Velocities of colliding particles (Kinetic and Potential Enegies)

In summary, the problem involves two beads, C and D, with diameters of 2.0mm and a distance of 9.0mm between their centers. Bead C has a mass of 1.0g and a charge of 2.5nC, while bead D has a mass of 1.5g and a charge of -1.0nC. The equation used is Kinitial + Uinitial = Kfinal + Ufinal, with both initial kinetic energies being zero and final potential being zero due to the distance between them being zero after collision. The Coulomb force is also taken into consideration for the final potential energy.
  • #1
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Homework Statement



Two 2.0mm diameter beads, C and D, are 9.0mm apart, measured between their centers. Bead C has mass 1.0g and charge 2.5nC . Bead D has mass 1.5g and charge -1.0nC .
K= 9*10^9 Nm^2/C^2
Vinitials =0
mass of C= 1 g
mass of D= 1.5 g
charge of C = 2.5nC
charge of D = -1nC
distance r between them= 9mm

Homework Equations



a) If the beads are released from rest, what is the speed vc at the instant the beads collide?
b) If the beads are released from rest, what is the speed vd at the instant the beads collide?

The Attempt at a Solution


I start by setting up my diagram of the two beads lying on the x-axis. They are 9mm apart from the center, We know initial velocities of both are zero. So I decided to use the conservation of energies formula. Kintial +Uinitial = Kfinal + Ufinal
C: 1/2mvi^2 + (K*C*D)/rinitial = 1/2mvf^2 + (K*C*D)/rfinal

Both Initial Kinetic is zero since they start as rest, as well as Final potential because the distance between them would be zero

So I have (K*C*D)/rinitial = 1/2(mvf^2)
But we have to take into account the kinetic energy of Bead D as well. When doing a similar problem I was told that both Final Kinetic energies must be added. But I would have two unknowns. Both vfinals. I don't know where to go from here. Any hint?
Thanks
 
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  • #2
Lets see if I understand you:
Both Initial Kinetic is zero since they start as rest, as well as Final potential because the distance between them would be zero
... I thought the potential changed as 1/d (d is the distance between their centers) ... but 1/0 is undefined.

But surely the beads stop before r=0?

Apart from that:
PE before the motion = PE after the motion + total kinetic energy.
But you also know something else that is conserved.
 
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  • #3
ok thanks for the reply/ so are you saying that the I should use the distance between the centers as rfinal for the PEfinal?
 
  • #4
You can answer that question for yourself by looking at how the Coulomb force works.

For two spheres touching - only the charges immediately next to each other have zero separation - the charges on the rest of the two spheres are further apart. But - classically - the area of the contact is zero so the charge in the contact area is also zero.

If you crunch the numbers you'll find that for r>R (where R is the radius of the bead/sphere), the force acts as if the entire charge is concentrated in the center at r=0. But for r<R, the actual charge distribution matters. Fortunately, two hard beads of charge won't get their centers that close together.
 
  • #5
for your question. It seems like you are on the right track with using the conservation of energy formula to solve this problem. Here are a few hints to help you continue:

1. Remember that the total energy (kinetic + potential) of the system is conserved. This means that the total energy before the collision must be equal to the total energy after the collision.

2. The kinetic energy of a particle is given by 1/2mv^2, where m is the mass of the particle and v is its velocity.

3. The potential energy between two charged particles is given by the Coulomb's law equation: U = (K*q1*q2)/r, where K is the Coulomb's constant, q1 and q2 are the charges of the particles, and r is the distance between them.

4. Since both beads are released from rest, their initial velocities are zero. This means that their initial kinetic energies are also zero.

5. The final potential energy will also be zero since the beads will be in contact with each other at the moment of collision.

6. You can set up two equations using the conservation of energy formula, one for each bead, and solve for the final velocities.

I hope these hints help you continue with your solution. Remember to use the given values for mass, charge, and distance and make sure to convert them to the appropriate units. Good luck!
 

1. What is the difference between kinetic and potential energy in a collision?

Kinetic energy is the energy an object possesses due to its motion, while potential energy is the energy an object possesses due to its position or configuration. In a collision between particles, kinetic energy is transferred from one particle to another, while potential energy remains constant.

2. How does the velocity of colliding particles affect their kinetic energy?

The kinetic energy of particles in a collision is directly proportional to their velocity. This means that as the velocity of the particles increases, so does their kinetic energy. Conversely, if the velocity decreases, the kinetic energy also decreases.

3. Can the potential energy of colliding particles change during a collision?

In most cases, the potential energy of colliding particles remains constant during a collision. This is because potential energy is dependent on the position or configuration of the particles, which does not typically change during a collision. However, in certain scenarios, such as an explosion, the potential energy of the particles may change due to a change in their position or configuration.

4. How is the total energy conserved in a collision between particles?

In a closed system, the total energy (kinetic + potential) is always conserved during a collision between particles. This means that the total energy before the collision is equal to the total energy after the collision. This is known as the law of conservation of energy.

5. How can the velocities of colliding particles be calculated?

The velocities of colliding particles can be calculated using the laws of conservation of momentum and energy. These equations take into account the masses and initial velocities of the particles, as well as the type of collision (elastic or inelastic). Additionally, computer simulations and experiments can also be used to measure the velocities of colliding particles.

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