Two particles have masses m_1 and m_2. If the | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{m_1}{m_2} d$

Vedclass · Answer

(D) The position of the center of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.For the center of mass to remain at the same position,the change in the position of the center of mass must be zero,i.e.,$\Delta R_{cm} = 0$.Therefore,$m_1 \Delta r_1 + m_2 \Delta r_2 = 0$.Let the first particle $(m_1)$ be moved by a distance $d$ towards the center of mass,so $\Delta r_1 = d$.Let the second particle $(m_2)$ be moved by a distance $x$ in the opposite direction,so $\Delta r_2 = -x$.Substituting these values into the equation: $m_1 d + m_2 (-x) = 0$.$m_1 d = m_2 x$.Solving for $x$,we get $x = \frac{m_1}{m_2} d$.

Two particles have masses $m_1$ and $m_2$. If the first particle $(m_1)$ is moved by a distance $d$ towards the center of mass,by what distance should the second particle $(m_2)$ be moved so that the center of mass of the system remains at the same position?

Similar Questions

Discuss the explosion of a projectile.

Two masses $m_1$ and $m_2$ $(m_1 > m_2)$ are connected by a massless,flexible,and inextensible string passed over a massless and frictionless pulley. The acceleration of the center of mass of the system is

Two bodies of masses $M$ and $4M$ initially at rest,start moving towards each other due to their mutual attraction. The velocity of their centre of mass when the first body attains a velocity $v_0$ is

In the given figure,if the speed given to block $A$ is $10\,m/s$ and block $B$ is initially at rest,then the speed of the centre of mass after $10\,s$ will be ....... $m/s$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module