Conservation of Energy with Metal Spheres

[ZEli]

Homework Statement

Two small metal spheres with masses 2.0g and 4.0g are tied together by a 5.4-cm-long massless string and are at rest on a frictionless surface. Each is charged to +2.1 μC .

The string is cut. What is the speed of each sphere when they are far apart?

Homework Equations

U = Kq1q2/r
F = K q1/q2/r^2
m1v1 = m2v2

The Attempt at a Solution

I'm not entirely sure how to approach this. I calculated that the energy in the system is .74 J and also the tension in the string is 14 N. How can I use that to my advantage?

 

[DukeLuke]

When they are far apart the potential energy -> 0, so they energy of the system that begins entirely as potential must now be all kinetic. This should give you one equation. To get a 2nd you know that momentum is conserved as well.

 

[ZEli]

So, .74 = 1/2mv^2. I picked the 2g sphere, so v = .86 m/s. This doesn't give me the right answer :(

 

[DukeLuke]

The initial energy is equal to the sum of the kinetic energies of the two spheres at the end. So it should be .73 = .5*m1*v1^2 + .5m2*v2^2. This leaves you with two unknowns (v1 and v2). To get a 2nd equation you can use conservation of momentum.

 

[ZEli]

Hmm, it's still not quite right. I have the two equations, .74 = .5*m1*v1^2 + .5m2*v2^2 and 2*v1 = 4*v2. I solved for v1, giving me v1 = 2*v2. Plugging into the first equation I end up with v2 = .35 and v1 = .70.

I understand that all potential energy turns into kinetic energy and that energy is conserved. Am I just using the wrong equation for conservation of energy?

 

[DukeLuke]

I get a different answer, maybe check your work? Remember when you plug v1=2v2 into the equation v1^2=4v2^2. Also don't forget mass is in grams.

 

[ZEli]

Ohh, I forgot to convert grams into kg. Thank you for all the help!