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Hello, I really need help with this problem, can you help me with part A,Part B, and Part C because I really need help and can also label which one is which thank you so much

Hello, I really need help with this problem, can you help me with part A,Part B, and Part C because I really need help and can also label which one is which thank you so much

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Hello, I really need help with this problem, can you help me with part A,Part B, and Part C because I really need help and can also label which one is which thank you so much

**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:** \(v_D = 4.9 \, \text{cm/s}\). **Figure 1 Description:** The diagram (Fig. 1) shows two beads, labeled C and D, separated by a distance \(r_0 = 10 \, \text{mm}\). Each bead has a radius \(R = 1 \, \text{mm}\). **Questions:** 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
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:** \(v_D = 4.9 \, \text{cm/s}\). **Figure 1 Description:** The diagram (Fig. 1) shows two beads, labeled C and D, separated by a distance \(r_0 = 10 \, \text{mm}\). Each bead has a radius \(R = 1 \, \text{mm}\). **Questions:** 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
Expert Solution
Introduction:

We are given the 2 charges and their masses. We are also given their separation distances.

We know that the potential energy of the 2 charge system is

U=kq1q2ro

We use conservation of energy here to relate energies. We then use conservation of momentum. Hence we find the velocity of each particle.

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