Chapter 26, Problem 63PQ

A glass sphere with radius 4.00 mm, mass 85.0 g, and total charge 4.00 μC is separated by 150.0 cm from a second glass sphere 2.00 mm in radius, with mass 300.0 g and total charge −5.00 μC. The charge distribution on both spheres is uniform. If the spheres are released from rest, what is the speed of each sphere the instant before they collide?

To determine

The speed of each sphere the instant before they collide.

Answer to Problem 63PQ

The speed of first sphere the instant before collision is $23.4\text{m/s}$ and speed of second sphere the instant before collision is $6.63\text{m/s}$.

Explanation of Solution

Write an expression for the conservation of momentum of teh spheres.

  $0=m_1{v}_1\widehat{i}+m_2{v}_2\left(-\widehat{i}\right)$                                                                                              (I)

Here, $m_1$ is the mass of first sphere, $m_2$ is the mass of second sphere, $v_1$ is the final velocity of first sphere and $v_2$ is the final velocity of the second sphere.

Rewrite the expression to find $v_2$.

  $v_2=\frac{m_1{v}_1}{m_2}$                                                                                                               (II)

Write an expression for conservation of energy of the spheres.

  $0=\frac{k\left(-q_1\right)q_2}{d}+\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2+\frac{k\left(-q_1\right)q_2}{r_1+r_2}$                                                       (III)

Here, $k$ is the culoumb constant, $q_1$ is the charge of first sphere, $q_2$ is the charge of second sphere, $r_1$ is the radius of first sphere, $r_2$ is the radius of second sphere and $d$ is the distance between teh spheres.

Rearrange equation (III) to find $v_1$.

  $v_1=\sqrt{\frac{2k{q}_1{q}_2}{m_{1}d}+\frac{2k{q}_1{q}_2}{m_1\left(r_1+r_2\right)}-m_2{v}_2^2}$                                                                       (IV)

Substitute equation (II) in equation (IV).

  $\begin{array}{c}{v}_1=\sqrt{\frac{2k{q}_1{q}_2}{m_{1}d}+\frac{2k{q}_1{q}_2}{m_1\left(r_1+r_2\right)}-m_2{\left(\frac{m_1{v}_1}{m_2}\right)}^2}\\ =\sqrt{\frac{2m_{2}k{q}_1{q}_2}{m_1\left(m_1+m_2\right)}\left(\frac{1}{r_1+r_2}-\frac{1}{d}\right)}\end{array}$                                                            (IV)

Conclusion:

Substitute $8.99\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2$ for $k$, $4.00\text{mm}$ for $r_1$, $85.0\text{g}$ for $m_1$, $4.00\text{μC}$ for $q_1$, $150.0\text{cm}$ for $d$, $2.00\text{mm}$ for $r_2$, $300.0\text{g}$ for $m_2$ and $5.00\text{μC}$ for $q_2$ in equation (V) to find $v_1$.

  $\begin{array}{c}{v}_1=\sqrt{\begin{array}{l}\left(\frac{2\left(\left(300.0\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(\left(4.00\text{μC}\right)\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)\right)\left(\left(5.00\text{μC}\right)\left(\frac{1\text{C}}{{10}^6\text{μC}}\right)\right)}{\left(\left(85.0\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(\left(\left(85.0\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)+\left(\left(300.0\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\right)}\right)\\ \left(\frac{1}{\left(\left(4.00\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)+\left(\left(2.00\text{mm}\right)\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}-\frac{1}{\left(\left(150.0\text{cm}\right)\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}\right)\end{array}}\\ =\sqrt{\begin{array}{l}\left(\frac{2\left(300.0\times {10}^{-3}\text{kg}\right)\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2{\text{/C}}^2\right)\left(4.00\times {10}^6\text{C}\right)\left(5.00\times {10}^6\text{C}\right)}{\left(85.0\times {10}^{-3}\text{kg}\right)\left(\left(85.0\times {10}^{-3}\text{kg}\right)+\left(300.0\times {10}^{-3}\text{kg}\right)\right)}\right)\\ \left(\frac{1}{\left(4.00\times {10}^{-3}\text{m}\right)+\left(2.00\times {10}^{-3}\text{m}\right)}-\frac{1}{\left(150.0\times {10}^{-2}\text{m}\right)}\right)\end{array}}\\ =23.4\text{m/s}\end{array}$

Substitute $85.0\text{g}$ for $m_1$, $300.0\text{g}$ for $m_2$ and $23.4\text{m/s}$ for $v_1$ in equation (II) to find $v_2$.

    $\begin{array}{c}{v}_2=\frac{\left(\left(85.0\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)\left(23.4\text{m/s}\right)}{\left(\left(300.0\text{g}\right)\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\right)}\\ =\frac{\left(85.0\times {10}^{-3}\text{kg}\right)\left(23.4\text{m/s}\right)}{\left(300.0\times {10}^{-3}\text{kg}\right)}\\ =6.63\text{m/s}\end{array}$

Thus, the speed of first sphere the instant before collision is $23.4\text{m/s}$ and speed of second sphere the instant before collision is $6.63\text{m/s}$.