Question

Figure shows position and velocities of two particles moving under mutual gravitational attraction in space at time t=0.
The position of centre of mass after one second is

• A. $x=4m$
• B. $x=6m$
• C. $x=8m$
• D. $x=10m$

Answer 1

Answer: (D)

$x=10m$

mutual force with not affect motion

the motion of center of mass can be supposed to be uniform

So initial velocity of center of mass

$u=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}=\frac{2kg\left(8\right)+3kg(-2)}{2+3}$

$\Rightarrow u=\frac{10}{5}=2m/s$ $+ve$

$\Downarrow$

towards increasing $x$

distance in second will be $x=xt=2x=2m$

$\Rightarrow$ New position $=oldposition+2$

or new position $=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}+2$

$x=\frac{2\times 2+3\times 12}{3}+2$

$x=10$ meters