Problem 92 - University Physics with Modern Physics

Question

Two point charges are moving to the right along the \(x\) -axis. Point charge 1 has charge \(q_{1}=2.00 \mu \mathrm{C},\) mass \(m_{1}=\) \(6.00 \times 10^{-5} \mathrm{kg},\) and speed \(v_{1} .\) Point charge 2 is to the right of \(q_{1}\) and has charge \(q_{2}=-5.00 \mu \mathrm{C},\) mass \(m_{2}=3.00 \times 10^{-5} \mathrm{kg},\) and speed \(v_{2} .\) At a particular instant, the charges are separated by a distance of 9.00 \(\mathrm{mm}\) and have speeds \(v_{1}=400 \mathrm{m} / \mathrm{s}\) and \(v_{2}=1300 \mathrm{m} / \mathrm{s} .\) The only forces on the particles are the forces they exert on each other. (a) Determine the speed \(v_{\mathrm{cm}}\) of the center of mass of the system. (b) The relative energy \(E_{\text { rel }}\) of the system is defined as the total energy minus the kinetic energy contributed by the motion of the center of mass: $$E_{\mathrm{rel}}=E-\frac{1}{2}\left(m_{1}+m_{2}\right) v_{\mathrm{cm}}^{2}$$ where \(E=\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r\) is the total energy of the system and \(r\) is the distance between the charges. Show that \(E_{\mathrm{rel}}=\frac{1}{2} \mu v^{2}+q_{1} q_{2} / 4 \pi \epsilon_{0} r, \quad\) where \(\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right)\) is called the reduced mass of the system and \(v=v_{2}-v_{1}\) is the relative speed of the moving particles. (c) For the numerical values given above, calculate the numerical value of \(E_{\text { rel. }}\) (d) Based on the result of part (c), for the conditions given above, will the particles escape from one another? Explain. (e) If the particles do escape, what will be their final relative speed when \(r \rightarrow \infty ?\) If the particles do not escape, what will be their distance of maximum separation? That is, what will be the value of \(r\) when \(v=0 ?\) (f) Repeat parts \((c)-(e)\) for \(v_{1}=400 \mathrm{m} / \mathrm{s}\) and \(v_{2}=1800 \mathrm{m} / \mathrm{s}\) when the separation is 9.00 \(\mathrm{mm} .\)

Step-by-Step Solution

Answer

Particles will escape; calculate final relative speed if they escape.

Step 1: Calculate Speed of Center of Mass

The formula to find the speed of the center of mass is given by \( v_{\rm cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} \). Substitute the given masses and velocities: \( v_{\rm cm} = \frac{(6.00 \times 10^{-5} \text{ kg}) (400 \text{ m/s}) + (3.00 \times 10^{-5} \text{ kg}) (1300 \text{ m/s})}{6.00 \times 10^{-5} \text{ kg} + 3.00 \times 10^{-5} \text{ kg}} \). After calculating, we find \( v_{\rm cm} \approx 700 \text{ m/s} \).

Step 2: Derive the Expression for Relative Energy

Start by noting the total energy \( E \) of the system: \[ E = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + \frac{q_1 q_2}{4 \pi \epsilon_0 r} \]. Also note that the relative energy \( E_{\rm rel} \) is \[ E_{\rm rel} = E - \frac{1}{2} (m_1 + m_2) v_{\rm cm}^2 \]. Substitute \( E \) to get \( E_{\rm rel} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + \frac{q_1 q_2}{4 \pi \epsilon_0 r} - \frac{1}{2} (m_1 + m_2) v_{\rm cm}^2 \). Recognizing that \( v = v_2 - v_1 \) and \( \mu = \frac{m_1 m_2}{m_1 + m_2} \), it simplifies to \( E_{\rm rel} = \frac{1}{2} \mu v^2 + \frac{q_1 q_2}{4 \pi \epsilon_0 r} \).

Step 3: Calculate Numerical Value of the Relative Energy

Using the expression \( E_{\rm rel} = \frac{1}{2} \, \frac{m_1 m_2}{m_1 + m_2} \, v^2 + \frac{q_1 q_2}{4 \pi \epsilon_0 r} \), plug in the values: \( r = 9.00 \times 10^{-3} \text{ m}, v = 900 \text{ m/s}, q_1 = 2.00 \times 10^{-6} \text{ C}, q_2 = -5.00 \times 10^{-6} \text{ C} \), and \( \epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \). Calculate the first term: \( \frac{1}{2} \, (2 \times 10^{-5} \text{ kg}) \, (900 \text{ m/s})^2 \). Calculate the second term: \( \frac{(2.00 \times 10^{-6} \text{ C}) (-5.00 \times 10^{-6} \text{ C})}{4 \pi (8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2) (9.00 \times 10^{-3} \text{ m})} \). Add these values to find \( E_{\rm rel} \).

Step 4: Determine if Particles Escape

To determine if the particles escape, compare the calculated \( E_{\rm rel} \) with zero. If \( E_{\rm rel} > 0 \), particles have enough kinetic energy to escape. If \( E_{\rm rel} \leq 0 \), they are bound. Based on Step 3, if \( E_{\rm rel} > 0 \), particles will escape each other's influence.

Step 5: Find Final Relative Speed or Maximum Distance

If particles escape, final relative speed \( v_{f} \) when \( r \rightarrow \infty \) is found as \( \sqrt{\frac{2 E_{\rm rel}}{\mu}} \). If they do not escape, set \( E_{\rm rel} = \frac{q_1 q_2}{4 \pi \epsilon_0 r_f} \) and solve for distance \( r_f \) when relative velocity \( v = 0 \).

Key Concepts

Point Charges

Point charges are small particles or objects that have an electrical charge. In physics, these are considered as idealized objects with the whole charge presumed to be concentrated at one point within its structure. This simplification helps in solving problems related to electric fields, forces, and potential.

• They can attract or repel each other depending on their charges.
• A positive charge will repel another positive and attract a negative charge.
• In physics exercises, point charges are often positioned along a specific axis, like the x-axis, to simplify calculations.

Understanding how point charges interact is crucial for electrostatics. Their mutual interaction via electric forces helps determine their motion and how they'll behave under different circumstances. By identifying the charge, mass, and position of each, we can predict their trajectories and energies involved.

Center of Mass

The center of mass is a point representing the average position of the entire mass of a system. It's similar to the balancing point if a set of masses were placed on a beam.

• In a system of particles, it's found by considering every particle's position and mass.
• For this example, the speed of the center of mass can be calculated using the masses and velocities of the point charges.

The formula for velocity of the center of mass, \[ v_{\rm cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}, \]shows us how mass and velocity of each component contribute to the system's overall motion. The center of mass provides us a simple way to analyze systems by focusing on this single point, especially in situations where internal forces are complex.

Relative Energy

Relative energy points to the energy of a system that isn't needed just to carry the system's center of mass motion. It's like isolating the energy involved in the movement of the particles relative to each other. The formula for relative energy is:\[ E_{\text{rel}} = E - \frac{1}{2}(m_1 + m_2)v_{\text{cm}}^2. \]This helps exclude the kinetic energy that would exist if the entire system moved as a whole without any internal movement. The original and derived forms of relative energy use different representations but convey the same idea. It helps us understand the interactions in terms of changes concerning only the involved particles, not the system frame itself. Whether particles escape or remain bound is often determined using relative energy.

Reduced Mass

Reduced mass is a concept that simplifies two-body problems, where both entities exert a force on each other. It allows us to treat the two-body problem as a one-body problem by providing an equivalent single mass:\[ \mu = \frac{m_1 m_2}{m_1 + m_2}. \]This concept simplifies complex systems into a more manageable form to understand energy and dynamic analysis.

• It highlights the interplay of mass between two objects related to forces they exert on each other.
• It becomes particularly useful in formulating the kinetic energy in the system where the system's motion reduces to an effective motion of one mass with the reduced mass value.

In electrostatics, especially with point charges, reduced mass directly relates to the relative kinetic motion of particles, giving insight into their interactive energy dynamics.