Electric potential energy finding velocities

AI Thread Summary
The discussion revolves around calculating the velocities of two charged metal balls after a force holding them in place is removed. Users are attempting to apply the conservation of momentum and energy equations, specifically using the formula for electric potential energy, E = (kq1q2)/d. There is confusion regarding the correct application of these equations, as initial calculations yield incorrect results for the velocities. The expected final velocities are v1=0.1 m/s and v2=0.07 m/s, indicating a need for clarification on the calculations. The thread highlights the challenges in solving physics problems involving electric forces and conservation laws.
jbutle54
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Two metal balls of mass m1=3.41g (diameter=3.41mm) and m2=7.88g (diameter=7.88mm) have positive charges of q1=1.41nC and q2=9.19nC. A force holds them in place so that their centers are separated by 8.79mm. What will their velocities be after the force is removed and they are separated by a large distance?

v1=?
v2=?



Can someone help me with which equations to use and any other info that would be helpful thanks



I've tried several equations but gotten wrong answers everytime.
 
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welcome to pf!

hi jbutle54! welcome to pf! :smile:
jbutle54 said:
Can someone help me with which equations to use and any other info

have you used conservation of momentum (ie centre of mass remains stationary)? :wink:
 
Yes I did. I end up with 2 variables and I plugged one into the other and still cannot figure it out.
 
should have worked …

show us your calculations :smile:​
 
wow ok

E= (kq1q2)/d k=8.99*10^9 q1=5 q2=8 d=.008

E ends up to be 4.495*10^13


m1v1=m2v2 .5(.0025)v1^2 + .5(.008)v2^2=E

ends up
v1=-1.06*10^8
v2=-1.579*10^6

which is wrong it should equal
v1=0.1m/s
v2=0.07m/s
 
jbutle54 said:
Two metal balls of mass m1=3.41g (diameter=3.41mm) and m2=7.88g (diameter=7.88mm) have positive charges of q1=1.41nC and q2=9.19nC.

jbutle54 said:
E= (kq1q2)/d k=8.99*10^9 q1=5 q2=8 d=.008

erm :redface: …

are we doing the same question? :smile:​
 

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