Problem 1: Two 2.0-mm-diameter beads, C and D, are ro = 10 mm apart, measured between their centers (see Fig.1). Bead C has mass 1.0 g and charge qc 2.0 nC. Bead D has mass mp = : 2.0 g C and charge qp = -1.0 nC. If the beads are released from rest, what qD mc = = are the speeds uc and up at the instant the beads collide? Partial answer: D = 4.9 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 uc and up, but you will not know yet how to compute them separately. 5=10mm R=1mm FIG. 1: The scheme for Problem 1 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 uc and up has to be satisfied to guarantee zero linear momentum (express up in terms of uc, but do not plug in the numerical values yet)? c) Use this expression for up in the energy conservation law from part a) and work out the symbolic formula for uc in terms of mc, mp, qc, qd, ro and R. Only after you have the final formula, plug in the numerical values of the parameters to compute the speed vc. Compute up from uc.

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Chapter1: Units, Trigonometry. And Vectors
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HI, I keep getting the wrong answer for part a, part b and part c I asked my professor for help but he won't help me to see why I got the answer wrong, is there any chance you can help me with part A,PART B AND PART C. Also can you label them so I can see which part is pART A, PART B AND PART C. Thank you.

Problem 1: Two 2.0-mm-diameter beads, C and D, are ro = 10 mm
apart, measured between their centers (see Fig.1). Bead C has mass
1.0 g and charge qc 2.0 nC. Bead D has mass mp =
: 2.0 g C
and charge qp = -1.0 nC. If the beads are released from rest, what
qD
mc =
=
are the speeds uc and up at the instant the beads collide?
Partial answer: D = 4.9 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 uc and up, but you will not know yet how to compute them
separately.
5=10mm
R=1mm
FIG. 1: The scheme for Problem 1
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 uc and up has to be satisfied to guarantee zero linear momentum (express up in terms
of uc, but do not plug in the numerical values yet)?
c) Use this expression for up in the energy conservation law from part a) and work out the symbolic
formula for uc in terms of mc, mp, qc, qd, ro and R. Only after you have the final formula, plug in the
numerical values of the parameters to compute the speed vc. Compute up from uc.
Transcribed Image Text:Problem 1: Two 2.0-mm-diameter beads, C and D, are ro = 10 mm apart, measured between their centers (see Fig.1). Bead C has mass 1.0 g and charge qc 2.0 nC. Bead D has mass mp = : 2.0 g C and charge qp = -1.0 nC. If the beads are released from rest, what qD mc = = are the speeds uc and up at the instant the beads collide? Partial answer: D = 4.9 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 uc and up, but you will not know yet how to compute them separately. 5=10mm R=1mm FIG. 1: The scheme for Problem 1 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 uc and up has to be satisfied to guarantee zero linear momentum (express up in terms of uc, but do not plug in the numerical values yet)? c) Use this expression for up in the energy conservation law from part a) and work out the symbolic formula for uc in terms of mc, mp, qc, qd, ro and R. Only after you have the final formula, plug in the numerical values of the parameters to compute the speed vc. Compute up from uc.
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