Question

Two small metal cubes with masses 2.0 g and 4.0 g are tied together by a 5.5-cm-long massless string and are at rest on a frictionless surface. Each is charged to +2.5 μC . NEED HELP WITH PART C! THANK YOU A AND B ARE GOOD!

Part A Part complete What is the energy of this system? Express your answer using two significant figures. Emech = 1.0   J   SubmitPrevious Answers Correct Part B Part complete What is the tension in the string? Express your answer using two significant figures. T = 19   N   SubmitPrevious Answers Correct Part C The string is cut. What is the speed of each sphere when they are far apart? Hint: There are two conserved quantities. Make use of both. Express your answers using two significant figures. Enter your answers numerically separated by a comma. v2g v2g , v4g = 14,−28   m/s

Answer 1

I am solving only part C as per request.

Energy of the system will be conserved as there is no frictional forces.

Initially only electrostatic potential energy will be there,

so initial energy = K q1q2/r = 1 joule

Now when they are far apart, their potential energy will be zero as they are not in the electrostatic field of each other.

SO, if the velocities are v1 and v2, they will be having only kinetic energies.

So, kinetic energy of blocks = initial potential energy = 1

1/2m1v1^2 + 1/2m2v2^2 = 1

Also there is no external force on them, so their momentum will aslo be conserved.

so m1v1 = m2v2

solving the equations we get:

v1 = 12.909

v2 = 25.819