Question

At $t=0$ the positions and velocities of two particles are as shown in the figure. They are kept on a smooth surface and being mutually attracted by gravitational force. Then the position of the centre of mass at $t=2 \, s \, $ is

• A. $X=5 \, m$
• B. $X=7 \, m$
• C. $X=3 \, m$
• D. $X=2 \, m$

Answer 1

Answer: (B)

Initial position of center of mass,
$x_{i}=\frac{1 \left(0\right) + 1 \left(10\right)}{1 + 1}=5 \, m$
Initial velocity of center of mass,
$v=\frac{1 \left(5\right) + 1 \left(- 3\right)}{1 + 1}=1 \, m$
This velocity remains constant as the internal forces during collision do not change the velocity of the center of mass. Therefore distance travelled by the center of mass in $2s$ ,
$\Delta x=v\Delta t$
$\Delta x=1\left(2\right)=2 \, m$
Final position of center of mass,
$x_{f}=x_{i}+\Delta x$
$x_{f}=5+2=7 \, m$