**Problem 1:** Two 2.0-mm-diameter beads, C and D, are \(r_0 = 10 \, \text{mm}\) apart, measured between their centers (see Fig. 1). Bead C has mass \(m_C = 1.0 \, \text{g}\) and charge \(q_C = 2.0 \, \text{nC}\). Bead D has mass \(m_D = 2.0 \, \text{g}\) and charge \(q_D = -1.0 \, \text{nC}\). If the beads are released from rest, what are the speeds \(v_C\) and \(v_D\) at the instant the beads collide? **Partial answer:** \(v_D = 4.9 \, \text{cm/s}\). **(a)** Write down the energy conservation law for this system. To compute potential energies, use the expression for the potential energy of two point charges (do not plug in the numbers at this step, but work only with symbols). Note that the final potential energy of the system is not zero, and it is determined by the distances between the centers of the beads when they collide. At the end of this step you will have an expression that contains both speeds \(v_C\) and \(v_D\), but you will not know yet how to compute them separately. **(b)** In addition to the energy conservation, in this problem you need to use the law of conservation of linear momentum. The linear momentum of this system as a whole is equal to zero. Which relation between the speeds \(v_C\) and \(v_D\) has to be satisfied to guarantee zero linear momentum (express \(v_D\) in terms of \(v_C\), but do not plug in the numerical values yet)? **(c)** Use this expression for \(v_D\) in the energy conservation law from part (a) and work out the symbolic formula for \(v_C\) in terms of \(m_C, m_D, q_C, q_D, r_0,\) and \(R\). Only after you have the final formula, plug in the numerical values of the parameters to compute the speed \(v_C\). Compute \(v_D\) from \(v_C\). --- **Diagram Explanation:** The figure (Fig. 1) represents a schematic
**Problem 1:** Two 2.0-mm-diameter beads, C and D, are \(r_0 = 10 \, \text{mm}\) apart, measured between their centers (see Fig. 1). Bead C has mass \(m_C = 1.0 \, \text{g}\) and charge \(q_C = 2.0 \, \text{nC}\). Bead D has mass \(m_D = 2.0 \, \text{g}\) and charge \(q_D = -1.0 \, \text{nC}\). If the beads are released from rest, what are the speeds \(v_C\) and \(v_D\) at the instant the beads collide? **Partial answer:** \(v_D = 4.9 \, \text{cm/s}\). **(a)** Write down the energy conservation law for this system. To compute potential energies, use the expression for the potential energy of two point charges (do not plug in the numbers at this step, but work only with symbols). Note that the final potential energy of the system is not zero, and it is determined by the distances between the centers of the beads when they collide. At the end of this step you will have an expression that contains both speeds \(v_C\) and \(v_D\), but you will not know yet how to compute them separately. **(b)** In addition to the energy conservation, in this problem you need to use the law of conservation of linear momentum. The linear momentum of this system as a whole is equal to zero. Which relation between the speeds \(v_C\) and \(v_D\) has to be satisfied to guarantee zero linear momentum (express \(v_D\) in terms of \(v_C\), but do not plug in the numerical values yet)? **(c)** Use this expression for \(v_D\) in the energy conservation law from part (a) and work out the symbolic formula for \(v_C\) in terms of \(m_C, m_D, q_C, q_D, r_0,\) and \(R\). Only after you have the final formula, plug in the numerical values of the parameters to compute the speed \(v_C\). Compute \(v_D\) from \(v_C\). --- **Diagram Explanation:** The figure (Fig. 1) represents a schematic
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Hi, I was wondering if you can help me with PART A,PART B, AND PART C because I am having trouble solving this problem, I was wondering if you can help me and can you also label which part is PART A,PART B AND C. Thank you
Transcribed Image Text:**Problem 1:**
Two 2.0-mm-diameter beads, C and D, are \(r_0 = 10 \, \text{mm}\) apart, measured between their centers (see Fig. 1). Bead C has mass \(m_C = 1.0 \, \text{g}\) and charge \(q_C = 2.0 \, \text{nC}\). Bead D has mass \(m_D = 2.0 \, \text{g}\) and charge \(q_D = -1.0 \, \text{nC}\). If the beads are released from rest, what are the speeds \(v_C\) and \(v_D\) at the instant the beads collide?
**Partial answer:** \(v_D = 4.9 \, \text{cm/s}\).
**(a)** Write down the energy conservation law for this system. To compute potential energies, use the expression for the potential energy of two point charges (do not plug in the numbers at this step, but work only with symbols). Note that the final potential energy of the system is not zero, and it is determined by the distances between the centers of the beads when they collide. At the end of this step you will have an expression that contains both speeds \(v_C\) and \(v_D\), but you will not know yet how to compute them separately.
**(b)** In addition to the energy conservation, in this problem you need to use the law of conservation of linear momentum. The linear momentum of this system as a whole is equal to zero. Which relation between the speeds \(v_C\) and \(v_D\) has to be satisfied to guarantee zero linear momentum (express \(v_D\) in terms of \(v_C\), but do not plug in the numerical values yet)?
**(c)** Use this expression for \(v_D\) in the energy conservation law from part (a) and work out the symbolic formula for \(v_C\) in terms of \(m_C, m_D, q_C, q_D, r_0,\) and \(R\). Only after you have the final formula, plug in the numerical values of the parameters to compute the speed \(v_C\). Compute \(v_D\) from \(v_C\).
---
**Diagram Explanation:**
The figure (Fig. 1) represents a schematic
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