a) Final velocity after docking: +0.094 m/s
b) Kinetic energy loss: 31.6 J
c) Final velocity after docking: -0.056 m/s
d) Kinetic energy loss: 31.6 J
Explanation:
a)
Since the system of two satellites is an isolated system, the total momentum is conserved. So we can write:
[tex]p_i = p_f[/tex]
Or
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2)v[/tex]
where, in the reference frame in which the first satellite was originally at rest, we have:
[tex]m_1 = 4.50\cdot 10^3 kg[/tex] is the mass of the 1st satellite
[tex]m_2 = 7.50\cdot 10^3 kg[/tex] is the mass of the 2nd satellite
[tex]u_1 = 0[/tex] is the initial velocity of the 1st satellite
[tex]u_2 = +0.150 m/s[/tex] is the initial velocity of the 2nd satellite
v is their final velocity after docking
Solving for v,
[tex]v=\frac{m_2 u_2}{m_1 +m_2}=\frac{(7.50\cdot 10^3)(0.150)}{4.50\cdot 10^3 + 7.50\cdot 10^3}=0.0938 m/s[/tex]
b)
The initial kinetic energy of the system is just the kinetic energy of the 2nd satellite:
[tex]K_i = \frac{1}{2}m_2 u_2^2 = \frac{1}{2}(7.50\cdot 10^3)(0.150)^2=84.4 J[/tex]
The final kinetic energy of the two combined satellites is:
[tex]K_f = \frac{1}{2}(m_1 +m_2)v^2=\frac{1}{2}(4.50\cdot 10^3+7.50\cdot 10^3)(0.0938)^2=52.8 J[/tex]
Threfore, the loss in kinetic energy during the collision is:
[tex]\Delta K = K_f - K_i = 52.8 - 84.4=-31.6 J[/tex]
c)
In this case, we are in the reference frame in which the second satellite is at rest. So, we have
[tex]u_2 = 0[/tex] (initial velocity of satellite 2 is zero)
[tex]u_1 = -0.150 m/s[/tex] (initial velocity of a satellite 1)
Therefore, by applying the equation of conservation of momentum,
[tex]m_1 u_1 + m_2 u_2 = (m_1 + m_2)v[/tex]
And solving for v,
[tex]v=\frac{m_1 u_1}{m_1 +m_2}=\frac{(4.50\cdot 10^3)(-0.150)}{4.50\cdot 10^3 + 7.50\cdot 10^3}=-0.0563 m/s[/tex]
d)
The initial kinetic energy of the system is just the kinetic energy of satellite 1, since satellite 2 is at rest:
[tex]K_i = \frac{1}{2}m_1 u_1^2 = \frac{1}{2}(4.50\cdot 10^3)(-0.150)^2=50.6 J[/tex]
The final kinetic energy of the system is the kinetic energy of the two combined satellites after docking:
[tex]K_f = \frac{1}{2}(m_1 + m_2)v^2=\frac{1}{2}(4.50\cdot 10^3+ 7.50\cdot 10^3)(-0.0563)^2=19.0 J[/tex]
Therefore, the kinetic energy lost in the collision is
[tex]\Delta K = K_f - K_i = 19.0 -50.6 = -31.6 J[/tex]
Learn more about collisions:
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