Answer: [tex]u_{1}=-0.075m/s[/tex] and [tex]u_{2}=0.500m/s[/tex]
Explanation:
An elastic collision is one in which both the total kinetic energy of the system and the linear momentum are conserved. That is, during the collision there is no sound, heat or permanent deformations in the bodies as a result of the impact.
Now, in the case of the satellites described here, we have:
[tex]m_{1}v_{1}+m_{2}v_{2}=m_{1}u_{1}+m_{2}u_{2}[/tex] (1) Conservation of momentum
[tex]\frac{1}{2}m_{1}v_{1}^{2} +\frac{1}{2}m_{2}v_{2}^{2} =\frac{1}{2}m_{1}u_{1}^{2} +\frac{1}{2}m_{2}u_{2}^{2} [/tex] (2) Conservation of kinetic energy
Where:
[tex]m_{1}=2.5(10)^{3}kg[/tex] is the mass of the first satellite
[tex]m_{2}=7.5(10)^{3}kg[/tex] is the mass of the second satellite
[tex]v_{1}=0.150m/s[/tex] is the initial velocity of the first satellite
[tex]v_{2}=0m/s[/tex] is the initial velocity of the second satellite (we are told it is at rest)
[tex]u_{1}[/tex] is the final relative velocity of the first satellite
[tex]u_{2}[/tex] is the final relative velocity of the second satellite
Now, as we know the second satellite is at rest before the collision, equations (1) and (2) change to:
[tex]m_{1}v_{1}=m_{1}u_{1}+m_{2}u_{2}[/tex] (3)
[tex]\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{1}u_{1}^{2} +\frac{1}{2}m_{2}u_{2}^{2}[/tex] (4)
Solving this system of equations we have the equations for [tex]u_{1}[/tex] and [tex]u_{2}[/tex]:
[tex]u_{1}=\frac{v_{1}(m_{1}-m_{2})}{m_{1}+m_{2}}[/tex] (5)
[tex]u_{2}=\frac{2m_{1}v_{1}}{m_{1}+m_{2}}[/tex] (6)
Substituting the known values on both equations:
[tex]u_{1}=\frac{0.150m/s(2.5(10)^{3}kg-7.5(10)^{3}kg)}{2.5(10)^{3}kg+7.5(10)^{3}kg}[/tex] (7)
[tex]u_{1}=-0.075m/s[/tex] (8) This is the final relative velocity of the first satellite
[tex]u_{2}=\frac{2(2.5(10)^{3}kg)(0.150m/s)}{2.5(10)^{3}kg+7.5(10)^{3}kg}[/tex] (9)
[tex]u_{2}=0.500m/s[/tex] (10) his is the final relative velocity of the second satellite
