91Ó°ÊÓ

Given data:

First sphere;

$\begin{array}{r}{m}_1=50.0\mathrm{g}=0.050\mathrm{kg}\\ q_1=−10.0\mathrm{μ}\mathrm{C}=−10.0×{10}^{−6}\mathrm{C}\\ R_1=30.0\mathrm{cm}=0.30\mathrm{m}\end{array}$

$\begin{array}{r}{m}_2=150.0\mathrm{g}=0.150\mathrm{kg}\\ q_2=−30.0×{10}^{−6}\mathrm{C}\\ R_2=20.0\mathrm{cm}=0.20\mathrm{m}\end{array}$

The kinetic energy is greatest at maximum speed, while the potential energy is zero. The potential energy is at its maximum at the contact, whereas the kinetic energy is zero. So, we have two instants at the contact and at the maximum speed; the constant is instant(a), while the instant after releasing is instant(b), and because the total mechanical energy is conservative during this motion, the equation is:

$\begin{array}{r}{K}_a+U_a=K_b+U_b\\ U_a=K_b\end{array}$

$U_a$is the potential energy between the two spheres at their point of contact, which is given by:

$U_a=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_{∘}}\frac{q_1{q}_2}{r}$

$K_b$ is the sum of kinetic energies of both spheres;

$K_b=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

The system's momentum is conservative, with the momentum of the first sphere equaling the momentum of the second sphere

$\begin{array}{r}{m}_1{v}_1=m_2{v}_2\\ v_1=v_2\left(\frac{m_2}{m_1}\right)\\ v_1=\left(\frac{150.0\mathrm{g}}{50.0\mathrm{g}}\right)v_2\\ v_1=3v_2\end{array}$

Because the speed of sphere one is three times that of sphere two, the kinetic energy is

$\begin{array}{r}{K}_b=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ K_b=\frac{1}{2}{m}_1{\left(3v_2\right)}^2+\frac{1}{2}{m}_2{v}_2^2\\ K_b=\frac{1}{2}{v}_2^2\left(9m_1+m_2\right)\end{array}$

Now comparing the equation $U_a=K_b$so the speed of pf the second sphere is

$\begin{array}{r}{U}_{\mathrm{a}}=K_b\\ \frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_{∘}}\frac{q_1{q}_2}{R_1+R_2}=\frac{1}{2}{v}_2^2\left(9m_1+m_2\right)\\ v_2=\sqrt{\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_{∘}}\frac{2q_1{q}_2}{\left(R_1+R_2\right)\left(9m_1+m_2\right)}}\end{array}$

Putting the values of $q_1,q_2,R_1,R_2,m_1,m_2$therefore the value of $V_2$is

$\begin{array}{r}{v}_2=\sqrt{\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_{∘}}\frac{2q_1{q}_2}{\left(R_1+R_2\right)\left(9m_1+m_2\right)}}\\ =\sqrt{\left(9.0×{10}^9\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{2\left(−10.0×−30.0{\left({10}^{−6}\mathrm{C}\right)}^2\right)}{(0.30\mathrm{m}+0.20\mathrm{m})\left(9\right[0.050\mathrm{kg}]+0.150\mathrm{kg})}}\\ =4.24\mathrm{m}/\mathrm{s}\end{array}$

Now the value of $V_1$is

$\begin{array}{r}{v}_1=3v_2\\ =3(4.24\mathrm{m}/\mathrm{s})\\ =12.70\mathrm{m}/\mathrm{s}\end{array}$

The electrostatic force between the two spheres caused them to move, hence the force exerted on each sphere is

$\begin{array}{r}F=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_{∘}}\frac{q_1{q}_2}{r^2}\\ F=\frac{1}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_{∘}}\frac{q_1{q}_2}{{\left(R_1+R_2\right)}^2}\\ F=\left(9.0×{10}^9\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{C}}^2\right)\frac{\left(−10.0×−30.0{\left({10}^{−6}\mathrm{C}\right)}^2\right)}{(0.30\mathrm{m}+0.20\mathrm{m})}\\ F=10.8\mathrm{N}\end{array}$

The motion of the sphere due to the force can be calculated using Newton's law,

So, the force is

$F=ma$

Calculating acceleration for the first sphere:

$\begin{array}{r}F=m_1{a}_1\\ a_1=\frac{F}{m_1}\\ a_1=\frac{10.8\mathrm{N}}{0.050\mathrm{kg}}\\ a_1=216\mathrm{m}/{\mathrm{s}}^2\end{array}$

Now for the second sphere,

$\begin{array}{c}{a}_2=\frac{F}{m_2}\\ a_2=\frac{10.8\mathrm{N}}{0.150\mathrm{kg}}\\ a_1=72\mathrm{m}/{\mathrm{s}}^2\end{array}$

Hence, the maximum speed achieved by the first and second spheres is

$12.7m/sand4.24m/s.$

And maximum acceleration achieved by the first and second spheres is

$216m/s\mathrm{and}72m/s.$