Question

N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have equal radii, = 0.10 . The spheres are released from rest with their centers a distance 41.0 apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres.
A) When their centers are a distance 26.0 apart, find the speed of the 27.0 sphere.

B) Find the speed of the sphere with mass 107.0 .

C) Find the magnitude of the relative velocity with which one sphere is approaching to the other.

D) How far from the initial position of the center of the 27.0 sphere do the surfaces of the two spheres collide?

Answer 1

Answer:

Explanation:

Apply the law of conservation of energy

$KE_i+PE_i=KE_f+PE_f$

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

from the law of conservation of the linear momentum

$m_1v_1=m_2v_2$

Therefore,

$Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)$

$=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]$

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

Substitute the values in the above result

$v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]$

$=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s$

B)  the speed of the sphere with mass 107.0 kg is

$v_2=\frac{m_1v_1}{m_2}$

$=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s$

C)  the magnitude of the relative velocity with which one sphere is

$v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s$

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

$\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962$

Thus,

$x_1=3.962x_2$

and

$x_2=0.252x_1$

When the sphere make contact with eachother

Therefore,

$x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r$

And

$x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r$

The point of contact of the sphere is

$32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m$