Motion of Centre of Mass Questions in English

(C) Given,$m_1 = 6 \,kg, m_2 = 4 \,kg$ and $\overrightarrow{v}_1 = 5 \hat{i}-2 \hat{j}+10 \hat{k}, \overrightarrow{v}_2 = 10 \hat{i}-2 \hat{j}+5 \hat{k}$.
The velocity of the centre of mass $\overrightarrow{v}_{cm}$ is given by the formula:
$\overrightarrow{v}_{cm} = \frac{m_1 \overrightarrow{v}_1 + m_2 \overrightarrow{v}_2}{m_1 + m_2}$
Substituting the values:
$\overrightarrow{v}_{cm} = \frac{6(5 \hat{i}-2 \hat{j}+10 \hat{k}) + 4(10 \hat{i}-2 \hat{j}+5 \hat{k})}{6 + 4}$
$\overrightarrow{v}_{cm} = \frac{(30 \hat{i}-12 \hat{j}+60 \hat{k}) + (40 \hat{i}-8 \hat{j}+20 \hat{k})}{10}$
$\overrightarrow{v}_{cm} = \frac{70 \hat{i}-20 \hat{j}+80 \hat{k}}{10}$
$\overrightarrow{v}_{cm} = 7 \hat{i}-2 \hat{j}+8 \hat{k}$

(C) The velocity of the center of mass $(v_{cm})$ is given by: $v_{cm} = \frac{M \cdot v_1 + m \cdot v_2}{M + m} = \frac{M(2) + 1(1)}{M + 1} = \frac{2M + 1}{M + 1}$.
The kinetic energy of the center of mass is given by: $K.E._{cm} = \frac{1}{2} (M + m) v_{cm}^2$.
Given $K.E._{cm} = \frac{4}{3} \ J$,we substitute the values:
$\frac{4}{3} = \frac{1}{2} (M + 1) \left( \frac{2M + 1}{M + 1} \right)^2$.
$\frac{4}{3} = \frac{1}{2} \frac{(2M + 1)^2}{M + 1}$.
$8(M + 1) = 3(4M^2 + 4M + 1)$.
$8M + 8 = 12M^2 + 12M + 3$.
$12M^2 + 4M - 5 = 0$.
Solving the quadratic equation using the formula $M = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$M = \frac{-4 \pm \sqrt{16 - 4(12)(-5)}}{2(12)} = \frac{-4 \pm \sqrt{16 + 240}}{24} = \frac{-4 \pm \sqrt{256}}{24} = \frac{-4 \pm 16}{24}$.
Since mass must be positive,$M = \frac{12}{24} = 0.5 \ kg$.

(A) Given: Masses of the balls $m_1 = 100 \ g = 0.1 \ kg$ and $m_2 = 200 \ g = 0.2 \ kg$.
At $t = 0.4 \ s$,the time of fall for the first ball is $t_1 = 0.4 \ s$.
The distance travelled by the first ball is $s_1 = \frac{1}{2} g t_1^2 = \frac{1}{2} \times 10 \times (0.4)^2 = 5 \times 0.16 = 0.8 \ m$.
At $t = 0.4 \ s$,the time of fall for the second ball is $t_2 = 0.4 - 0.2 = 0.2 \ s$.
The distance travelled by the second ball is $s_2 = \frac{1}{2} g t_2^2 = \frac{1}{2} \times 10 \times (0.2)^2 = 5 \times 0.04 = 0.2 \ m$.
The distance of the center of mass from the release point is given by $s_{cm} = \frac{m_1 s_1 + m_2 s_2}{m_1 + m_2}$.
Substituting the values: $s_{cm} = \frac{0.1 \times 0.8 + 0.2 \times 0.2}{0.1 + 0.2} = \frac{0.08 + 0.04}{0.3} = \frac{0.12}{0.3} = 0.4 \ m$.

(A) The centre of mass of a system moves according to the net external force acting on it,as given by the equation $F_{ext} = M a_{cm}$.
Since the particles $A$ and $B$ are moving under a mutual force of attraction,there is no external force acting on the system $(F_{ext} = 0)$.
Initially,both particles are at rest,so the initial velocity of the centre of mass is $v_{cm, initial} = 0$.
Since $F_{ext} = 0$,the acceleration of the centre of mass is $a_{cm} = 0$,which implies that the velocity of the centre of mass remains constant over time.
Therefore,the velocity of the centre of mass at any instant remains equal to its initial velocity,which is $0$.

(C) Let the masses be $m_1 = 2 \text{ kg}$ and $m_2 = 1 \text{ kg}$.
The acceleration of the system is given by $a = \frac{(m_1 - m_2)g}{m_1 + m_2} = \frac{(2 - 1) \times 10}{2 + 1} = \frac{10}{3} \text{ m/s}^2$.
The $2 \text{ kg}$ block moves downwards with acceleration $a_1 = 10/3 \text{ m/s}^2$ and the $1 \text{ kg}$ block moves upwards with acceleration $a_2 = 10/3 \text{ m/s}^2$.
Taking the downward direction as positive,the acceleration of the center of mass is $a_{cm} = \frac{m_1 a_1 - m_2 a_2}{m_1 + m_2} = \frac{2(10/3) - 1(10/3)}{2 + 1} = \frac{10/3}{3} = 10/9 \text{ m/s}^2$.
The distance traversed by the center of mass in $t = 2 \text{ s}$ is $d = \frac{1}{2} a_{cm} t^2 = \frac{1}{2} \times \frac{10}{9} \times (2)^2 = \frac{1}{2} \times \frac{10}{9} \times 4 = \frac{20}{9} \approx 2.22 \text{ m}$.

(C) The acceleration of the centre of mass is given by $a_{cm} = \frac{m_A a_1 + m_B a_2}{m_A + m_B}$.
Since $m_A = m_B = m$,we have $a_{cm} = \frac{m(6\hat{i} + 6\hat{j}) + m(0)}{2m} = 3\hat{i} + 3\hat{j} \text{ m/s}^2$.
The initial velocity of the centre of mass is $v_{cm} = \frac{m_A v_1 + m_B v_2}{m_A + m_B} = \frac{m(4\hat{i}) + m(4\hat{j})}{2m} = 2\hat{i} + 2\hat{j} \text{ m/s}$.
Since the acceleration vector $a_{cm} = 3\hat{i} + 3\hat{j}$ and the initial velocity vector $v_{cm} = 2\hat{i} + 2\hat{j}$ are parallel (specifically,$a_{cm} = 1.5 v_{cm}$),the motion of the centre of mass is along a straight line.