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

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

Tardigrade Admin · Accepted Answer

As vecF text ext =0, ∴ vecaC M=0 vecv CM =(2 × 8 hati-3 × 2 hati/2+3)=2 hati, X CM =(2 × 2+3 × 12/2+3)=8 As vecv CM is constant, so centre of mass will move 2 m in one second. ∴ x=8+2=10 m

Q. 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

$x = 4 m$

$x = 6 m$

$x = 8 m$

$x = 10 m$

As $F_{ext} = 0, ∴ a_{CM} = 0$
$v_{CM} = \frac{2 \times 8 i ^ - 3 \times 2 i ^}{2 + 3} = 2 \hat{i}, X_{CM} = \frac{2 \times 2 + 3 \times 12}{2 + 3} = 8$
As $v_{CM}$ is constant, so centre of mass will move $2 m$ in one second.
$∴ x = 8 + 2 = 10 m$

Q. 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

x=4m

x=6m

x=8m

x=10m