Respuesta :
Answer:
Explanation:
Apply the law of conservation of energy
[tex]KE_i+PE_i=KE_f+PE_f[/tex]
[tex]Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)[/tex]
from the law of conservation of the linear momentum
[tex]m_1v_1=m_2v_2[/tex]
Therefore,
[tex]Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)[/tex]
[tex]=\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} ][/tex]
[tex]v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ][/tex]
Substitute the values in the above result
[tex]v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ][/tex]
[tex]=[\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[/tex]
B) the speed of the sphere with mass 107.0 kg is
[tex]v_2=\frac{m_1v_1}{m_2}[/tex]
[tex]=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s[/tex]
C) the magnitude of the relative velocity with which one sphere is
[tex]v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s[/tex]
D) the distance of the centre is proportional to the acceleration
[tex]\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962[/tex]
Thus,
[tex]x_1=3.962x_2[/tex]
and
[tex]x_2=0.252x_1[/tex]
When the sphere make contact with eachother
Therefore,
[tex]x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r[/tex]
And
[tex]x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r[/tex]
The point of contact of the sphere is
[tex]32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m[/tex]
(A) The speed of the sphere of 27 kg is [tex]1.2664 \times 10^{-5} \;\rm m/s[/tex].
(B) The speed of sphere of mass 107 kg is [tex]3.195 \times 10^{-6} \;\rm m/s[/tex].
(C) The magnitude of the relative velocity with which one sphere is approaching to the other is [tex]15.85 \times 10^{-6} \;\rm m/s[/tex].
(D) The distance from the initial position of the center of the 27.0 sphere is 20.506 m.
Given data:
The mass of sphere 1 is, [tex]m_{1} = 27.0 \;\rm kg[/tex].
The mass of sphere 2 is, [tex]m_{2} = 107.0 \;\rm kg[/tex].
The radius of each spheres are, r = 0.10 m.
The distance between the centers of each sphere is, d = 41.0 m.
(A)
In this part, we can apply the conservation of energy to find the speed at given distance of 26.0 m. So,
Total energy at initial = Total energy at final
[tex]\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{1}{2}(m_{1}v^{2}_{1}+m_{2}v^{2}_{2})[/tex]
Now, as per the conservation of momentum,
[tex]m_{1}v_{1}=m_{2}v_{2}\\\\v_{2}=\dfrac{m_{1}v_{1}}{m_{2}}[/tex]
Therefore,
[tex]\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{1}{2}(m_{1}v^{2}_{1}+m_{2} \times [m_{1}v_{1}/m_{2}]^{2})\\\\\\\dfrac{Gm_{1}m_{2}}{r_{f}}-\dfrac{Gm_{1}m_{2}}{r_{i}}=\dfrac{m_{1}v^{2}_{1}}{2} \times (\dfrac{m_{1}+m_{2}}{m_{2}})[/tex]
Modifying as,
[tex]v^{2}_{1}=[\dfrac{2Gm^{2}_{2}}{m_{1}+m_{2}}] \times [\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}}][/tex]
Substitute the values in the above result
[tex]v^{2}_{1}=[\dfrac{2 \times 6.67 \times 10^{-11} \times 107^{2}}{27+107}] \times [\dfrac{1}{26}-\dfrac{1}{41}]\\\\v_{1}=\sqrt{1.6038 \times 10^{-5}}\\\\v_{1}=1.2664 \times 10^{-5} \;\rm m/s[/tex]
Thus, the speed of the sphere of 27 kg is [tex]1.2664 \times 10^{-5} \;\rm m/s[/tex].
(B)
The sphere with mass 107 kg is calculated as,
[tex]v_{2}=\dfrac{m_{1}v_{1}}{m_{2}}\\\\v_{2}=\dfrac{27 \times 1.2664 \times 10^{-5} \;\rm m/s }{107}\\\\v_{2}=3.195 \times 10^{-6} \;\rm m/s[/tex]
Thus, the speed of sphere of mass 107 kg is [tex]3.195 \times 10^{-6} \;\rm m/s[/tex].
(C)
The magnitude of the relative velocity with which one sphere is
[tex]v_{r} = v_{1} +v_{2}\\\\v_{r} =( 1.2664 \times 10^{-5})+ (3.195 \times 10^{-6})\\\\v_{r}=15.85 \times 10^{-6} \;\rm m/s[/tex]
Thus, the magnitude of the relative velocity with which one sphere is approaching to the other is [tex]15.85 \times 10^{-6} \;\rm m/s[/tex].
(D)
The distance of the centre is proportional to the acceleration,
[tex]\dfrac{x_{1}}{x_{2}}=\dfrac{m_{2}}{m_{1}}\\\\\\\dfrac{x_{1}}{x_{2}}=\dfrac{107}{27}\\\\\\\dfrac{x_{1}}{x_{2}}=3.962\\\\\\x_{1}=3.962 \times x_{2}[/tex]
As per the given problem,
[tex]x_{1}+x_{2}+2R = d\\\\3.962x_{2}+x_{2}+(2R) = 41 \\\\x_{2} = 8.262-0.403R[/tex]
And,
[tex]x_{1}=0.252x_{2}\\\\x_{1}=0.252 \times (8.262-0.403R)\\\\x_{1}=32.747-1.597R[/tex]
Then for point of contact of the sphere:
[tex]x_{1} = x_{2}\\\\32.747-1.597R = 8.262-0.403R\\\\R =20.506 \;\rm m[/tex]
Thus, the distance from the initial position of the center of the 27.0 sphere is 20.506 m.
Learn more about the conservation of linear momentum here:
from the initial position of the center of the 27.0 sphere
