v_t$.
By the law of conservation of momentum: $m v_m + M v_t = 0$.
Substituting the values: $80(1 + v_t) + 320 v_t = 0$.
$80 + 80 v_t + 320 v_t = 0 \implies 400 v_t = -80 \implies v_t = -0.2 \, m/s$.
The negative sign indicates the trolley moves in the opposite direction of the man.
The velocity of the man relative to the ground is $v_m = 1 + (-0.2) = 0.8 \, m/s$.
The displacement of the man relative to the ground in $t = 4 \, s$ is $d = v_m \times t = 0.8 \times 4 = 3.2 \, m$.

(B) Since the surface is frictionless,no external horizontal force can be exerted to move the person.
According to the law of conservation of linear momentum,if the total external force on a system is zero,the total linear momentum remains constant.
By spitting,sneezing,or throwing an object in one direction,the person imparts momentum to that matter.
To conserve the total momentum of the system (which is initially zero),the person must acquire an equal and opposite momentum,causing them to recoil or move in the opposite direction.
Thus,by spitting or sneezing,the person can move off the frictionless plane.

(D) According to Newton's third law,for every action,there is an equal and opposite reaction.
When two objects interact,the force exerted by object $A$ on object $B$ $(F_{AB})$ is equal in magnitude and opposite in direction to the force exerted by object $B$ on object $A$ $(F_{BA})$,i.e.,$F_{AB} = -F_{BA}$.
Since force is the rate of change of momentum $(F = \frac{dp}{dt})$,this implies that the change in momentum of object $A$ is equal and opposite to the change in momentum of object $B$.
Therefore,the total change in momentum of the system is zero,meaning the total momentum remains conserved in the absence of any external force.
Thus,Newton's third law of motion leads to the law of conservation of momentum.

(B) According to the law of conservation of linear momentum,if the net external force on a system is zero,the total momentum of the system remains conserved.
Since the pond is frictionless,the man can create a change in his momentum by throwing an object or mass in the opposite direction of the shore.
By spitting horizontally in the direction away from the shore,the man exerts a force on the saliva,and according to Newton's third law of motion,the saliva exerts an equal and opposite force on the man.
This force provides the man with a small velocity towards the shore,allowing him to reach it.

(C) According to Newton's second law of motion,the force $F$ acting on a particle is given by the rate of change of its linear momentum $P$,which is $F = \frac{dP}{dt}$.
If the momentum $P$ of the particle is constant,then its derivative with respect to time is zero,i.e.,$\frac{dP}{dt} = 0$.
Therefore,the force $F$ required to maintain this state is zero.
Thus,option $(c)$ is correct.

(C) According to the principle of conservation of linear momentum,the total momentum before the mass is dropped equals the total momentum after the mass is dropped.
Initial momentum $P_i = m_1 \times v_1 = 1000\, kg \times 50\, km/h = 50000\, kg \cdot km/h$.
Final mass $M = m_1 + m_2 = 1000\, kg + 250\, kg = 1250\, kg$.
Let the final velocity be $v$.
Final momentum $P_f = M \times v = 1250\, kg \times v$.
Equating initial and final momentum: $50000 = 1250 \times v$.
$v = \frac{50000}{1250} = 40\, km/h$.

(A) Given: Mass $m = 100\, g = 0.1\, kg$,velocity $v = 10\, m/s$,time of contact $\Delta t = 0.1\, s$,angle with the wall $\theta = 30^\circ$.
The component of momentum perpendicular to the wall changes during the collision. The angle of the velocity vector with the normal to the wall is $\phi = 90^\circ - 30^\circ = 60^\circ$.
The change in momentum $\Delta p$ is given by the change in the component of momentum perpendicular to the wall:
$\Delta p = p_f - p_i = (-mv \cos \phi) - (mv \cos \phi) = -2mv \cos \phi$.
The magnitude of the change in momentum is $|\Delta p| = 2mv \cos 60^\circ$.
Force $F = \frac{|\Delta p|}{\Delta t} = \frac{2mv \cos 60^\circ}{\Delta t}$.
Substituting the values:
$F = \frac{2 \times 0.1\, kg \times 10\, m/s \times \cos 60^\circ}{0.1\, s} = \frac{2 \times 0.1 \times 10 \times 0.5}{0.1} = 10\, N$.

(B) The momentum $p$ of a body is defined as the product of its mass $m$ and velocity $v$,given by $p = mv$.
If the momentum $p$ is constant and the mass $m$ of the body is assumed to be constant,then the velocity $v = p/m$ must also be constant.
Since velocity is constant,the acceleration $a = dv/dt$ must be zero.
According to Newton's second law,the net force $F = ma$ acting on the body must also be zero.
Therefore,if momentum is constant,the velocity must be constant.

(C) The motion of a rocket is governed by Newton's $3^{rd}$ law of motion and the principle of conservation of linear momentum.
As the rocket expels exhaust gases at high velocity in the backward direction,the rocket gains an equal and opposite momentum in the forward direction.
Since no external force acts on the system (rocket + fuel),the total linear momentum of the system remains conserved.

(C) According to the law of conservation of linear momentum,the total momentum before firing is equal to the total momentum after firing.
Initial momentum = $0$.
Final momentum = $m_B v_B + m_G v_G = 0$.
Here,$m_B = 200 \,g = 0.2 \,kg$,$v_B = 5 \,m/s$,and $m_G = 1 \,kg$.
$0.2 \times 5 + 1 \times v_G = 0$.
$1 + v_G = 0$.
$v_G = -1 \,m/s$.
The negative sign indicates that the gun recoils in the opposite direction to the bullet.
Therefore,the magnitude of the recoil velocity is $1 \,m/s$.

(A) According to the law of conservation of linear momentum,the total momentum before firing is equal to the total momentum after firing.
Initial momentum of the system = $0$.
Final momentum of the system = $m_B v_B + m_G v_G = 0$.
Here,$m_B = 5 \,g = 5 \times 10^{-3} \,kg$,$v_B = 500 \,m/s$,and $m_G = 5 \,kg$.
$5 \times 10^{-3} \times 500 + 5 \times v_G = 0$.
$2.5 + 5 \times v_G = 0$.
$v_G = -\frac{2.5}{5} = -0.5 \,m/s$.
The magnitude of the recoil velocity is $0.5 \,m/s$.

(C) Initial momentum of the body is $0$ because it is at rest.
Let the masses of the three pieces be $m_1 = M/4$,$m_2 = M/4$,and $m_3 = M - (M/4 + M/4) = M/2$.
The momenta of the first two pieces are $p_1 = (M/4) \times 3$ and $p_2 = (M/4) \times 4$ in perpendicular directions.
The resultant momentum of these two pieces is $p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{(\frac{3M}{4})^2 + (\frac{4M}{4})^2} = \sqrt{\frac{9M^2}{16} + \frac{16M^2}{16}} = \sqrt{\frac{25M^2}{16}} = \frac{5M}{4}$.
According to the law of conservation of linear momentum,the total momentum must remain zero. Therefore,the third piece must have a momentum $p_3$ equal and opposite to $p_{12}$.
$p_3 = \frac{5M}{4} = m_3 \times v = \frac{M}{2} \times v$.
Solving for $v$: $v = \frac{5M/4}{M/2} = \frac{5}{4} \times 2 = 2.5 \, m/s$.

(C) The law of conservation of momentum states that the total momentum of a closed system remains constant over time (is conserved).
An isolated or closed system is defined as a system where the mass remains constant and the net external force acting on the system is $0$.
Therefore,the total momentum of a system is conserved if and only if there is no net external force acting on the system.

(D) According to the law of conservation of linear momentum,the total momentum before projection must equal the total momentum after projection.
Initially,both the tank and the body are at rest,so the initial momentum is $0$.
Let $M = 100 \, kg$ be the mass of the tank and $m = 0.25 \, kg$ be the mass of the body.
Let $V$ be the recoil velocity of the tank and $v = 100 \, m/s$ be the velocity of the body.
$M \times V + m \times v = 0$
$100 \times V + 0.25 \times 100 = 0$
$100 \times V = -25$
$V = -25 / 100 = -0.25 \, m/s$.
The magnitude of the recoil velocity is $0.25 \, m/s$.

(A) According to the principle of conservation of linear momentum,the total momentum before firing is equal to the total momentum after firing.
Initially,both the gun and the bullet are at rest,so the initial momentum is $0$.
Let $m_B = 0.1\,kg$ be the mass of the bullet,$v_B = 100\,m/s$ be the velocity of the bullet,$m_G = 50\,kg$ be the mass of the gun,and $v_G$ be the recoil velocity of the gun.
According to the conservation of momentum: $m_G v_G + m_B v_B = 0$.
$v_G = -\frac{m_B v_B}{m_G}$.
Substituting the values: $v_G = -\frac{0.1 \times 100}{50} = -\frac{10}{50} = -0.2\,m/s$.
The magnitude of the recoil velocity is $0.2\,m/s$.

(D) According to the law of conservation of linear momentum,the initial momentum of the system is zero.
Therefore,the magnitude of the momentum of the gun must be equal to the magnitude of the momentum of the bullet.
$m_G v_G = m_B v_B$
Here,$m_G = 1\, kg$ (mass of the gun),$v_G = 5\, m/s$ (recoil velocity of the gun),and $m_B = 10\, g = 10 \times 10^{-3}\, kg = 0.01\, kg$ (mass of the bullet).
Substituting the values:
$1 \times 5 = 0.01 \times v_B$
$v_B = \frac{5}{0.01} = 500\, m/s$.
Thus,the muzzle velocity of the bullet is $500\, m/s$.

(C) jet engine operates based on $Newton's$ third law of motion and the principle of conservation of linear momentum.
In a jet engine,air is taken in,compressed,mixed with fuel,and ignited. The resulting high-pressure,high-temperature gases are expelled at high velocity through the exhaust nozzle.
According to the law of conservation of linear momentum,the momentum of the system (engine + gases) remains constant in the absence of external forces. The backward momentum of the expelled gases creates an equal and opposite forward momentum (thrust) on the engine,propelling the aircraft forward.

(A) According to the law of conservation of linear momentum,if no external force acts on the system,the total momentum remains constant.
Initially,the body is at rest,so the initial momentum is $P_{initial} = 0$.
After the explosion,if the two pieces have masses $m_1$ and $m_2$ and velocities $v_1$ and $v_2$ respectively,then $m_1 v_1 + m_2 v_2 = 0$.
This implies $m_1 v_1 = -m_2 v_2$.
Taking the magnitude,$|p_1| = |p_2|$.
Thus,both parts will have numerically equal momentum,but in opposite directions.

(B) According to the law of conservation of linear momentum,the initial momentum of the system is equal to the final momentum of the system.
Since the tank is initially stationary,the initial momentum is $0$.
Let $M = 125000 \ lb$ be the mass of the tank,$m = 25 \ lb$ be the mass of the shell,$v = 1000 \ ft/sec$ be the velocity of the shell,and $V$ be the recoil velocity of the tank.
By conservation of momentum: $M \times V + m \times v = 0$.
$125000 \times V + 25 \times 1000 = 0$.
$125000 \times V = -25000$.
$V = -\frac{25000}{125000} = -0.2 \ ft/sec$.
The magnitude of the recoil velocity is $0.2 \ ft/sec$.

(D) Since the initial momentum of the bomb was zero,after the explosion,the two parts must possess equal and opposite momentum.
Let $m_A = 4 \, kg$ and $m_B = 8 \, kg$.
Given $v_B = 6 \, m/s$.
According to the law of conservation of momentum:
$m_A v_A = m_B v_B$
$4 \times v_A = 8 \times 6$
$4 \times v_A = 48$
$v_A = 12 \, m/s$
The kinetic energy of the $4 \, kg$ mass is given by:
$K.E. = \frac{1}{2} m_A v_A^2$
$K.E. = \frac{1}{2} \times 4 \times (12)^2$
$K.E. = 2 \times 144 = 288 \, J$.

(A) The kinetic energy $E$ of a body is given by the formula $E = \frac{P^2}{2m}$,where $P$ is the momentum and $m$ is the mass of the body.
According to the law of conservation of linear momentum,when a bullet is fired from a rifle,the magnitude of the momentum of the rifle is equal to the magnitude of the momentum of the bullet $(P_{rifle} = P_{bullet} = P)$.
Since $P$ is constant for both,the kinetic energy is inversely proportional to the mass $(E \propto \frac{1}{m})$.
Because the mass of the rifle $(M)$ is much greater than the mass of the bullet $(m)$,the kinetic energy of the rifle will be significantly less than the kinetic energy of the bullet.

(A) The total mass of the bomb is $M = 12 \ kg$. It divides into two parts $m_1$ and $m_2$ such that $m_1 + m_2 = 12$ and $m_1/m_2 = 1/3$.
Solving these,we get $m_1 = 3 \ kg$ (smaller part) and $m_2 = 9 \ kg$ (bigger part).
The kinetic energy of the smaller part is $K_1 = \frac{1}{2} m_1 v_1^2 = 216 \ J$.
Substituting $m_1 = 3 \ kg$,we have $\frac{1}{2} \times 3 \times v_1^2 = 216$,which gives $v_1^2 = 144$,so $v_1 = 12 \ m/s$.
The momentum of the smaller part is $p_1 = m_1 v_1 = 3 \times 12 = 36 \ kg \cdot m/s$.
Since the bomb was initially at rest,the total momentum must be zero. Therefore,the magnitudes of the momenta of the two parts must be equal: $p_1 = p_2 = 36 \ kg \cdot m/s$.
Thus,the momentum of the bigger part is $36 \ kg \cdot m/s$.

(C) Since the bomb was initially at rest,the initial momentum of the system is $0.$
According to the law of conservation of linear momentum,the final momentum of the system must also be $0.$
Let $m_1 = 3\,kg$ and $m_2 = 6\,kg.$ Let $v_1 = 1.6\,m/s$ be the velocity of the $3\,kg$ mass and $v_2$ be the velocity of the $6\,kg$ mass.
$m_1v_1 + m_2v_2 = 0$
$3 \times (-1.6) + 6 \times v_2 = 0$
$-4.8 + 6v_2 = 0$
$6v_2 = 4.8$
$v_2 = 0.8\,m/s$
The kinetic energy of the $6\,kg$ mass is given by $K.E. = \frac{1}{2}m_2v_2^2.$
$K.E. = \frac{1}{2} \times 6 \times (0.8)^2 = 3 \times 0.64 = 1.92\,J.$

(D) According to the law of conservation of linear momentum,since the initial particle is stationary,the total initial momentum is $0$.
Therefore,the magnitudes of the momenta of the two particles must be equal: $p_1 = p_2 = p$.
The kinetic energy $E$ of a particle is given by the formula $E = \frac{p^2}{2m}$.
Since the momentum $p$ is the same for both particles,the kinetic energy is inversely proportional to the mass: $E \propto \frac{1}{m}$.
Thus,the ratio of their kinetic energies is $\frac{E_1}{E_2} = \frac{m_2}{m_1}$.

(D) According to the law of conservation of linear momentum,the total momentum before the explosion is zero,so the two fragments must have equal and opposite linear momenta: $m_1 v_1 = m_2 v_2$.
Given $m_1 = 1.0 \, kg$,$v_1 = 80 \, m/s$,and $m_2 = 2.0 \, kg$.
$1.0 \times 80 = 2.0 \times v_2 \implies v_2 = 40 \, m/s$.
The total kinetic energy imparted to the fragments is $K = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$.
$K = \frac{1}{2} \times 1.0 \times (80)^2 + \frac{1}{2} \times 2.0 \times (40)^2$.
$K = \frac{1}{2} \times 6400 + 1.0 \times 1600 = 3200 + 1600 = 4800 \, J$.
Converting to $kJ$,$K = 4.8 \, kJ$.

(C) Initial momentum of the $3\,m$ mass is $0$ because it is at rest.
Due to the explosion,this mass splits into three fragments of equal mass $m$.
Let the velocity of the third fragment be $\vec{V}$.
Final momentum of the system = $m\vec{V} + m(v\hat{i}) + m(v\hat{j})$.
According to the law of conservation of linear momentum,the total initial momentum must equal the total final momentum:
$0 = m\vec{V} + mv\hat{i} + mv\hat{j}$
$m\vec{V} = -mv\hat{i} - mv\hat{j}$
$\vec{V} = -v(\hat{i} + \hat{j})$

(B) Let the initial mass of the body be $m$.
Initial linear momentum = $mv$.
When it breaks into two equal masses,each part has mass $\frac{m}{2}$.
One fragment moves backward with velocity $v$,so its velocity is $-v$.
Let the velocity of the other fragment be $v_2$.
According to the law of conservation of linear momentum:
Initial momentum = Final momentum
$mv = \frac{m}{2}(-v) + \frac{m}{2}(v_2)$
$mv = -\frac{mv}{2} + \frac{mv_2}{2}$
$mv + \frac{mv}{2} = \frac{mv_2}{2}$
$\frac{3mv}{2} = \frac{mv_2}{2}$
$v_2 = 3v$.
Since the result is positive,the other fragment moves with velocity $3v$ in the forward direction.

(A) When a shell explodes in mid-air,the internal chemical energy stored in the explosive material is converted into mechanical energy.
This release of energy results in an increase in the total kinetic energy of the fragments.
However,since the explosion is caused by internal forces,there is no external force acting on the system.
According to the law of conservation of linear momentum,the total momentum of the system remains constant.

(B) According to the law of conservation of linear momentum,the initial momentum of the bomb is zero.
Therefore,the final momentum of the two pieces must be equal in magnitude and opposite in direction.
Let $m_A = 18 \, kg$,$v_A = 6 \, m/s$ and $m_B = 12 \, kg$.
$m_A v_A = m_B v_B$
$18 \times 6 = 12 \times v_B$
$v_B = \frac{108}{12} = 9 \, m/s$
The kinetic energy of the $12 \, kg$ mass is given by:
$K.E. = \frac{1}{2} m_B v_B^2$
$K.E. = \frac{1}{2} \times 12 \times (9)^2$
$K.E. = 6 \times 81 = 486 \, J$

(A) During a collision,external impulsive forces may act on the system. According to the impulse-momentum theorem,the change in momentum is equal to the impulse,which is given by $\int F_{ext} dt$. If the time of impact $\Delta t$ is extremely small,the impulse provided by external forces becomes negligible compared to the internal forces of the collision. Therefore,the total linear momentum of the system remains conserved. Thus,the correct option is $A$.

(C) According to the law of conservation of linear momentum,the total initial momentum of the shell is $0$ because it is initially at rest.
After the explosion,the total momentum must remain $0$.
Let the masses of the two pieces be $m_1 = m_2 = m$.
Let their velocities be $\vec{v}_1$ and $\vec{v}_2$.
By the law of conservation of momentum: $m\vec{v}_1 + m\vec{v}_2 = 0$.
This implies $\vec{v}_1 = -\vec{v}_2$.
Since the masses are equal,the magnitudes of their velocities must be equal $(|v_1| = |v_2|)$,and the negative sign indicates that they move in exactly opposite directions.

(B) Let the mass of the cannon ball be $m$. At the highest point,the vertical component of velocity is $0$,and the horizontal component is $u_x = u \cos \theta = 200 \cos 60^{\circ} = 200 \times 0.5 = 100 \, m/s$.
Thus,the momentum of the cannon ball at the highest point is $\vec{P} = m \times 100 \hat{i}$.
After the explosion,the mass splits into $3$ equal fragments of mass $m/3$ each.
Let the velocity of the third fragment be $\vec{V} = V_x \hat{i} + V_y \hat{j}$.
The velocities of the fragments are $\vec{v}_1 = 100 \hat{j}$,$\vec{v}_2 = -100 \hat{j}$,and $\vec{v}_3 = V_x \hat{i} + V_y \hat{j}$.
By the law of conservation of linear momentum: $\vec{P}_{initial} = \vec{P}_{final}$.
$m(100 \hat{i}) = \frac{m}{3}(100 \hat{j}) + \frac{m}{3}(-100 \hat{j}) + \frac{m}{3}(V_x \hat{i} + V_y \hat{j})$.
$100 \hat{i} = \frac{1}{3}(V_x \hat{i} + V_y \hat{j})$.
Comparing components: $V_x = 300 \, m/s$ and $V_y = 0$.
Therefore,the third fragment moves with a velocity of $300 \, m/s$ in the horizontal direction.

(D) Given total mass $M = 5 \,kg$. The masses of the fragments are in the ratio $1:1:3$,so the masses are $m_1 = 1 \,kg$,$m_2 = 1 \,kg$,and $m_3 = 3 \,kg$.
The two fragments of mass $1 \,kg$ move in mutually perpendicular directions with speed $v = 21 \,m/s$.
The momentum of the first fragment is $P_x = m_1 v = 1 \times 21 = 21 \,kg \cdot m/s$ along the $x$-axis.
The momentum of the second fragment is $P_y = m_2 v = 1 \times 21 = 21 \,kg \cdot m/s$ along the $y$-axis.
The resultant momentum of these two fragments is $P_{res} = \sqrt{P_x^2 + P_y^2} = \sqrt{21^2 + 21^2} = 21\sqrt{2} \,kg \cdot m/s$.
According to the law of conservation of linear momentum,the initial momentum of the body at rest is zero. Therefore,the momentum of the third (heaviest) fragment must be equal and opposite to the resultant momentum of the first two fragments.
$P_3 = P_{res} = 21\sqrt{2} \,kg \cdot m/s$.
Since $P_3 = m_3 v_3$,we have $3 \times v_3 = 21\sqrt{2}$.
$v_3 = \frac{21\sqrt{2}}{3} = 7\sqrt{2} \,m/s$.

(C) According to the law of conservation of linear momentum,the total momentum before firing is equal to the total momentum after firing.
Initial momentum of the system (gun + bullet) = $0$.
Final momentum = $m_G v_G + m_B v_B = 0$.
Therefore,the magnitude of the momentum of the gun equals the magnitude of the momentum of the bullet: $m_G v_G = m_B v_B$.
Given:
Mass of bullet $(m_B)$ = $50 \, g = 0.05 \, kg$.
Velocity of bullet $(v_B)$ = $30 \, m/s$.
Velocity of gun $(v_G)$ = $1 \, m/s$.
$m_G = \frac{m_B v_B}{v_G} = \frac{0.05 \times 30}{1} = 1.5 \, kg$.

(D) According to the law of conservation of linear momentum,the total momentum before the explosion must equal the total momentum after the explosion.
Initial momentum $P_i = mv$.
After the explosion,the shell breaks into two pieces: one of mass $m_1 = m/4$ with velocity $v_1 = 0$,and the other of mass $m_2 = m - m/4 = 3m/4$ with velocity $v_2$.
Final momentum $P_f = m_1 v_1 + m_2 v_2 = (m/4)(0) + (3m/4)v_2 = (3m/4)v_2$.
Equating initial and final momentum:
$mv = (3m/4)v_2$
$v = (3/4)v_2$
$v_2 = \frac{4}{3}v$.
Therefore,the velocity of the other shell is $\frac{4}{3}v$.

(C) Since the body is at rest and explodes into two equal fragments,by the law of conservation of linear momentum,the fragments must move in opposite directions with equal speeds. Let the speed of each fragment be $v = 10 \, m/s$.
Let the point of explosion be $A$. After time $t$,the fragments are at positions where their radius vectors from $A$ make an angle of $90^o$. Due to symmetry,each radius vector makes an angle of $45^o$ with the vertical line passing through $A$.
Let $D$ be the point directly below $A$ at the same vertical level as the fragments at time $t$. The vertical distance fallen by each fragment is $h = \frac{1}{2}gt^2 = 5t^2$ (taking $g = 10 \, m/s^2$).
The horizontal distance traveled by one fragment is $x = vt = 10t$.
In the right-angled triangle formed by the vertical displacement and horizontal displacement,the angle with the vertical is $45^o$. Thus,$\tan(45^o) = \frac{\text{horizontal distance}}{\text{vertical distance}} = \frac{x}{h}$.
Since $\tan(45^o) = 1$,we have $x = h$.
$10t = 5t^2$
$5t^2 - 10t = 0 \Rightarrow 5t(t - 2) = 0$.
Since $t \neq 0$,we get $t = 2 \, s$.

(A) The initial velocity of the $50 \, kg$ mass is $u = 100 \, m/s$. Using $v = u - gt$ with $g = 10 \, m/s^2$ (standard approximation),the velocity after $t = 5 \, s$ is $v = 100 - 10 \times 5 = 50 \, m/s$.
Initial momentum $P_i = m \times v = 50 \times 50 = 2500 \, kg \cdot m/s$ (upwards).
Let the velocity of the $30 \, kg$ piece be $V$. According to the law of conservation of momentum,$P_i = P_f$.
$2500 = (20 \times 150) + (30 \times V)$.
$2500 = 3000 + 30V$.
$30V = 2500 - 3000 = -500$.
$V = -500 / 30 = -16.67 \, m/s$.
Note: Using $g = 9.8 \, m/s^2$ as per the original prompt: $v = 100 - 9.8 \times 5 = 51 \, m/s$.
$P_i = 50 \times 51 = 2550 \, kg \cdot m/s$.
$2550 = (20 \times 150) + 30V$.
$2550 = 3000 + 30V$.
$30V = -450$.
$V = -15 \, m/s$.
The negative sign indicates the $30 \, kg$ piece moves downwards at $15 \, m/s$.

(A) According to the law of conservation of linear momentum,the total momentum before the explosion must equal the total momentum after the explosion.
Initial momentum of the spacecraft = $MV$.
After the explosion,the spacecraft splits into two parts: one of mass $m$ and the other of mass $(M - m)$.
The part of mass $m$ comes to rest,so its velocity is $0$.
Let the velocity of the remaining part of mass $(M - m)$ be $v$.
Final momentum = $m \times 0 + (M - m) \times v = (M - m)v$.
Equating initial and final momentum:
$MV = (M - m)v$
$v = \frac{MV}{M - m}$.

(D) According to the law of conservation of linear momentum,the total momentum before firing is equal to the total momentum after firing.
Initial momentum of the system = $0$.
Final momentum = $m_B v_B + m_G v_G = 0$,where $m_B$ is the mass of the bullet,$v_B$ is the velocity of the bullet,$m_G$ is the mass of the gun,and $v_G$ is the recoil velocity of the gun.
Given: $m_B = 50 \, g = 0.05 \, kg$,$v_B = 1 \, km/s = 1000 \, m/s$,$m_G = 5 \, kg$.
Substituting the values: $(0.05 \, kg) \times (1000 \, m/s) + (5 \, kg) \times v_G = 0$.
$50 + 5 v_G = 0$.
$5 v_G = -50$.
$v_G = -10 \, m/s$.
The speed of recoil is the magnitude of the velocity,which is $10 \, m/s$.

(D) Initially,the body is at rest,so the initial momentum is $0$. By the law of conservation of linear momentum,the final momentum must also be $0$.
Let the two parts of mass $m$ move along the $x$ and $y$ axes with velocities $\vec{v}_1 = 12\hat{i} \, m/s$ and $\vec{v}_2 = 12\hat{j} \, m/s$.
The resultant momentum of these two parts is $\vec{P}_{12} = m(12\hat{i}) + m(12\hat{j}) = 12m(\hat{i} + \hat{j})$.
The magnitude of this resultant momentum is $P_{12} = \sqrt{(12m)^2 + (12m)^2} = 12m\sqrt{2}$.
Let the third part have mass $3m$ and velocity $\vec{V}$. For the total momentum to be zero,the momentum of the third part $\vec{P}_3$ must be equal and opposite to $\vec{P}_{12}$.
So,$3m \times V = 12m\sqrt{2}$.
$V = \frac{12m\sqrt{2}}{3m} = 4\sqrt{2} \, m/s$.
The direction of this velocity is opposite to the resultant of the first two parts,which makes an angle of $135^\circ$ with each of the first two parts.

(A) According to the principle of conservation of linear momentum,the total initial momentum must equal the total final momentum.
Initial momentum $P_i = Mv$.
After the explosion,the spacecraft breaks into two pieces of mass $m = M/2$. One piece is stationary $(v_1 = 0)$,and the other piece moves with velocity $v_2$.
Final momentum $P_f = m(0) + m(v_2) = m v_2$.
Since $m = M/2$,we have $P_f = (M/2)v_2$.
Equating initial and final momentum: $Mv = (M/2)v_2$.
Solving for $v_2$: $v_2 = 2v$.
However,based on the provided options where $m$ is treated as a variable mass component,the conservation equation is $Mv = m(0) + (M - m)v_2$.
Thus,$v_2 = \frac{Mv}{M - m}$.

(A) Let the two pieces have equal mass $m$ and the third piece have a mass of $3m$.
According to the law of conservation of linear momentum,since the initial momentum of the system was zero,the final momentum must also be zero.
Let the velocities of the two pieces be $\vec{v}_1 = 30\hat{i} \, m/s$ and $\vec{v}_2 = 30\hat{j} \, m/s$.
The momentum of these two pieces is $\vec{p}_1 = m(30\hat{i})$ and $\vec{p}_2 = m(30\hat{j})$.
The resultant momentum of these two pieces is $\vec{p}_{12} = \vec{p}_1 + \vec{p}_2 = 30m\hat{i} + 30m\hat{j}$.
The magnitude of this resultant momentum is $p_{12} = \sqrt{(30m)^2 + (30m)^2} = 30m\sqrt{2}$.
For the total momentum to be zero,the momentum of the third piece $\vec{p}_3$ must be equal and opposite to $\vec{p}_{12}$.
Thus,$3m \cdot V = 30m\sqrt{2}$,where $V$ is the magnitude of the velocity of the third piece.
$V = \frac{30m\sqrt{2}}{3m} = 10\sqrt{2} \, m/s$.
The direction of the third piece must be opposite to the resultant of the first two pieces. Since the resultant of the two perpendicular pieces makes an angle of $45^{\circ}$ with each,the third piece must make an angle of $180^{\circ} - 45^{\circ} = 135^{\circ}$ with each of the first two pieces.

(B) According to the law of conservation of linear momentum,the total initial momentum of the system must be equal to the total final momentum.
Since the bomb is initially at rest,the initial momentum $P_i = 0$.
Let the masses of the two fragments be $m_1$ and $m_2$,and their velocities be $v_1$ and $v_2$ respectively.
$m_1 v_1 + m_2 v_2 = 0$
Given that one fragment moves vertically upwards with velocity $v_0$,let $v_1 = v_0$ (upward direction is positive).
$m_1 v_0 + m_2 v_2 = 0$
$v_2 = -(m_1 / m_2) v_0$
The negative sign indicates that the second fragment must move in the opposite direction to the first fragment,which is vertically downwards.

(C) According to Newton's third law,internal forces always exist in equal and opposite pairs.
Since the net internal force $\vec{F}_{int} = \sum \vec{F}_{ij} = 0$,the rate of change of linear momentum $\frac{d\vec{P}}{dt} = \vec{F}_{ext} + \vec{F}_{int} = \vec{F}_{ext}$.
If only internal forces act on the system,$\vec{F}_{ext} = 0$,which implies $\frac{d\vec{P}}{dt} = 0$.
Therefore,the linear momentum $\vec{P}$ of the system remains constant,meaning it is conserved.

(D) The law of conservation of momentum states that in the absence of any external force acting on a system,the total momentum of the system remains constant.
During a collision between two objects,the internal forces are much larger than any external forces (like friction or gravity) acting over the short duration of the collision.
Therefore,the total linear momentum of the system is always conserved during a collision,regardless of whether the collision is elastic or inelastic.
Thus,the correct option is $D$.

(D) The total mass of the bomb is $M = 60 \ kg$. It is at rest,so the initial momentum is $P_i = 0$.
By the law of conservation of linear momentum,the final momentum must also be zero: $m_1 v_1 + m_2 v_2 = 0$,which implies $m_1 v_1 = -m_2 v_2$.
Taking magnitudes,$p_1 = p_2 = p$.
The kinetic energy $K$ is related to momentum $p$ by $K = \frac{p^2}{2m}$.
Thus,$\frac{K_1}{K_2} = \frac{p^2 / 2m_1}{p^2 / 2m_2} = \frac{m_2}{m_1}$.
Given $m_1 = 40 \ kg$ and $m_2 = 60 \ kg - 40 \ kg = 20 \ kg$.
Given $K_1 = 96 \ J$.
Substituting the values: $\frac{96}{K_2} = \frac{20}{40} = \frac{1}{2}$.
Therefore,$K_2 = 96 \times 2 = 192 \ J$.

(A) According to the law of conservation of linear momentum,the total momentum before firing is equal to the total momentum after firing.
Initial momentum = $0$.
Final momentum = $m_b v_b + m_g v_g = 0$.
Here,$m_b = 50 \ g = 0.05 \ kg$,$v_b = 30 \ m \ s^{-1}$,and $v_g = -1 \ m \ s^{-1}$ (recoil velocity).
$0.05 \times 30 + m_g \times (-1) = 0$.
$1.5 - m_g = 0$.
$m_g = 1.5 \ kg$.

(D) According to the law of conservation of momentum,the momentum of the rifle and the bullet must be equal and opposite,i.e.,$m_b v_b = M_r V_r$,where $m_b$ and $v_b$ are the mass and velocity of the bullet,and $M_r$ and $V_r$ are the mass and recoil velocity of the rifle.
Since the mass of the rifle $M_r$ is much greater than the mass of the bullet $m_b$ $(M_r \gg m_b)$,the recoil velocity of the rifle $V_r$ is much smaller than the velocity of the bullet $v_b$.
The kinetic energy $K$ is given by $K = \frac{p^2}{2m}$,where $p$ is the momentum.
Since the magnitudes of their momenta are equal $(p_b = p_r = p)$,the kinetic energy is inversely proportional to the mass $(K \propto \frac{1}{m})$.
Because $M_r > m_b$,it follows that $K_r < K_b$.
Therefore,the kinetic energy of the recoiling rifle is less than the kinetic energy of the bullet.

(A) Let the mass of the shell be $2m$. At the highest point,the velocity of the shell is $v_x = v \cos \theta$ in the horizontal direction.
According to the law of conservation of linear momentum,the total momentum before the explosion must equal the total momentum after the explosion.
Initial momentum $P_i = (2m)(v \cos \theta)$.
After the explosion,one piece of mass $m$ retraces its path,meaning its velocity is $-v \cos \theta$.
Let the velocity of the other piece of mass $m$ be $v'$.
Applying conservation of momentum: $(2m)(v \cos \theta) = m(-v \cos \theta) + m(v')$.
Dividing by $m$: $2v \cos \theta = -v \cos \theta + v'$.
Therefore,$v' = 3v \cos \theta$.

(C) In a gravity-free room,the center of mass of the system remains at rest because there are no external forces acting on the system.
Let the man move upwards with velocity $v$ when he throws the ball downwards with velocity $u$.
By the law of conservation of linear momentum,the initial momentum is zero:
$m_1 v - m_2 u = 0 \implies v = \frac{m_2 u}{m_1}$.
The time taken by the ball to reach the floor is $t = \frac{h}{u}$.
During this time,the man moves upwards by a distance $d = v \times t = \left( \frac{m_2 u}{m_1} \right) \times \left( \frac{h}{u} \right) = \frac{m_2}{m_1} h$.
The initial height of the man was $h$. Since he moves upwards by $\frac{m_2}{m_1} h$,his new distance from the floor is $h + \frac{m_2}{m_1} h = \left( 1 + \frac{m_2}{m_1} \right) h$.