System of particles momentum and Energy Questions in English

(D) Statement $1$: Linear momentum $\vec{P} = \sum m_i \vec{v}_i = 0$. This does not imply that the velocity of each particle is zero. For example,in a system of two particles moving in opposite directions with equal momentum,the total momentum is zero,but the kinetic energy $K = \sum \frac{1}{2} m_i v_i^2$ is non-zero.
Statement $2$: Kinetic energy $K = \sum \frac{1}{2} m_i v_i^2 = 0$. Since mass $m_i > 0$ and $v_i^2 \ge 0$,the only way for the sum to be zero is if each $v_i = 0$. If all velocities are zero,then the total linear momentum $\vec{P} = \sum m_i (0) = 0$.
Therefore,$1$ does not imply $2$,but $2$ implies $1$.

(C) The kinetic energy of a system of particles is given by $K = \sum \frac{1}{2} m_i v_i^2$. If $K = 0$,then each individual particle must have a velocity $v_i = 0$,which implies the system is at rest. If the system is at rest,the total linear momentum $P = \sum m_i v_i = 0$. Thus,$B$ implies $A$.
However,if the linear momentum $P = 0$,the particles could be moving in opposite directions such that their vector sum is zero (e.g.,two particles of equal mass moving with velocities $v$ and $-v$). In this case,the kinetic energy $K = \frac{1}{2}mv^2 + \frac{1}{2}m(-v)^2 = mv^2 \neq 0$. Thus,$A$ does not imply $B$.

(C) The total momentum of a system is the vector sum of the momenta of individual particles: $\vec{P}_{total} = \sum \vec{p}_i = 0$ (since no external force acts on the system).
However,the sum of the magnitudes of the momenta is given by $S = \sum |\vec{p}_i|$.
Even if the vector sum is zero,the individual momenta $\vec{p}_i$ can change over time due to internal forces acting between the particles.
Since the individual momenta change,their magnitudes $|\vec{p}_i|$ change,and consequently,their sum $S$ is generally non-zero and not necessarily constant.

(A) Let the mass of the plank be $M = 40 \ kg$,mass of $A$ be $m_A = 40 \ kg$,and mass of $B$ be $m_B = 60 \ kg$.
The length of the plank is $L = 120 \ cm$.
Initially,$A$ is at $x = 0$ and $B$ is at $x = 60 \ cm$ (middle).
Since the surface is smooth and there are no external horizontal forces,the center of mass of the system remains fixed.
Let the displacement of the plank be $S_P$ and the displacement of $A$ relative to the ground be $S_A$.
Since $B$ is fixed with respect to the ground,its displacement $S_B = 0$.
The center of mass equation is: $m_A S_A + m_B S_B + M S_P = 0$.
Substituting the values: $40 S_A + 60(0) + 40 S_P = 0$,which simplifies to $S_A = -S_P$.
$A$ moves towards $B$,so $A$ covers the distance of $60 \ cm$ relative to the plank. Let $x$ be the distance $A$ moves relative to the ground. Then $x = 60 + S_P$ (since the plank moves left by $|S_P|$).
Also,$S_A = x = 60 + S_P$.
Substituting $S_A = -S_P$: $-S_P = 60 + S_P \implies 2S_P = -60 \implies S_P = -30 \ cm$.
Thus,$S_A = 30 \ cm$.
$A$ meets $B$ at a position $30 \ cm$ from the original left end of the plank,which is the middle of the plank.

(A) Let $m_{i}$ be the mass and $v_{i}$ be the velocity of the $i^{th}$ particle. The velocity relative to the centre of mass is $v_{i}^{\prime} = v_{i} - V$,so $v_{i} = v_{i}^{\prime} + V$. The total momentum $p = \sum m_{i} v_{i} = \sum m_{i}(v_{i}^{\prime} + V) = \sum m_{i} v_{i}^{\prime} + V \sum m_{i} = \sum p_{i}^{\prime} + MV$. Since $\sum m_{i} r_{i}^{\prime} = 0$,differentiating gives $\sum m_{i} v_{i}^{\prime} = \sum p_{i}^{\prime} = 0$.
$(b)$ $K = \sum \frac{1}{2} m_{i} v_{i}^{2} = \sum \frac{1}{2} m_{i} (v_{i}^{\prime} + V) \cdot (v_{i}^{\prime} + V) = \sum \frac{1}{2} m_{i} v_{i}^{\prime 2} + \sum m_{i} v_{i}^{\prime} \cdot V + \sum \frac{1}{2} m_{i} V^{2} = K^{\prime} + V \cdot (\sum p_{i}^{\prime}) + \frac{1}{2} M V^{2}$. Since $\sum p_{i}^{\prime} = 0$,$K = K^{\prime} + \frac{1}{2} M V^{2}$.
$(c)$ $L = \sum r_{i} \times p_{i} = \sum (R + r_{i}^{\prime}) \times m_{i} (V + v_{i}^{\prime}) = \sum (R \times m_{i} V + R \times m_{i} v_{i}^{\prime} + r_{i}^{\prime} \times m_{i} V + r_{i}^{\prime} \times m_{i} v_{i}^{\prime}) = R \times MV + R \times (\sum p_{i}^{\prime}) + (\sum m_{i} r_{i}^{\prime}) \times V + L^{\prime}$. Since $\sum p_{i}^{\prime} = 0$ and $\sum m_{i} r_{i}^{\prime} = 0$,$L = L^{\prime} + R \times MV$.
$(d)$ $\frac{d L^{\prime}}{d t} = \frac{d}{d t} \sum (r_{i}^{\prime} \times p_{i}^{\prime}) = \sum (v_{i}^{\prime} \times p_{i}^{\prime}) + \sum (r_{i}^{\prime} \times \frac{d p_{i}^{\prime}}{d t})$. Since $v_{i}^{\prime} \times (m_{i} v_{i}^{\prime}) = 0$,$\frac{d L^{\prime}}{d t} = \sum r_{i}^{\prime} \times F_{i}^{\prime} = \tau_{ext}^{\prime}$.

(C) The total momentum $P$ of a system of $n$ particles is defined as the vector sum of the individual momenta of all particles in the system: $P = \sum_{i=1}^{n} p_i = \sum_{i=1}^{n} m_i v_i$.
By the definition of the centre of mass,the total momentum of a system is also equal to the product of the total mass $M$ of the system and the velocity of the centre of mass $V_{cm}$:
$P = M V_{cm}$,where $M = \sum m_i$ and $V_{cm} = \frac{\sum m_i v_i}{M}$.
Therefore,both the sum of individual momenta and the product of total mass and velocity of the centre of mass represent the total momentum of the system.

(B) According to the principle of conservation of linear momentum,the total momentum of the system before the collision must equal the total momentum after the collision.
Before the collision,the particle of mass $m$ has velocity $v$,and the rod of mass $M$ is at rest. Thus,the initial momentum is $P_i = mv$.
After the collision,the particle comes to rest,and the rod moves with a velocity $V_{cm}$ of its centre of mass. Thus,the final momentum is $P_f = MV_{cm}$.
Equating the two: $mv = MV_{cm}$.
Solving for $V_{cm}$: $V_{cm} = mv/M$.

(B) According to the law of conservation of linear momentum for the system (rod + particle):
Initial momentum = Final momentum
$mv + 0 = 0 + Mv_{cm}$
Here,the particle comes to rest after the collision,so its final velocity is $0$.
The rod acquires a velocity $v_{cm}$ for its centre of mass.
Therefore,$mv = Mv_{cm}$
$v_{cm} = \frac{mv}{M}$

(A) Statement $I$ is true by definition: the total kinetic energy of a system of particles is the scalar sum of the kinetic energies of individual particles,$K = \sum \frac{1}{2} m_i v_i^2$.
Statement $II$ is also true based on the theorem of kinetic energy for a system of particles. If $\vec{v}_i$ is the velocity of the $i$-th particle relative to the origin,$\vec{V}_{cm}$ is the velocity of the center of mass,and $\vec{v}_i'$ is the velocity of the $i$-th particle relative to the center of mass,then $\vec{v}_i = \vec{V}_{cm} + \vec{v}_i'$.
The total kinetic energy is $K = \sum \frac{1}{2} m_i v_i^2 = \sum \frac{1}{2} m_i (\vec{V}_{cm} + \vec{v}_i')^2 = \frac{1}{2} M V_{cm}^2 + \sum \frac{1}{2} m_i (v_i')^2$,where $M = \sum m_i$.
Thus,the total kinetic energy is the sum of the kinetic energy of the center of mass and the kinetic energy relative to the center of mass.