91Ó°ÊÓ

• The mass of one of the sphere is$m_1=50kg$.
• The mass of other sphere is$m_2=100kg$.
• The radii is r=20m.
• The distance between their centres is d=40m .
• The distance is $d^{\text{'}}=20m$.

In this problem, to evaluate the speed and relative speed of the sphere, conservation of energy will be used. Also, conservation of momentum as well as conservation of energy will be conserved

According to the question, the two-sphere doesn’t experience any kind of external force, and also there is no outside force on the entire system. So, if there is no available external force, then the linear momentum of the system will be conserved.

Thus, the linear momentum is conserved.

The relation of gravitational potential energy can be written as,

$$\begin{gathered} ∆U=\left(-\frac{G{m}_1{m}_2}{d}\right)-\left(-\frac{G{m}_1{m}_2}{d^{\text{'}}}\right) \\ ∆U=G{m}_1{m}_2\left(-\frac{1}{d}\right)-\left(-\frac{1}{d^{\text{'}}}\right) \\ =-G{m}_1{m}_2\left\{\left(\frac{1}{d}\right)-\left(\frac{1}{d^{\text{'}}}\right)\right\} \end{gathered}$$

Here, Gis the universal gravitational constant.

Substitute the values in the above equation.

$$\begin{gathered} ∆U=\left(6.67×{10}^{-11}{\mathrm{m}}^3/\mathrm{kg}-{\mathrm{s}}^2\right)\left(25\mathrm{kg}\right)\left(100\mathrm{kg}\right)\left[\frac{1}{40\mathrm{m}}-\frac{1}{20\mathrm{m}}\right]\left(\frac{1\mathrm{J}}{1\mathrm{kg}-{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ ∆U=-4.16×{10}^{-9}\mathrm{J} \end{gathered}$$

The relation from conservation of energy can be written as,

$$\begin{gathered} K_1=-∆U \\ \frac{1}{2}{m}_1{v}_1^2=-\left(-4.16×{10}^{-9}\mathrm{J}\right) \end{gathered}$$

Here,$v_1$is the velocity of one of the sphere.

Substitute the values in the above equation.

$$\begin{gathered} \frac{1}{2}\left(25kg\right)v_1^2=4.16×{10}^{-9}\mathrm{J} \\ {\mathrm{v}}_1=\left(\frac{4.16×{10}^{-9}\mathrm{J}×2}{25\mathrm{kg}}\right) \\ {\mathrm{v}}_1=1.82×{10}^{-5}\mathrm{m}/\mathrm{s} \end{gathered}$$

The relation from conservation of energy can be written as,

$$\begin{gathered} k_2=∆U \\ \frac{1}{2}{m}_2{v}_2^2=-\left(-4.16×{10}^{-9}\mathrm{J}\right) \end{gathered}$$

Here,$v_2$ is the velocity of other sphere.

Substitute the values in the above equation.

$$\begin{gathered} \frac{1}{2}\left(100kg\right)v_2^2=4.16×{10}^{-9}J \\ {v}_2=\left(\frac{4.16×{10}^{-9}\mathrm{J}×2}{100\mathrm{kg}}\right) \\ {\mathrm{v}}_2=9.12×{10}^{-6}\mathrm{m}/\mathrm{s} \end{gathered}$$

The relation of relative velocity can be written as,

$v_R=v_1+v_2$

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{v}}_{\mathrm{R}}=1.82×{10}^{-5}\mathrm{m}/\mathrm{s}+9.12×{10}^{-6}\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{R}}=2.73×{10}^{-5}\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the velocity of both the spheres are $1.82×{10}^{-5}\mathrm{m}/\mathrm{s}$and $9.12×{10}^{-6}m/s$,

and the relative velocity is$2.73×{10}^{-5}\mathrm{m}/\mathrm{s}$ .

The relation to find the distance can be written as,

$\mathrm{x}=\frac{{\mathrm{m}}_1{\mathrm{d}}_1+{\mathrm{m}}_2\mathrm{d}}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

Here,$d_1$is the distance of sphere 1 from the initial positon whose value is zero.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{x}=\frac{\left(25\mathrm{kg}\right)\left(0\right)+\left(100\mathrm{kg}\right)\left(40\mathrm{m}\right)}{\left(25\mathrm{kg}\right)+\left(100\mathrm{kg}\right)} \\ \mathrm{x}=32\mathrm{m} \end{gathered}$$

Thus, the required distance is 32m .