**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 a mass \( m_C = 1.0 \, \text{g} \) and charge \( q_C = 2.0 \, \text{nC} \). Bead D has a 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 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 \( v_C \). --- **Figure 1:** - An illustration

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Chapter1: Units, Trigonometry. And Vectors
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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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I am struggling really bad with this problem I was wondering if you can help me because I am having a really bad time figuring it out, is there a chance that you can help me with PART A, Part b, and part c and also can you label which part is PART A, PART B AND PART C. Thank you

**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 a mass \( m_C = 1.0 \, \text{g} \) and charge \( q_C = 2.0 \, \text{nC} \). Bead D has a 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 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 \( v_C \).

---

**Figure 1:**

- An illustration
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 a mass \( m_C = 1.0 \, \text{g} \) and charge \( q_C = 2.0 \, \text{nC} \). Bead D has a 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 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 \( v_C \). --- **Figure 1:** - An illustration
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