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

Answer 1

Answer: (D)

As $\vec{F}_{\text {ext }}=0, \therefore \vec{a}_{C M}=0$
$\vec{v}_{ CM }=\frac{2 \times 8 \hat{i}-3 \times 2 \hat{i}}{2+3}=2 \hat{i}, X_{ CM }=\frac{2 \times 2+3 \times 12}{2+3}=8$
As $\vec{v}_{ CM }$ is constant, so centre of mass will move $2 m$ in one second.
$\therefore x=8+2=10 m$