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steposvetlana [31]

3 years ago

9

The 0.100 kg sphere in (Figure 1) is released from rest at the position shown in the sketch, with its center 0.400 m from the ce

nter of the 5.00 kg mass. Assume that the only forces on the 0.100 kg sphere are the gravitational forces exerted by the other two spheres and that the 5.00 kg and 10.0 kg spheres are held in place at their initial positions.

1 answer:

slega [8]3 years ago

6 0

Answer:

the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : $3.3337*10^{-5} \ m/s$

Explanation:

Using the expression of the Change in Gravitational Potential Energy:

$U= -(\frac{ Gm_1m_2 }{r_2} - \frac{ Gm_1m_2 }{r_1}) \\ \\ U=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})$

So when the sphere exerts just 10 kg mass; the change in the gravitational potential energy is :

$U_1=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})$

$U_1=6.67*10^{-11}*0.1 \ kg*10 \ kg(\frac{ 1 }{0.6 \ m} - \frac{ 1 }{0.8 \ m})$

$U_1 = 2.778*10^{-11} J$

the change in the gravitational potential energy when the sphere exerts just 5 kg mass is ;

$U_2=Gm_1m_2 (\frac{ 1 }{r_2} - \frac{ 1 }{r_1})$

$U_2=6.67*10^{-11}*0.1 \ kg*5 \ kg(\frac{ 1 }{0.4 \ m} - \frac{ 1 }{0.2 \ m})$

$U_2 = -8.335*10^{-11} J$

The net total change is:

$U_{total } = U_1 +U_2$

$U_{total} = 2.778*10^{-11} + (-8.335*10^{-11})$

$U_{total} = -5.557*10^{-11}$

We all know that for there to be a balance ;loss of gravitational potential energy must be equal to the gain in kinetic energy .

SO;

K.E = $5.557*10^{-11}$

$\frac{1}{2}mv^2 = 5.557*10^{-11}$

$v^2 = \frac{2*5.557*10^{-11} \ J}{m_1}$

$v=\sqrt{ \frac{2*5.557*10^{-11} \ J}{0. 1 \kg}$

v = $3.3337*10^{-5} \ m/s$

Thus, the speed of 0.10 sphere when it is moved to 0.20 kg to the left is : $3.3337*10^{-5} \ m/s$

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1) $+2.19\mu C$

The electrostatic force between two charges is given by

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Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

$F = -72.1 mN = -0.0721 N$ (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

$q_2 = Q-q_1$

So we can rewrite eq.(1) as

$F=k \frac{q_1 (Q-q_1)}{r^2}$

Solving for q1,

$Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0$

Since $Q=4.37\cdot 10^{-6} C$ , we can substituting all numbers into the equation:

$8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0$

which gives two solutions:

$q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C$

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

$q_1 = +6.70 \mu C$

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