Two small particles each of mass m carrying positive charge each are attached to the ends of non conducting light thread of length of 2l. a third particle of mass 2m is attached at mid point of the thread. The whole system is placed on a smooth horizontal floor and the particle of mass 2m is given a velocity V.

q. the velocity centre of mass of the system is a)V b)V/2 c)V/4 PLEASE EXPLAIN.

Considering the three particles connected with string as a system .

Here the velocity of the particle of mass 2m is given a velocity v, and the other two particles' velocity at the instant is zero

As we have taken the three particles connected with string as a system , so all the forces exerted by the particles on each other, tension force by the string on the particles are taken as internal forces . The system is placed on the smooth horizontal surface ,so there is no external force on the system .
Hence , the total momentum of the system will remain conserved and the velocity of centre of mass will also remain constant .


For a system consisting of n particles of masses m₁ m_{2 .......} m_n moving with the velocities v₁, v₂,.......v_n respectively, the velocity of centre of mass is given by:

$v_{cm}=\frac{m_1{v}_1+m_2{v}_2+......+m_{\mathrm{n}}{v}_{\mathrm{n}}}{m_1+m_2+........+m_{\mathrm{n}}}$

Calculating the velocity of centre of mass of the system

$$\begin{gathered} v_{cm}=\frac{m\left(0\right)+m\left(0\right)+2m\left(v\right)}{m+m+2m}= \frac{2mv}{4m} \\ \Rightarrow v_{cm}= \frac{v}{2} \end{gathered}$$