Physics, asked by sanjanaraj6699, 5 months ago

8.
The figure shows the positions and velocities of two particles. If the particles move under the mutual
of each other, then the position of centre of mass at t =1 s is :-

(A) x=5 m (B) x=6 m
(C) x=3 m
(D) x=2 m
horizontal plane ly along the x-axis, at a certain height​

Attachments:

Answers

Answered by aryan073
5

Given :

•Time period =1 second.

x=2m •______________________•x=8m

•m1=1kg , v1=5m/s

•m2=1kg ,v2=3m/s

________________________________________

To Find :

• The position of centre of mass at t=1 sec =?

________________________________________

Solution :

➡ No other force is applied on the two particles system therefore their centre of mass position is at

\\ \implies\sf{x_{com} =\dfrac{m_{1}x_{1}+m_{2}x_{2}}{m_{1}+m_{2}}}

\\ \implies\sf{x_{com} =\dfrac{1 \times 2 +1 \times 8}{2}}

\\ \implies\boxed{\sf{x_{com} =5}}

\\ \implies\sf{velocity \: of \: COM = \dfrac{m_{1} v_{1}+m_{2}v_{2}}{m_{1}+m_{2}}}

\\ \implies\sf{velocity \: of \: COM=\dfrac{ 1 \times 5-1 \times 3}{2}}

\\ \implies\sf{velocity \:of \: COM= 1m/s}

\\ \implies\boxed{\sf{ Velocity \: of \: COM=1m/s}}

After 1 second center of mass move 1m right

new center of mass at =5+1=6 m

Similar questions