Conservation of Linear Momentum Questions in English

(D) The system consists of the trolley and the sandbag. Since the track is frictionless,there is no external horizontal force acting on the system.
According to the law of conservation of linear momentum,the total momentum of the system remains constant.
Since the sand is falling inside the trolley,the mass of the system (trolley + sandbag + sand) remains constant as a whole.
Because there is no external force and the total mass of the system moving at the initial velocity remains constant,the velocity of the trolley does not change.
Therefore,the speed of the trolley remains $27 \ km/hr$.

(C) According to the law of conservation of linear momentum,the total momentum before the split equals the total momentum after the split.
Initial momentum: $\vec{P}_i = m v_0 \hat{i}$.
Final momentum: $\vec{P}_f = (m/3)(2v_0 \hat{j}) + (2m/3)\vec{v}$,where $\vec{v}$ is the velocity of the remaining part.
Equating $\vec{P}_i = \vec{P}_f$:
$m v_0 \hat{i} = \frac{2m v_0}{3} \hat{j} + \frac{2m}{3} \vec{v}$.
Dividing by $m$ and rearranging:
$v_0 \hat{i} - \frac{2}{3} v_0 \hat{j} = \frac{2}{3} \vec{v}$.
Multiplying by $3/2$:
$\vec{v} = \frac{3}{2} v_0 \hat{i} - v_0 \hat{j}$.

(A) Let the velocities of the fragments be $v_1$ and $v_2$ for the $3 \ kg$ and $6 \ kg$ masses respectively. Since the initial momentum is zero,by the law of conservation of linear momentum: $3v_1 = 6v_2$,which implies $v_1 = 2v_2$.
The kinetic energy of the $6 \ kg$ fragment is given by $KE_2 = \frac{1}{2} \times 6 \times v_2^2 = 120 \ J$.
This simplifies to $3v_2^2 = 120$,so $v_2^2 = 40 \ (m/s)^2$.
The kinetic energy of the $3 \ kg$ fragment is $KE_1 = \frac{1}{2} \times 3 \times v_1^2$.
Substituting $v_1 = 2v_2$,we get $KE_1 = \frac{1}{2} \times 3 \times (2v_2)^2 = \frac{1}{2} \times 3 \times 4v_2^2 = 6v_2^2$.
Since $v_2^2 = 40$,$KE_1 = 6 \times 40 = 240 \ J$.

(B) According to the law of conservation of linear momentum,the total initial momentum of the system is zero because both the gun and the bullet are initially at rest.
Initial momentum $P_i = 0$.
Final momentum $P_f = M V_G + m v$,where $V_G$ is the recoil velocity of the gun and $v$ is the velocity of the bullet.
Since $P_i = P_f$,we have $0 = M V_G + mv$.
Rearranging the equation to solve for the recoil velocity of the gun $V_G$:
$M V_G = -mv$
$V_G = -\frac{mv}{M}$.
The magnitude of the recoil velocity is $\frac{mv}{M}$.

(B) According to the law of conservation of linear momentum,the initial momentum of the bomb is zero since it is at rest.
$m_1 v_1 + m_2 v_2 = 0$
Given $m_1 = 4 \ kg$,$m_2 = 12 \ kg$,and $v_2 = 4 \ m/s$.
$4 \times v_1 + 12 \times 4 = 0$
$4 v_1 = -48$
$v_1 = -12 \ m/s$ (The negative sign indicates the direction is opposite to the $12 \ kg$ fragment).
The kinetic energy of the $4 \ kg$ fragment is given by $K = \frac{1}{2} m_1 v_1^2$.
$K = \frac{1}{2} \times 4 \times (-12)^2$
$K = 2 \times 144 = 288 \ J$.

(D) According to the law of conservation of linear momentum,the total initial momentum of the system is equal to the total final momentum.
Initial momentum: $\vec{P}_i = m_1\vec{u}_1 + m_2\vec{u}_2 + m_3\vec{u}_3$
$\vec{P}_i = 10(10\hat{i}) + 20(10\hat{j}) + 40(10\hat{k}) = 100\hat{i} + 200\hat{j} + 400\hat{k}$
Final momentum: $\vec{P}_f = m_1\vec{v}_1 + m_2\vec{v}_2 + m_3\vec{v}_3$
Given $\vec{v}_1 = 0$ and $\vec{v}_2 = (3\hat{i} + 4\hat{j})\;m/s$.
$\vec{P}_f = 10(0) + 20(3\hat{i} + 4\hat{j}) + 40\vec{v}_3 = 60\hat{i} + 80\hat{j} + 40\vec{v}_3$
Equating $\vec{P}_i = \vec{P}_f$:
$100\hat{i} + 200\hat{j} + 400\hat{k} = 60\hat{i} + 80\hat{j} + 40\vec{v}_3$
$40\vec{v}_3 = (100 - 60)\hat{i} + (200 - 80)\hat{j} + 400\hat{k}$
$40\vec{v}_3 = 40\hat{i} + 120\hat{j} + 400\hat{k}$
$\vec{v}_3 = \hat{i} + 3\hat{j} + 10\hat{k}\;m/s$.

(A) Given: Total mass $M = 12 \ kg$. Mass ratio $m_1:m_2 = 1:3$.
Thus,$m_1 = 3 \ kg$ and $m_2 = 9 \ kg$.
Since the bomb is initially at rest,the initial momentum is zero. By the law of conservation of momentum,the magnitudes of the momenta of the two fragments must be equal: $p_1 = p_2 = p$.
The kinetic energy $K$ is related to momentum $p$ by $K = \frac{p^2}{2m}$,so $p = \sqrt{2mK}$.
For the smaller fragment $(m_1 = 3 \ kg)$: $p = \sqrt{2 \times 3 \times 216} = \sqrt{1296} = 36 \ kg \cdot m/s$.
Since the magnitudes of the momenta are equal,the momentum of the larger fragment is also $36 \ kg \cdot m/s$.

(A) Let the velocity of the platform relative to the ice be $V$ in the backward direction and the velocity of the man relative to the ice be $w$ in the forward direction.
Given that the velocity of the man relative to the platform is $v$,we have $w + V = v$,which implies $w = v - V$.
Since there is no external horizontal force acting on the system (man + platform),the linear momentum of the system is conserved.
Initially,both are at rest,so the initial momentum is $0$.
Applying the law of conservation of linear momentum: $MV - mw = 0$.
Substituting $w = v - V$ into the equation: $MV - m(v - V) = 0$.
$MV - mv + mV = 0$.
$V(M + m) = mv$.
Therefore,the velocity of the platform relative to the ice is $V = \frac{mv}{M + m}$.

(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 $P_i = MV$.
After the explosion,one part of mass $m$ is at rest (velocity $= 0$),and the remaining part has mass $(M - m)$ moving with velocity $v$.
Final momentum $P_f = m(0) + (M - m)v = (M - m)v$.
Equating $P_i = P_f$:
$MV = (M - m)v$.
Therefore,the velocity of the other part is $v = \frac{MV}{M - m}$.

(C) The mass of the third fragment is $M - (M/4 + M/4) = M/2$.
Let the velocities of the first two fragments be $\vec{v}_1 = 3\hat{i} \ m/s$ and $\vec{v}_2 = 4\hat{j} \ m/s$.
The momentum of the first fragment is $\vec{p}_1 = (M/4)(3\hat{i}) = (3M/4)\hat{i}$.
The momentum of the second fragment is $\vec{p}_2 = (M/4)(4\hat{j}) = M\hat{j}$.
According to the law of conservation of linear momentum,the total initial momentum is zero,so $\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0$.
Therefore,$\vec{p}_3 = -(\vec{p}_1 + \vec{p}_2) = -(3M/4)\hat{i} - M\hat{j}$.
The magnitude of the momentum of the third fragment is $p_3 = \sqrt{(3M/4)^2 + M^2} = \sqrt{9M^2/16 + M^2} = \sqrt{25M^2/16} = 5M/4$.
Since $p_3 = (M/2)v_3$,we have $(M/2)v_3 = 5M/4$.
Solving for $v_3$,we get $v_3 = (5M/4) \times (2/M) = 2.5 \ m/s$.

(A) Let the mass of each of the two fragments be $m$ and their speed be $u = 30 \; m/s$. The mass of the third fragment is $3m$. Let its velocity be $v$ at an angle $\theta$ with the direction opposite to the resultant momentum of the first two fragments.
By the law of conservation of linear momentum,the initial momentum is zero,so the final momentum must also be zero.
The resultant momentum of the two fragments moving perpendicularly is $P_r = \sqrt{(mu)^2 + (mu)^2} = mu\sqrt{2}$.
This resultant momentum is directed at $45^\circ$ to each of the two fragments.
The third fragment must move in the opposite direction to balance this,so its momentum $3mv$ must be equal in magnitude to $mu\sqrt{2}$.
$3mv = mu\sqrt{2} \Rightarrow v = \frac{u\sqrt{2}}{3} = \frac{30\sqrt{2}}{3} = 10\sqrt{2} \; m/s$.
The direction is $180^\circ - 45^\circ = 135^\circ$ relative to the direction of each of the first two fragments.

(D) The total mass is $5 \ kg$. The masses of the fragments are $m_1 = 1 \ kg$,$m_2 = 1 \ kg$,and $m_3 = 3 \ kg$.
Since the body is initially at rest,the initial momentum is zero. By the law of conservation of linear momentum,the final vector sum of the momenta must be zero: $\vec{P}_1 + \vec{P}_2 + \vec{P}_3 = 0$.
Let the two $1 \ kg$ fragments move along the $X$ and $Y$ axes respectively: $\vec{P}_1 = (1 \ kg)(21 \ m/s) \hat{i} = 21 \hat{i} \ kg \cdot m/s$ and $\vec{P}_2 = (1 \ kg)(21 \ m/s) \hat{j} = 21 \hat{j} \ kg \cdot m/s$.
The momentum of the third fragment must be $\vec{P}_3 = -(\vec{P}_1 + \vec{P}_2) = -(21 \hat{i} + 21 \hat{j}) \ kg \cdot m/s$.
The magnitude of the momentum of the third fragment is $|\vec{P}_3| = \sqrt{21^2 + 21^2} = 21\sqrt{2} \ kg \cdot m/s$.
Since $|\vec{P}_3| = m_3 \cdot v_3$,we have $3 \cdot v_3 = 21\sqrt{2}$.
$v_3 = 7\sqrt{2} \approx 7 \times 1.414 = 9.898 \ m/s$.

(B) The mass of one piece is $m$. Therefore,the mass of the other piece is $(M - m)$.
Let the velocity of the other piece be $v'$. According to the law of conservation of linear momentum:
Initial momentum of the spacecraft = Final momentum of the pieces
$Mv = m(0) + (M - m)v'$
$\therefore v' = \frac{M}{M - m} v$

(B) According to the law of conservation of linear momentum,the total initial momentum is zero since the body is stationary.
$0 = 2p\,\hat{i} + p\,\hat{j} + \vec{p_3}$
Therefore,the momentum of the third piece is:
$\vec{p_3} = -2p\,\hat{i} - p\,\hat{j}$
The magnitude of the momentum of the third piece is:
$|\vec{p_3}| = \sqrt{(-2p)^2 + (-p)^2} = \sqrt{4p^2 + p^2} = \sqrt{5p^2} = \sqrt{5}p$
Thus,the correct option is $B$.

(C) Let the mass of the person be $m_p = 80 \ kg$ and the mass of the trolley be $m_t = 320 \ kg$.
Since there is no external horizontal force on the system,the total momentum of the system remains zero.
Let $v_p$ be the velocity of the person relative to the trolley $(1 \ m/s)$ and $v_t$ be the velocity of the trolley relative to the ground.
The velocity of the person relative to the ground is $v_g = v_p - v_t$.
According to the conservation of linear momentum: $m_p v_g + m_t v_t = 0$.
$80(1 - v_t) + 320 v_t = 0$.
$80 - 80 v_t + 320 v_t = 0$.
$240 v_t = -80 \implies v_t = -1/3 \ m/s$ (The trolley moves backward).
The velocity of the person relative to the ground is $v_g = 1 - 1/3 = 2/3 \ m/s$.
Displacement relative to the ground after $t = 4 \ s$ is $d = v_g \times t = (2/3) \times 4 = 8/3 \approx 2.67 \ m$.
Wait,re-evaluating the standard interpretation: If the person walks at $1 \ m/s$ relative to the trolley,the velocity of the center of mass remains zero. The displacement of the person relative to the ground is $d_p = v_g \times t$. Given the options provided,the calculation $0.8 \times 4 = 3.2$ suggests the velocity of the trolley is $0.2 \ m/s$ derived from $80 \times 1 = (80+320)v_t$. Thus,$v_g = 1 - 0.2 = 0.8 \ m/s$. Displacement = $0.8 \times 4 = 3.2 \ m$.

(C) Since the bomb is initially at rest,the initial momentum is zero.
According to the law of conservation of linear momentum,the final momentum must also be zero.
Let the mass of each fragment be $m$. The total mass of the bomb is $3m$.
Let the velocities of the three fragments be $\vec{v}_1 = 9\hat{i} \ m s^{-1}$,$\vec{v}_2 = 12\hat{j} \ m s^{-1}$,and $\vec{v}_3$.
Applying the conservation of momentum: $m\vec{v}_1 + m\vec{v}_2 + m\vec{v}_3 = 0$.
Dividing by $m$,we get $\vec{v}_1 + \vec{v}_2 + \vec{v}_3 = 0$.
$\vec{v}_3 = -(\vec{v}_1 + \vec{v}_2) = -(9\hat{i} + 12\hat{j})$.
The magnitude of the velocity of the third fragment is $|\vec{v}_3| = \sqrt{(-9)^2 + (-12)^2}$.
$|\vec{v}_3| = \sqrt{81 + 144} = \sqrt{225} = 15 \ m s^{-1}$.

(C) According to the law of conservation of linear momentum,the initial momentum is equal to the final momentum.
Initial mass $m_1 = 1000\, kg$,initial velocity $u_1 = 50\, km/h$.
Final mass $m_2 = 1000\, kg + 250\, kg = 1250\, kg$.
Let the new velocity be $v$.
$m_1 u_1 = m_2 v$
$1000 \times 50 = 1250 \times v$
$v = \frac{50000}{1250} = 40\, km/h$.

(C) Let $m = 0.2\, kg$ be the mass of the shell and $M = 4\, kg$ be the mass of the gun. Let $v$ be the velocity of the shell and $V$ be the recoil velocity of the gun.
By the law of conservation of momentum,$mv = MV$,so $V = (m/M)v$.
The total energy generated by the explosion is equal to the sum of the kinetic energies of the shell and the gun:
$E = \frac{1}{2}mv^2 + \frac{1}{2}MV^2$
Substituting $V = (m/M)v$:
$E = \frac{1}{2}mv^2 + \frac{1}{2}M(\frac{m}{M}v)^2 = \frac{1}{2}mv^2 (1 + \frac{m}{M})$
Given $E = 1.05\, kJ = 1050\, J$,$m = 0.2\, kg$,and $M = 4\, kg$:
$1050 = \frac{1}{2} \times 0.2 \times v^2 \times (1 + \frac{0.2}{4})$
$1050 = 0.1 \times v^2 \times (1 + 0.05)$
$1050 = 0.1 \times 1.05 \times v^2$
$1050 = 0.105 \times v^2$
$v^2 = \frac{1050}{0.105} = 10000$
$v = 100\, m/s$.

(D) According to the law of conservation of linear momentum,the initial momentum of the rock is zero. Therefore,the vector sum of the momenta of the three parts must be zero.
Let the momenta of the three parts be $\vec{p}_1$,$\vec{p}_2$,and $\vec{p}_3$.
$\vec{p}_1 + \vec{p}_2 + \vec{p}_3 = 0 \implies \vec{p}_3 = -(\vec{p}_1 + \vec{p}_2)$.
The magnitude of the momentum of the first part is $p_1 = m_1 v_1 = 1 \, kg \times 12 \, m s^{-1} = 12 \, kg \, m s^{-1}$.
The magnitude of the momentum of the second part is $p_2 = m_2 v_2 = 2 \, kg \times 8 \, m s^{-1} = 16 \, kg \, m s^{-1}$.
Since the two parts move at right angles to each other,the magnitude of the resultant momentum of these two parts is $p_{12} = \sqrt{p_1^2 + p_2^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, kg \, m s^{-1}$.
The third part must have a momentum equal in magnitude and opposite in direction to this resultant momentum to satisfy the conservation law.
Thus,$p_3 = p_{12} = 20 \, kg \, m s^{-1}$.
Given the speed of the third part $v_3 = 4 \, m s^{-1}$,its mass $m_3$ is:
$m_3 = \frac{p_3}{v_3} = \frac{20 \, kg \, m s^{-1}}{4 \, m s^{-1}} = 5 \, kg$.

(C) According to the law of conservation of linear momentum,since the initial momentum is zero,the magnitudes of the momenta of the two pieces must be equal: $p_1 = p_2 = p$.
The kinetic energy $K$ of a body is related to its momentum $p$ and mass $m$ by the formula $K = \frac{p^2}{2m}$.
Since $p$ is the same for both pieces,the kinetic energy is inversely proportional to the mass: $K \propto \frac{1}{m}$.
Let $E_1$ be the kinetic energy of the piece with mass $m_1 = 2M$ and $E_2$ be the kinetic energy of the piece with mass $m_2 = 3M$.
Then,$\frac{E_1}{E_2} = \frac{m_2}{m_1} = \frac{3M}{2M} = \frac{3}{2}$.
Given the total kinetic energy $E = E_1 + E_2$,we can write:
$E_1 = \left( \frac{m_2}{m_1 + m_2} \right) E = \left( \frac{3M}{2M + 3M} \right) E = \left( \frac{3M}{5M} \right) E = \frac{3E}{5}$.

(A) Initially,the bomb 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 $v_A$ be the velocity of mass $m$ and $v_B$ be the velocity of mass $2m$.
$m v_A + 2m v_B = 0$
Given $v_A = 16 \text{ m/s}$,we have $m(16) + 2m v_B = 0$.
$2m v_B = -16m \implies v_B = -8 \text{ m/s}$.
The negative sign indicates that the two pieces move in opposite directions.
The total kinetic energy $K$ released is the sum of the kinetic energies of the two pieces:
$K = \frac{1}{2} m v_A^2 + \frac{1}{2} (2m) v_B^2$
$K = \frac{1}{2} m (16)^2 + \frac{1}{2} (2m) (-8)^2$
$K = \frac{1}{2} m (256) + m (64)$
$K = 128m + 64m = 192m \text{ J}$.
Thus,the total kinetic energy released is $192m \text{ J}$.

(B) Let the total mass of the spaceship be $m$. The initial momentum is $\vec{P}_i = m v_0 \hat{j}$.
After the explosion,the spaceship splits into two parts: $m_1 = \frac{m}{4}$ and $m_2 = \frac{3m}{4}$.
The velocity of the first part is $\vec{v}_1 = 2v_0 \hat{i}$.
Let the velocity of the remaining part be $\vec{v}_2 = v_x \hat{i} + v_y \hat{j}$.
By the law of conservation of linear momentum,$\vec{P}_i = \vec{P}_f$,so $m v_0 \hat{j} = \frac{m}{4}(2v_0 \hat{i}) + \frac{3m}{4}(v_x \hat{i} + v_y \hat{j})$.
Equating the components:
Along $x$-axis: $0 = \frac{2m v_0}{4} + \frac{3m v_x}{4} \Rightarrow 3v_x = -2v_0 \Rightarrow v_x = -\frac{2}{3}v_0$.
Along $y$-axis: $m v_0 = \frac{3m v_y}{4} \Rightarrow v_y = \frac{4}{3}v_0$.
The magnitude of the velocity of the remaining part is $v_2 = \sqrt{v_x^2 + v_y^2} = \sqrt{(-\frac{2}{3}v_0)^2 + (\frac{4}{3}v_0)^2} = \sqrt{\frac{4}{9}v_0^2 + \frac{16}{9}v_0^2} = \sqrt{\frac{20}{9}}v_0 = \frac{\sqrt{20}}{3}v_0$.

(C) Case $I$: Simultaneous jump.
Let $V_1$ be the velocity of the buggy. The velocity of each man relative to the ground is $(10 - V_1)$.
Using conservation of linear momentum: $0 = 100 V_1 - 2 \times 50(10 - V_1)$.
$100 V_1 = 100(10 - V_1) \implies V_1 = 10 - V_1 \implies 2 V_1 = 10 \implies V_1 = 5\, m/s$.
Case $II$: Sequential jump.
First man jumps: $0 = (100 + 50) v_a - 50(10 - v_a) \implies 150 v_a = 500 - 50 v_a \implies 200 v_a = 500 \implies v_a = 2.5\, m/s$.
Second man jumps from the buggy (now mass $150\, kg$): $150 v_a = 100 V_2 - 50(10 - V_2)$.
$375 = 100 V_2 - 500 + 50 V_2 \implies 150 V_2 = 875 \implies V_2 = 875 / 150 = 35 / 6 \approx 5.83\, m/s$.
Ratio $V_1 : V_2 = 5 : (35/6) = 30 : 35 = 6 : 7$. The ratio of recoil velocities (sequential to simultaneous) is $7 : 6$.

(C) Given: Initial mass of the wagon,$m = 5 \times 10^3 \text{ kg}$.
Initial velocity of the wagon,$v = 1.2 \text{ m/s}$.
After collecting rain drops,the total mass of the wagon becomes $m' = m + 10^3 \text{ kg} = 6 \times 10^3 \text{ kg}$.
Since the rain drops fall vertically,they have no horizontal component of momentum. Therefore,the horizontal momentum of the system is conserved.
According to the law of conservation of linear momentum: $m \times v = m' \times v'$.
Substituting the values: $(5 \times 10^3) \times 1.2 = (6 \times 10^3) \times v'$.
$v' = \frac{5 \times 10^3 \times 1.2}{6 \times 10^3}$.
$v' = \frac{5 \times 1.2}{6} = 1 \text{ m/s}$.

(C) The system is on a frictionless table,so there is no external force acting in the $x$-direction. The initial velocity of the system in the $x$-direction is zero,so the total momentum in the $x$-direction must remain zero throughout the motion.
Let $v_{x_A}$ and $v_{x_B}$ be the $x$-components of the velocities of masses $A$ and $B$ respectively.
Applying the Conservation of Linear Momentum in the $x$-direction:
$M v_{x_A} + (2M) v_{x_B} = 0$
$M v_{x_A} = -2M v_{x_B}$
$v_{x_A} = -2 v_{x_B}$
Given that the $x$-component of the velocity of mass $B$ is $V_x$,we have $v_{x_B} = V_x$.
Therefore,$v_{x_A} = -2 V_x$.
The velocity in the $z$-direction remains constant at $u \hat{k}$ for both masses because there are no external forces in the $z$-direction.
Thus,the velocity of mass $A$ is $\vec{V}_A = -2 V_x \hat{i} + u \hat{k}$.

(B) Since there are no external horizontal forces acting on the system (ball + spring gun),the linear momentum of the system is conserved.
Let $V$ be the final velocity of the combined system (ball + spring gun) after the ball stops relative to the barrel.
According to the law of conservation of linear momentum:
$m v_0 + M(0) = (m + M) V$
Given:
$m = 60 \text{ g} = 0.06 \text{ kg}$
$M = 240 \text{ g} = 0.24 \text{ kg}$
$v_0 = 22 \text{ m/s}$
Substituting the values:
$0.06 \times 22 + 0 = (0.06 + 0.24) V$
$1.32 = 0.30 V$
$V = \frac{1.32}{0.30} = 4.4 \text{ m/s}$
Thus,the speed of the spring gun after the ball stops relative to the barrel is $4.4 \text{ m/s}$.

(B) According to Newton's Second Law,the rate of change of total momentum $\vec{P}$ of a system is equal to the net external force acting on it: $\frac{d\vec{P}}{dt} = \vec{F}_{ext}$.
Since the system is free from any external forces,$\vec{F}_{ext} = 0$.
Therefore,$\frac{d\vec{P}}{dt} = 0$,which implies that the total momentum $\vec{P}$ of the system is constant.
Internal forces act in action-reaction pairs and cancel each other out,so they do not change the total momentum of the system.
Thus,the total momentum could be non-zero (if the system has an initial velocity),but it must remain constant over time.

(C) To determine if momentum is conserved,we analyze the external forces acting on the system consisting of the rail car and the bowling balls.
$1$. In the horizontal direction (parallel to the track),there are no external forces acting on the system because the track is frictionless and the rail car is isolated.
$2$. According to Newton's Second Law,if the net external force on a system is zero,the total linear momentum of the system remains constant.
$3$. Therefore,the horizontal component of the momentum is conserved.
$4$. In the vertical direction,there is an external force acting on the system: the gravitational force (weight) and the normal force from the track. As the balls are dropped,the system experiences a change in vertical momentum due to gravity and the impact with the car floor.
$5$. Since the horizontal component of the net external force is zero,the momentum component parallel to the track is conserved.

(B) Since the track is frictionless and there is no external horizontal force acting on the system (rail car $+$ bowling balls) in the direction of motion,the linear momentum of the system is conserved in that direction.
Let $V$ be the final velocity of the system.
Initial momentum of the system $= M \times v_0 + (N \times m) \times 0 = Mv_0$.
Final momentum of the system $= (M + Nm) \times V$.
By the law of conservation of linear momentum:
$Mv_0 = (M + Nm) \times V$.
Solving for $V$,we get $V = \frac{Mv_0}{M + Nm}$.

(B) Let $v_1$ be the velocity of the man with respect to the ground and $v_2$ be the velocity of the trolley with respect to the ground.
Since the system is on a smooth horizontal surface,there is no external horizontal force acting on the system.
Therefore,the momentum of the system is conserved.
Initially,the system is at rest,so the initial momentum is $0$.
$m v_1 + M v_2 = 0$
$v_1 = -\frac{M}{m} v_2$
We are given the relative velocity of the man with respect to the trolley as $u_{rel} = v_1 - v_2$.
Substituting $v_1$ in the relative velocity equation:
$u_{rel} = -\frac{M}{m} v_2 - v_2$
$u_{rel} = -v_2 \left( \frac{M}{m} + 1 \right)$
$u_{rel} = -v_2 \left( \frac{M + m}{m} \right)$
$v_2 = -\frac{m u_{rel}}{M + m}$
The magnitude of the velocity of the trolley is $\frac{m u_{rel}}{m + M}$.

(A) Let $v_1$ be the velocity of the man and $v_2$ be the velocity of the trolley with respect to the ground.
Since the system is on a smooth horizontal surface,there is no external horizontal force acting on the system. Therefore,the momentum of the system is conserved.
Initially,the system is at rest,so the initial momentum is $0$.
$m v_1 + M v_2 = 0 \implies v_2 = -\frac{m}{M} v_1$.
The relative velocity of the man with respect to the trolley is given by $u_{rel} = v_1 - v_2$.
Substituting $v_2$ in the equation: $u_{rel} = v_1 - (-\frac{m}{M} v_1) = v_1 (1 + \frac{m}{M}) = v_1 (\frac{M + m}{M})$.
Solving for $v_1$,we get $v_1 = \frac{M u_{rel}}{M + m}$.

(C) Since the system (man + trolley) is on a smooth horizontal surface,there is no external horizontal force acting on it. Therefore,the center of mass of the system remains at rest.
Let $v_m$ be the velocity of the man and $v_t$ be the velocity of the trolley with respect to the ground.
According to the conservation of linear momentum,$m v_m + M v_t = 0$,which implies $v_t = -\frac{m}{M} v_m$.
The velocity of the man with respect to the trolley is given as $u_{rel} = v_m - v_t$.
Substituting $v_t$,we get $u_{rel} = v_m - (-\frac{m}{M} v_m) = v_m (1 + \frac{m}{M}) = v_m \left( \frac{M+m}{M} \right)$.
Thus,the velocity of the man with respect to the ground is $v_m = u_{rel} \left( \frac{M}{M+m} \right)$.
The time taken by the man to cover the length $L$ of the trolley is $t = \frac{L}{u_{rel}}$.
Since the trolley also moves,the man covers the distance $L$ relative to the trolley at a relative speed $u_{rel}$.
Therefore,the time taken is $t = \frac{L}{u_{rel}}$.

(D) Let the velocity of the trolley (along with the person at $B$) be $v$ towards the right after the person at $A$ jumps. The velocity of the person at $A$ with respect to the ground is $v_1 = v - u_{rel}$ (towards the left).
Since there is no external horizontal force on the system,the linear momentum of the system is conserved.
Initial momentum $P_i = 0$.
Final momentum $P_f = m_1(v - u_{rel}) + (m_2 + M)v = 0$.
$m_1 v - m_1 u_{rel} + m_2 v + M v = 0$.
$(m_1 + m_2 + M)v = m_1 u_{rel}$.
$v = \frac{m_1 u_{rel}}{m_1 + m_2 + M}$.
Since $v > 0$,the trolley moves towards the right.
Also,since no external force acts on the system,the centre of mass remains stationary.
Thus,all the given statements are correct.

(D) Initially,the system (trolley + two persons) is at rest,so the initial momentum of the system is zero.
Since no external horizontal force acts on the system,the total momentum of the system must remain conserved (i.e.,zero).
When the person of mass $m_2$ jumps towards the right with a relative velocity $u_{rel}$ with respect to the trolley,let the velocity of the trolley (and the person $m_1$) be $v$ towards the left.
The velocity of the person $m_2$ with respect to the ground is $v_{m2} = u_{rel} - v$ towards the right.
Applying the conservation of linear momentum: $0 = (m_1 + M)(-v) + m_2(u_{rel} - v)$.
$0 = -m_1 v - Mv + m_2 u_{rel} - m_2 v$.
$m_2 u_{rel} = (m_1 + m_2 + M)v$.
$v = \frac{m_2 u_{rel}}{m_1 + m_2 + M}$.
Since $v$ is towards the left,the trolley moves towards the left.
Also,as there is no external force,the centre of mass of the system remains stationary.
Thus,all the given statements are correct.

(A) Let the velocity of the trolley be $v$ in the direction of the person with mass $m_1$.
Since the system is initially at rest,the total external force on the system (trolley + two persons) is zero. Thus,the linear momentum of the system is conserved.
Let the velocity of the trolley be $v$ (towards the right).
The velocity of person $m_1$ with respect to the ground is $v - u_{rel}$ (assuming they jump towards the left).
The velocity of person $m_2$ with respect to the ground is $v + u_{rel}$ (assuming they jump towards the right).
Applying the law of conservation of linear momentum:
$0 = Mv + m_1(v - u_{rel}) + m_2(v + u_{rel})$
$0 = Mv + m_1v - m_1u_{rel} + m_2v + m_2u_{rel}$
$0 = (M + m_1 + m_2)v + (m_2 - m_1)u_{rel}$
$(M + m_1 + m_2)v = (m_1 - m_2)u_{rel}$
$v = \frac{(m_1 - m_2)u_{rel}}{M + m_1 + m_2}$
Taking the magnitude,the velocity of the trolley is $\frac{|m_1 - m_2|u_{rel}}{M + m_1 + m_2}$.

(A) $1$. Initial velocity components: $u_x = 50 \cos 53^{\circ} = 50 \times (3/5) = 30 \, m/s$ and $u_y = 50 \sin 53^{\circ} = 50 \times (4/5) = 40 \, m/s$.
$2$. At the highest point,the velocity is $v_x = 30 \, m/s$ and $v_y = 0$. The horizontal distance from the origin to the highest point is $R/2 = (u_x^2 \sin 2\theta) / (2g) = (30 \times 40) / 10 = 120 \, m$.
$3$. At the highest point,the mass $m$ splits into two parts $m/2$ and $m/2$. One part comes to rest,so its velocity $v_1 = 0$. By conservation of momentum: $m v_x = (m/2) v_1 + (m/2) v_2 \implies m(30) = 0 + (m/2) v_2 \implies v_2 = 60 \, m/s$.
$4$. The part at rest falls vertically to the ground at $x_1 = 120 \, m$. The second part moves horizontally with $v_2 = 60 \, m/s$ and falls from height $H = u_y^2 / (2g) = 40^2 / 20 = 80 \, m$. Time to fall $t = \sqrt{2H/g} = \sqrt{160/10} = 4 \, s$.
$5$. Horizontal distance of the second part from the highest point is $d = v_2 \times t = 60 \times 4 = 240 \, m$. Its final position is $x_2 = 120 + 240 = 360 \, m$.
$6$. The distance between the two pieces is $|x_2 - x_1| = |360 - 120| = 240 \, m$.

(C) The system consists of the rail car and all the sand. Since the track is frictionless and the rail car is isolated (no external horizontal forces are acting on the system),the net external force on the system in the horizontal direction is zero. According to the law of conservation of linear momentum,if the net external force on a system is zero,the total linear momentum of the system remains constant. As the sand pours out vertically,it carries no horizontal momentum relative to the ground because its horizontal velocity is the same as the rail car at the moment it leaves. Therefore,the total momentum of the rail car plus all the sand (inside and outside) is conserved. Option $(C)$ is correct.

(C) Let the mass and velocity of the first piece be $m_1 = 4 \ kg$ and $v_1,$ and the mass and velocity of the second piece be $m_2 = 12 \ kg$ and $v_2 = 4 \ m s^{-1}.$
Since the bomb is initially at rest,the initial momentum is $0.$
By the law of conservation of linear momentum,the final momentum must also be $0.$
$m_1 v_1 + m_2 v_2 = 0$
$4 v_1 + 12 \times 4 = 0$
$4 v_1 = -48$
$v_1 = -12 \ m s^{-1}$
The magnitude of the velocity is $12 \ m s^{-1}.$
The kinetic energy of the $4 \ kg$ mass is:
$K.E. = \frac{1}{2} m_1 v_1^2$
$K.E. = \frac{1}{2} \times 4 \times (12)^2$
$K.E. = 2 \times 144 = 288 \ J.$

(NONE) During the collision,the impulsive forces acting on the system are:
$1$. The normal force between the ball and the wedge,which acts along the $Y'$ direction.
$2$. The normal force from the ground on the wedge,which acts along the $Y$ direction.
$3$. The impulsive friction force from the ground on the wedge,which acts along the $X$ direction.
For momentum to be conserved in a specific direction,the net external impulsive force in that direction must be zero.
- Along $Y'$: The normal force between the ball and the wedge is an internal force for the $(M+m)$ system,but it is an external impulsive force for the individual systems $m$ or $M$. Thus,momentum is not conserved for $m$ or $M$ along $Y'$.
- Along $X$: The impulsive friction force from the ground acts on the wedge,so momentum is not conserved for the $(M+m)$ system along $X$.
- Along $Y$: The impulsive normal force from the ground acts on the wedge,so momentum is not conserved for the $(M+m)$ system along $Y$.
Since none of the given conditions result in a net zero external impulsive force,none of the listed options are correct. However,if we consider the direction perpendicular to the impulsive forces,momentum might be conserved. Given the options provided,there is no correct choice among them.

(B) Since the shell is initially at rest,the initial momentum is zero. According to the law of conservation of linear momentum,the final vector sum of the momenta of the fragments must be zero.
Let $\vec{p}_1$,$\vec{p}_2$,and $\vec{p}_m$ be the momenta of the three fragments.
$\vec{p}_1 = 1 \times 5 \hat{i} = 5 \hat{i} \ kg \cdot m/s$
$\vec{p}_2 = 2 \times 6 \hat{j} = 12 \hat{j} \ kg \cdot m/s$
$\vec{p}_m = m \vec{v}_m$
Since $\vec{p}_1 + \vec{p}_2 + \vec{p}_m = 0$,we have $\vec{p}_m = -(\vec{p}_1 + \vec{p}_2) = -(5 \hat{i} + 12 \hat{j})$.
The magnitude of the momentum of the $m \ kg$ piece is $|\vec{p}_m| = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \ kg \cdot m/s$.
Given the speed of the $m \ kg$ piece is $6.5 \ m/s$,we have $m \times 6.5 = 13$.
Thus,$m = \frac{13}{6.5} = 2 \ kg$.
The total mass of the shell is $M = 1 + 2 + m = 1 + 2 + 2 = 5 \ kg$.

(A) Let the mass of the block be $m$ and the mass of the wedge be $m$. Initially, the block has velocity $u$ and the wedge is at rest.
Since the surface is smooth and there are no external horizontal forces, the linear momentum of the system is conserved in the horizontal direction.
When the block reaches the maximum height $h$ relative to the wedge, both the block and the wedge move with the same horizontal velocity $v$.
By conservation of momentum: $mu = (m + m)v$, which gives $v = u/2$.
However, the question asks for the final velocity of the wedge after the block has moved back down to the horizontal plane.
Since the collision is elastic (smooth surfaces, no energy loss), the system behaves like a one-dimensional elastic collision between two equal masses.
In an elastic collision between two bodies of equal mass where one is initially at rest, they exchange their velocities.
Therefore, the block comes to rest and the wedge moves with the initial velocity $u$ of the block.

(C) Let the mass of the bomb be $3m$. At the topmost point,its velocity is $0$. After the explosion,the center of mass continues to follow the original trajectory of the bomb,which is a free fall from height $h$ under gravity $g$.
Let $v_1, v_2, v_3$ be the velocities of the three fragments. By conservation of momentum,$m v_1 + m v_2 + m v_3 = 0$. Since two fragments take $20 \ s$ to reach the ground,they must have been projected downwards with the same speed $v'$. Let the third fragment be projected upwards with speed $v$. Then $m v - m v' - m v' = 0$,so $v = 2v'$.
For the fragment projected upwards: $h = -vt_1 + \frac{1}{2}gt_1^2$ (where $t_1 = 10 \ s$ and $h$ is displacement downwards).
For the fragments projected downwards: $h = v't_2 + \frac{1}{2}gt_2^2$ (where $t_2 = 20 \ s$).
Substituting $v = 2v'$: $h = -2v'(10) + \frac{1}{2}g(10)^2 = -20v' + 500$.
Also $h = v'(20) + \frac{1}{2}g(20)^2 = 20v' + 2000$.
Adding the two equations: $2h = 2500 \implies h = 1250 \ m$.

(C) Let the initial mass of the cannon be $M = 10m$. The initial velocity is $v_0 = 10 \ m/s$.
After firing $n$ shells, the mass of the cannon becomes $M' = 10m - nm$.
Let the final velocity of the cannon be $v_f$.
By the law of conservation of linear momentum:
$P_{initial} = P_{final}$
$(10m)(10) = (10m - nm)v_f + (nm)u$
$100m = (10 - n)m v_f + nmu$
$100 = (10 - n)v_f + nu$
If the velocity of the cannon becomes $u$, then $v_f = u$.
Substituting $v_f = u$ into the equation:
$100 = (10 - n)u + nu$
$100 = 10u - nu + nu$
$100 = 10u$
$u = 10 \ m/s$.
Therefore, if $u = 10 \ m/s$, the velocity of the cannon will remain $10 \ m/s$ regardless of the number of shells fired.

(B) Let the left plank be $P_1$ and the right plank be $P_2$. Initially,the system is at rest.
When the cat leaps from $P_1$ to $P_2$ with speed $v_{cg}$ (to the right),by conservation of momentum for the system (cat + $P_1$):
$0 = m_c v_{cg} + m_s v_{P1} \implies v_{P1} = -\frac{m_c v_{cg}}{m_s}$ (to the left).
When the cat lands on $P_2$,the cat and $P_2$ move together. By conservation of momentum for the system (cat + $P_2$):
$m_c v_{cg} = (m_c + m_s) v_{P2} \implies v_{P2} = \frac{m_c v_{cg}}{m_c + m_s}$.
When the cat leaps back from $P_2$ to $P_1$ with speed $v_{cg}$ (to the left),let the final velocity of $P_2$ be $v'_{P2}$. By conservation of momentum for the system (cat + $P_2$):
$(m_c + m_s) v_{P2} = m_c (-v_{cg}) + m_s v'_{P2}$.
Substituting $v_{P2}$:
$(m_c + m_s) \left( \frac{m_c v_{cg}}{m_c + m_s} \right) = -m_c v_{cg} + m_s v'_{P2}$.
$m_c v_{cg} = -m_c v_{cg} + m_s v'_{P2} \implies 2m_c v_{cg} = m_s v'_{P2} \implies v'_{P2} = \frac{2m_c v_{cg}}{m_s}$.

(D) Let the velocity of the man be $v_m$ and the velocity of the cart be $v_c$ with respect to the ground after the jump.
By the law of conservation of linear momentum, the initial momentum is zero, so $m v_m + 2m v_c = 0$, which implies $v_m = -2v_c$.
The relative velocity of the man with respect to the cart is $u = v_m - v_c$.
Substituting $v_m = -2v_c$, we get $u = -2v_c - v_c = -3v_c$, so $v_c = -u/3$.
Then $v_m = -2(-u/3) = 2u/3$.
The work done by internal forces is equal to the change in kinetic energy of the system: $W = \Delta K = K_f - K_i$.
Since the system was initially at rest, $K_i = 0$.
$K_f = \frac{1}{2} m v_m^2 + \frac{1}{2} (2m) v_c^2 = \frac{1}{2} m (2u/3)^2 + m (-u/3)^2$.
$K_f = \frac{1}{2} m (4u^2/9) + m (u^2/9) = \frac{2mu^2}{9} + \frac{mu^2}{9} = \frac{3mu^2}{9} = \frac{mu^2}{3}$.
Thus, the work done is $\frac{mu^2}{3}$.

(A) At the highest point,the initial momentum of the projectile is $P_i = (3m)(40) = 120m$ in the horizontal direction.
Let the velocity of the part with mass $m$ be $\vec{v} = v_x \hat{i} + v_y \hat{j}$.
The part with mass $2m$ moves vertically up with velocity $25\, m/s$,so its momentum is $\vec{P}_{2m} = (2m)(25\hat{j}) = 50m\hat{j}$.
By the law of conservation of linear momentum,$\vec{P}_i = \vec{P}_{2m} + \vec{P}_m$.
$120m\hat{i} = 50m\hat{j} + m(v_x \hat{i} + v_y \hat{j})$.
Equating the components:
Horizontal: $120m = mv_x \implies v_x = 120\, m/s$.
Vertical: $0 = 50m + mv_y \implies v_y = -50\, m/s$.
The speed of the part with mass $m$ is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{120^2 + (-50)^2} = \sqrt{14400 + 2500} = \sqrt{16900} = 130\, m/s$.

(D) Let the total mass be $M = 1 \; kg$. The masses of the three parts are $m_1 = 0.2 \; kg$,$m_2 = 0.2 \; kg$,and $m_3 = 0.6 \; kg$ (ratio $1:1:3$).
Since the bomb is initially stationary,the initial momentum is $0$.
By the law of conservation of linear momentum,the final momentum must also be $0$.
Let the velocities of the two smaller parts be $\vec{v}_1 = 30\hat{i} \; m/s$ and $\vec{v}_2 = 30\hat{j} \; m/s$.
The momentum of these two parts is $\vec{p}_{12} = m_1\vec{v}_1 + m_2\vec{v}_2 = 0.2(30\hat{i} + 30\hat{j}) = 6\hat{i} + 6\hat{j} \; kg \cdot m/s$.
The magnitude of this momentum is $p_{12} = \sqrt{6^2 + 6^2} = 6\sqrt{2} \; kg \cdot m/s$.
For the total momentum to be zero,the third part must have momentum $\vec{p}_3 = -\vec{p}_{12}$.
Thus,$m_3 v_3 = 6\sqrt{2}$.
$0.6 \cdot v_3 = 6\sqrt{2}$.
$v_3 = \frac{6\sqrt{2}}{0.6} = 10\sqrt{2} \; m/s$.

(B) The initial velocity of the ball is $u = 200 \, m/s$ at an angle $\theta = 60^{\circ}$.
At the maximum height,the vertical component of velocity is $0$,and the horizontal component is $v_x = u \cos 60^{\circ} = 200 \times 0.5 = 100 \, m/s$.
The total mass of the ball is $m$. At maximum height,the momentum is $\vec{P}_{initial} = m \times 100 \hat{i}$.
The ball explodes into $3$ equal fragments of mass $m/3$ each.
Let the velocities of the fragments be $\vec{v}_1 = 100 \hat{j}$,$\vec{v}_2 = -100 \hat{j}$,and $\vec{v}_3 = \vec{V}$.
By the law of conservation of linear momentum:
$\vec{P}_{initial} = \vec{P}_{final}$
$m(100 \hat{i}) = \frac{m}{3}(100 \hat{j}) + \frac{m}{3}(-100 \hat{j}) + \frac{m}{3}(\vec{V})$
$100 \hat{i} = \frac{1}{3}(100 \hat{j} - 100 \hat{j} + \vec{V})$
$100 \hat{i} = \frac{1}{3} \vec{V}$
$\vec{V} = 300 \hat{i} \, m/s$.
Thus,the third part moves at $300 \, m/s$ in the horizontal direction.

(A) According to the law of conservation of linear momentum,if the net external force acting on a system is zero,the total linear momentum of the system remains constant.
In the case of two balls colliding in free space,there are no external forces acting on the system.
Therefore,the total linear momentum of the system is conserved regardless of the type of collision (elastic or inelastic) or the nature of the collision (head-on or oblique).