Velocity of two masses due to electric potential energy

Click For Summary
The discussion focuses on calculating the velocity of two masses influenced by electric potential energy, determined by their potential difference. The minimum distance between the masses is 10 cm, while the maximum is 30 cm due to their repulsion when the connecting string is cut. The potential energy is derived from the formula involving the charges and their distances, leading to the equation for kinetic energy, 1/2 mv^2. The initial calculation yielded a velocity of approximately 6.9 m/s, which was questioned for accuracy. It is clarified that four masses are involved, all connected, and they move away from each other upon the string being cut.
Jaccobtw
Messages
163
Reaction score
32
Homework Statement
Four masses 10g each are tied together by 10cm strings to make a square as shown. Two of the masses carry a charge of 2μC. The string between the two charged masses is cut and the system begins to move. What is the maximum speed of the masses in m/s? Do not consider gravity or friction. You can imagine the masses to be on a horizontal frictionless table.
Relevant Equations
U = kq/r
KE = 1/2mv^2
Screenshot (96).png

We can find the potential energy by finding the potential difference between the two masses. the minimum distance between the two masses is 10 cm. The maximum is 30 cm because they can be 3 string lengths apart as they repulse each other once the string is cut.

So, to get potential difference $$k_e q (\frac{1}{0.1} -\frac{1}{0.3})$$

Multiply by the other charge to get potential energy:

$$k_e q_1 q_2 (\frac{1}{0.1} -\frac{1}{0.3})$$

Set equal to 1/2 mv^2 and solve for velocity

I get about 6.9 m/s but this was wrong. Was my reasoning incorrect?
 
Physics news on Phys.org
How many masses move and how do they move relative to each other?
 
PeroK said:
How many masses move and how do they move relative to each other?
Two masses move away from each other
 
Jaccobtw said:
Two masses move away from each other
There are four masses in the problem. All tied to each other.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
Replies
15
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
5
Views
1K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 7 ·
Replies
7
Views
3K