Two small metal cubes with masses 2.0 g and 4.0 g are tied together by a 5.5-cm-

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 +2.0 mu C . What is the energy of this system? Express your answer using two significant figures. What is the tension in the string? Express your answer using two significant figures. The string is cut. What is the speed of each sphere when they are far apart? Express your answers using two significant figures. Enter your answers numerically separated by a comma.

Explanation / Answer

a)
Initially Energy of this system , E = Potential Energy
P.E = K*q1q2/r
P.E = 8.9*10^9 * (2*10^-6)^2 / (5.5*10^-2) J
P.E = 0.647 J
Energy of this system = 0.647 J

(b)
Tension in the string = Force between the two charges
F = q*E
T = k*q^2/r
T = 8.9*10^9 *  (2*10^-6)^2 / (5.5*10^-2)^2
T = 11.76 N
Tension in the string, T = 11.76 N

(c)
When they are far apart all the electric potential energy has been converted to Kinetic energy.
So KE (2 g ) + KE (4 g ) = 0.647 J
0.5*(2*10^-3)v2g^2 + 0.5*(4*10^-3)v4g^2 = 0.647 -----------1

Momentum is also conserved ,
m1*v1 = m2*v2
2 *v2g = 4*v4g
v2g = 2*v4g ----------------2

Using eq 1 and 2 -

0.5*(2*10^-3)(2*v4g)^2 + 0.5*(4*10^-3)v4g^2 = 0.647
Caluculating v4g -
v4g = 10.38 m/s

v2g = 2*v4g
v2g = 2*10.38 m/s
v2g = 20.76 m/s

v2g = 20.76 m/s
v4g = 10.38 m/s

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