Final repulsed velocity of two different charged masses

In summary, when two small metal spheres with masses 2.0g and 4.0g are tied together by a 5cm long massless string and have a (mue)c charge, the velocities of the spheres when they are far apart can be determined by looking at the total energy of the system before and after the string is cut. The total kinetic energy before is zero and the total potential energy is kqQ/R, while after the total kinetic energy is 0.5*m1*v1^2 + 0.5*m2*v2^2 and the total potential energy is zero. Using the equation kqQ/R = 0.5*m1*v1^2 + 0.5*m
  • #1
Dester
3
0
1. Two small metal spheres with masses 2.0g and 4.0g are tied together by a 5cm long massless string and are at rest on a frictionless table. Each sphere has a (mue)c charge.
The string is cut, what are the velocities of the sphere when they are far apart

I know two conservative qualities are at play, electric energy, and I'm not sure of the other
I tried using change under the curve with r being the variable (kQq/r) being equal to work and then said mv^2/2. However I can't articulate how I am sure this is the wrong approach.

What is the right way/approach?
 
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  • #2
Dester said:
What is the right way/approach?

You have the right idea.

You want to look at the total energy of the system before and after.

Before:

total kinetic energy is zero (both spheres are at rest)
total potential energy is kqQ/R


After:

total kinetic energy is 0.5*m1*v1^2 + 0.5*m2*v2^2
total potential energy is zero


So you get this equation as a result:

total energy before = total energy after
kqQ/R = 0.5*m1*v1^2 + 0.5*m2*v2^2

but that's two unknowns (v1, v2) with only one equation.


You need another bit of information and that is the force each sphere experiences is the same for all time (action / reaction).

Look at each of the spheres:

F = m1*a1
F = m2*a2

In terms of velocity:

F*dt = m1*dv1
F*dt = m2*dv2

You should be able to integrate that, equate them, and come up with another relationship between v1 and v2.


It might be worthwhile to check the answer by doing it the hard way, if only for the exercise. F = ma, F=kqQ/R^2, find position of each sphere as function of time or perhaps velocity as a function of x. In either case you take the limit as t or x goes to infinity to find the final velocities.
 

1. What is "final repulsed velocity of two different charged masses"?

Final repulsed velocity of two different charged masses refers to the velocity at which two objects with opposite charges separate from each other after being in close proximity.

2. How is the final repulsed velocity of two different charged masses calculated?

The final repulsed velocity of two different charged masses can be calculated using the formula v = (q1q2/r)m where v is the final velocity, q1 and q2 are the charges of the two objects, r is the distance between them, and m is a constant.

3. What factors affect the final repulsed velocity of two different charged masses?

The final repulsed velocity of two different charged masses is affected by the magnitude of the charges of the two objects, the distance between them, and the strength of the electric field between them.

4. Can the final repulsed velocity of two different charged masses be negative?

No, the final repulsed velocity of two different charged masses cannot be negative because it represents the speed at which the two objects are moving away from each other.

5. How does the final repulsed velocity of two different charged masses relate to Coulomb's Law?

The final repulsed velocity of two different charged masses is directly related to Coulomb's Law, which states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This means that as the final repulsed velocity increases, the force between the two objects also increases.

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