Consider a two-particle system with particles | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The position of the centre of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.For the centre of mass to remain at the sam

Vedclass · Accepted Answer

$\left( \frac{m_1}{m_2} \right) d$

Vedclass · Answer

(D) The position of the centre of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.For the centre of mass to remain at the same position,the change in the position of the centre of mass must be zero,i.e.,$\Delta R_{cm} = 0$.Therefore,$m_1 \Delta r_1 + m_2 \Delta r_2 = 0$.Given that the first particle is moved by a distance $d$ towards the centre of mass,let $\Delta r_1 = d$.Substituting this into the equation: $m_1 d + m_2 \Delta r_2 = 0$.Solving for $\Delta r_2$: $\Delta r_2 = -\left( \frac{m_1}{m_2} \right) d$.The negative sign indicates that the second particle must be moved in the opposite direction to the first particle,also towards the centre of mass,by a distance of $\left( \frac{m_1}{m_2} \right) d$.

Consider a two-particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$,by what distance should the second particle be moved,so as to keep the centre of mass at the same position?

Similar Questions

$A$ ball of mass $100 \ g$ is dropped at time $t=0$. $A$ second ball of mass $200 \ g$ is dropped from the same point at $t=0.2 \ s$. The distance between the center of mass of the two balls and the release point at $t=0.4 \ s$ is: (Assume $g=10 \ m \ s^{-2}$) (in $m$)

$A$ body of mass $m_1 = 4 \text{ kg}$ moves at $5 \hat{i} \text{ m/s}$ and another body of mass $m_2 = 2 \text{ kg}$ moves at $10 \hat{i} \text{ m/s}$. The kinetic energy of the centre of mass is:

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod at an angle $\theta$ with the rod. The acceleration of the centre of mass of the rod is:

In the figure shown,the blocks have equal masses. Friction,mass of the string,and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity $= g$).

Discuss the motion of binary (double) stars in astronomy.

Vedclass Test Series

Exam Paper Generator

Online Exam Module