Answer:
Explanation:
Apply the law of conservation of energy

![Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)](https://tex.z-dn.net/?f=Gm_1m_2%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1v_1%5E2%2Bm_2v_2%5E2%29)
from the law of conservation of the linear momentum

Therefore,
![Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)](https://tex.z-dn.net/?f=Gm_1m_2%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D%3D%5Cfrac%7B1%7D%7B2%7D%20%28m_1v_1%5E2%2Bm_2v_2%5E2%29)
![=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%20%5Bm_1v_1%5E2%2Bm_2%5B%5Cfrac%7Bm_1v_1%7D%7Bm_2%7D%20%5D%5E2%5D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B2%7D%20%5Bm_1v_1%5E2%2B%5Cfrac%7Bm_1%5E2v_1%5E2%7D%7Bm_2%7D%20%5D%5C%5C%5C%5C%3D%5Cfrac%7Bm_1v_1%5E2%7D%7B2%7D%20%5B%5Cfrac%7Bm_1%2Bm_2%7D%7Bm_2%7D%20%5D)
![v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]](https://tex.z-dn.net/?f=v_1%5E2%3D%5B%5Cfrac%7B2Gm_2%5E2%7D%7Bm_1%2Bm_2%7D%20%5D%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D)
Substitute the values in the above result
![v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]](https://tex.z-dn.net/?f=v_1%5E2%3D%5B%5Cfrac%7B2Gm_2%5E2%7D%7Bm_1%2Bm_2%7D%20%5D%5B%5Cfrac%7B1%7D%7Br_f%7D%20-%5Cfrac%7B1%7D%7Br_1%7D%20%5D)
![=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B2%286.67%5Ctimes%2010%5E-%5E1%5E1%29%28107%29%5E2%7D%7B27%2B107%7D%20%5D%5B%5Cfrac%7B1%7D%7B26%7D%20-%5Cfrac%7B1%7D%7B41%7D%5D%20%5C%5C%5C%5C%3D1.6038%5Ctimes%2010%5E-%5E1%5E0%5C%5C%5C%5Cv_1%3D%5Csqrt%7B1.6038%5Ctimes%20106-%5E1%5E0%7D%20%5C%5C%5C%5C%3D1.2664%20%5Ctimes%2010%5E-%5E5m%2Fs)
B) the speed of the sphere with mass 107.0 kg is

\\\\=3.195\times 10^-^6m/s](https://tex.z-dn.net/?f=%3D%5B%5Cfrac%7B27%7D%7B107%7D%20%5D%281.2664%20%5Ctimes%2010%5E-%5E5%29%5C%5C%5C%5C%3D3.195%5Ctimes%2010%5E-%5E6m%2Fs)
C) the magnitude of the relative velocity with which one sphere is

D) the distance of the centre is proportional to the acceleration

Thus,

and

When the sphere make contact with eachother
Therefore,

And

The point of contact of the sphere is

The potential energy of the spring is 6.75 J
The elastic potential energy stored in the spring is given by the equation:

where;
k is the spring constant
x is the compression/stretching of the string
In this problem, we have the spring as follows:
k = 150 N/m is the spring constant
x = 0.3 m is the compression
Substituting in the equation, we get


Therefore. the elastic potential energy stored in the spring is 6.75J .
Learn more about potential energy here:
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