Two small metal spheres with masses 2.0g and 4.0g are tied together by a 5.1--lo

Question

Two small metal spheres with masses 2.0g and 4.0g are tied together by a 5.1--long massless string and are at rest on a frictionless surface. Each is charged to +2.3uC u=mule sign.. .

question have 3 parts, i answered the first two, stuck in the second part,, can anyone help? thanks..

A) What is the energy of this system?
Express your answer using two significant figures.
=0.93 J
Correct

B)What is the tension in the string?
Express your answer using two significant figures.
=18 N
Correct


the part i am stuck at....is part c
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. answer should be in m/s...please someone attempt to help me, or give me a head start, or just solve it, whichever one, its making me not sleep at night..thank you.

Explanation / Answer

C) The two conserved quantities are the total energy energy and the momentum. The initial momentum is zero, so

0 = Pf = p1f + p2f, so

p1f = -p2f, and

m1v1f = -m2v2f, so

v1f = -(m2/m1)v2f

In addition, the Coulomb potential energy initially equals the kinetic energy final, or

0.93J = (1/2)m1(v1f)^2 + (1/2)m2(v2f)^2 (0.93J was the answer to part A)

From the conservation of momentum,

0.93J = (1/2)(m2^2/m1)(v2f)^2 + (1/2)m2(v2f)^2
= (1/2)( (4*10^-3 kg)^2/(2*10^-3 kg) + (4*10^-3 kg) ) (v2f)^2
= 0.006 kg (v2f)^2, so

v2f = (0.93J/0.006 kg) = 12.45 m/s

And, the final velocity of the other sphere is

v1f = - (m2/m1)v2f = - (4*10^-3 kg/2*10^-3 kg)(12.45 m/s) = 24.9 m/s

So, the two velocities are 24.9 m/s for the lighter sphere and 12.5 m/s for the heavier sphere.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.