Question

Two metal balls of mass \(m_{1}=5.00 \mathrm{~g}\) (diameter \(=5.00 \mathrm{~mm}\) ) and \(m_{2}=8.00 \mathrm{~g}\) (diameter \(=8.00 \mathrm{~mm}\) ) have positive charges of \(q_{1}=5.00 \mathrm{nC}\) and \(q_{2}=8.00 \mathrm{nC}\), respectively. A force holds them in place so that their centers are separated by \(8.00 \mathrm{~mm}\). What will their velocities be after the force has been removed and they are separated by a large distance?

Short answer

Answer: The final velocities of the two metal balls are approximately \(v_1 \approx -6.37 \frac{m}{s}\) and \(v_2 \approx 3.98 \frac{m}{s}\).

Step-by-step solution

Calculate the initial potential energy stored in the system due to electrostatic repulsion.

First, we need to determine the initial potential energy stored in the system due to the electrostatic repulsion between the two metal balls. This potential energy is given by the formula: \[U = \frac{k \cdot q_1 \cdot q_2}{r}\] Where: - \(U\) is the potential energy stored in the system. - \(k\) is the electrostatic constant, which is approximately \(8.99 \cdot 10^9 \frac{Nm^2}{C^2}\). - \(q_1\) and \(q_2\) are the charges of the two balls. - \(r\) is the distance between the two balls' centers. Given the problem's data, we can plug the values into the formula: \[U = \frac{8.99 \cdot 10^9 \frac{Nm^2}{C^2} \cdot (5.00 \cdot 10^{-9} C) \cdot (8.00 \cdot 10^{-9} C)}{8.00 \cdot 10^{-3} m}\]

Calculate the initial potential energy stored in the system.

Now, let's perform the calculation: \[U = \frac{359.6 \frac{Jm}{C^2} \cdot 40.00 \cdot 10^{-18} C^2}{8.00 \cdot 10^{-3} m} = 1.80 \times 10^{-6} J\] So the initial potential energy stored in the system due to electrostatic repulsion is \(1.80 \times 10^{-6} \mathrm{J}\).

Apply the conservation of energy principle.

When the force holding the balls in place is removed, the initial potential energy stored in the system is converted into the kinetic energy of the two balls. In this case, we can assume that there is no loss of energy due to air friction or other external forces, so the total mechanical energy is conserved: \[K_1 + K_2 = U\] Where: - \(K_1\) and \(K_2\) are the final kinetic energies of balls 1 and 2, respectively. The kinetic energy of each ball can be expressed as: \[K_1 = \frac{1}{2} m_1 v_1^2\] \[K_2 = \frac{1}{2} m_2 v_2^2\] So we can write the conservation of energy equation as: \[\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = U\]

Apply Newton's third law.

According to Newton's third law, every action has an equal and opposite reaction. In this case, the balls experience an equal and opposite force when they repel due to their charges. The momentum of an object can be expressed as the product of its mass and velocity: \[p_1 = m_1 v_1\] \[p_2 = m_2 v_2\] As the two balls experience equal and opposite forces, their final momenta are equal and opposite: \[m_1 v_1 = -m_2 v_2\] From this equation, we can find the relationship between the final velocities of the two balls: \[v_1 = - \frac{m_2}{m_1} v_2\]

Solve for the final velocities.

Now we have two equations: \[\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = U\] \[v_1 = - \frac{m_2}{m_1} v_2\] Substitute the second equation into the first one to eliminate \(v_1\): \[\frac{1}{2} m_1 \left(- \frac{m_2}{m_1} v_2\right)^2 + \frac{1}{2} m_2 v_2^2 = U\] Now, we can solve for \(v_2\): \[v_2^2 = \frac{2 U}{m_1 + m_2}\] Plug in the values of \(U\), \(m_1\), and \(m_2\): \[v_2^2 = \frac{2 \cdot 1.80 \times 10^{-6} J}{(5.00 \cdot 10^{-3} kg) + (8.00 \cdot 10^{-3} kg)}\] Calculate \(v_2\): \[v_2 \approx 3.98 \frac{m}{s}\] Finally, we can find \(v_1\) using the relation between \(v_1\) and \(v_2\): \[v_1 = - \frac{m_2}{m_1} v_2 \approx -6.37 \frac{m}{s}\] So, the final velocities of the two metal balls after the force has been removed and they are separated by a large distance are approximately \(v_1 \approx -6.37 \frac{m}{s}\) and \(v_2 \approx 3.98 \frac{m}{s}\).