Question

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 center 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.

Answer 1

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 :   $3.3337*10^{-5} \ m/s$