11-098-satellite_and_junk.jpg
A spherical satellite of radius 4.6 m and mass M = 195 kg is originally moving with velocity vsatellite,i = < 3000, 0, 0 > m/s, and is originally rotating with an angular speed ?1 = 2 radians/second, in the direction shown in the diagram. A small piece of space junk of mass m = 4.3 kg is initially moving toward the satellite with velocity vjunk,i = < -2300, 0, 0 > m/s. The space junk hits the edge of the satellite at location C as shown in the diagram, and moves off with a new velocity vjunk,f = < -1100, 500, 0 > m/s. Both before and after the collision, the rotation of the space junk is negligible.
MOMENTUM
1) After the collision, what is the final momentum of the satellite?
psatellite,f = kg·m/s
2) What is the final velocity of the satellite?
vsatellite,f = m/s
ANGULAR MOMENTUM
3) Assuming the density of the satellite is uniform throughout, what is the moment of inertia of the satellite? (The moment of inertia of a uniform-density sphere is (2/5)MR2.)
I = kg·m2
4) What is the initial rotational angular momentum of the satellite, around location D (its center of mass)? (Be sure your signs are correct).
Lsatellite,i = kg·m2/s
5) Which of the following statements about the translational angular momentum of the space junk, about location D, are true? Check all that apply:
Because the space junk is traveling in a straight line, its angular momentum is zero.The translational angular momentum of the space junk is the same when the space junk is at locations A, B, and just before getting to C.(r_perp) is the same when the space junk is at locations A, B, and just before getting to C.xp_y - yp_x is the same when the space junk is at locations A, B, and just before getting to C.The translational angular momentum of the space junk is in the -z direction.
6) An instant before the collision, when the space junk is almost at location C, what is the translational angular momentum of the space junk about location D?
Ljunk,i
= kg·m2/s
7) An instant after the collision, when the space junk is just slightly beyond location C, what is the translational angular momentum of the space junk about location D?
Ljunk,f
= kg·m2/s
8) At the same instant after the collision, what is the rotational angular momentum of the satellite?
L,f
= kg·m2/s
9) What is the new angular speed of the satellite?
omega_2 = radians/s
ENERGY
Consider a system consisting of both objects (space junk and satellite).
10) Before the collision, what was the total kinetic energy (translational + rotational) of the system?
Ki
= joules
11) After the collision, what is the total kinetic energy of the system?
Kf
= joules
12) What was the rise in thermal energy for the space junk and satellite combined? (Your answer will be judged correct if it is consistent with earlier work, even if there were mistakes in the earlier work -- "propagation of errors".)
?Ethermal
= joules
To solve these questions, we need to apply the laws of conservation of momentum, angular momentum, and energy. We'll go through each question step by step.
1) After the collision, the final momentum of the satellite can be found using the equation:
psatellite,f = M * vsatellite,f
where M is the mass of the satellite and vsatellite,f is the final velocity of the satellite. The mass of the satellite is given as M = 195 kg, and we can calculate the final velocity of the satellite using the conservation of momentum equation.
2) The final velocity of the satellite can be found using the conservation of momentum equation:
M * vsatellite,i = M * vsatellite,f + m * vjunk,f
where m is the mass of the space junk and vjunk,f is the final velocity of the space junk. The mass of the space junk is given as m = 4.3 kg. Rearranging the equation, we can solve for vsatellite,f.
3) The moment of inertia of the satellite can be calculated using the formula for a uniform-density sphere:
I = (2/5) * M * R^2
where M is the mass of the satellite and R is the radius of the satellite. The mass of the satellite is given as M = 195 kg and the radius of the satellite is given as 4.6 m.
4) The initial rotational angular momentum of the satellite around location D can be calculated using the angular momentum formula:
Lsatellite,i = I * ?1
where I is the moment of inertia of the satellite and ?1 is the initial angular velocity. Use the calculated moment of inertia from question 3 and the given value of ?1 = 2 radians/second.
5) To answer this question, we need to consider the properties of translational angular momentum. Determine which statements are true based on the given information.
6) The translational angular momentum of the space junk at location C can be calculated using the equation:
Ljunk,i = m * r_perp * vjunk,i
where m is the mass of the space junk, r_perp is the perpendicular distance between location D and C, and vjunk,i is the initial velocity of the space junk. The mass of the space junk is given as m = 4.3 kg, the initial velocity of the space junk is given as vjunk,i = < -2300, 0, 0 > m/s, and the perpendicular distance r_perp can be calculated using the given radius of the satellite and the fact that the space junk hits the edge of the satellite at location C.
7) The translational angular momentum of the space junk just slightly beyond location C can be calculated using the equation:
Ljunk,f = m * r_perp * vjunk,f
where m is the mass of the space junk, r_perp is the perpendicular distance between location D and C, and vjunk,f is the final velocity of the space junk. The mass of the space junk is given as m = 4.3 kg, the final velocity of the space junk is given as vjunk,f = < -1100, 500, 0 > m/s, and the perpendicular distance r_perp can be calculated using the given radius of the satellite and the fact that the space junk hits the edge of the satellite at location C.
8) The rotational angular momentum of the satellite after the collision can be calculated using the equation:
L,f = I * ?2
where I is the moment of inertia of the satellite and ?2 is the new angular velocity of the satellite. Use the calculated moment of inertia from question 3 and solve for ?2.
9) The new angular speed of the satellite can be calculated by dividing the new angular velocity ?2 by the radius of the satellite.
10) Before the collision, the total kinetic energy of the system (translational + rotational) can be calculated using the equations:
Ki = (1/2) * M * |vsatellite,i|^2 + (1/2) * I * ?1^2
where M is the mass of the satellite, |vsatellite,i| is the magnitude of the initial velocity of the satellite, I is the moment of inertia of the satellite, and ?1 is the initial angular velocity of the satellite.
11) After the collision, the total kinetic energy of the system can be calculated using the equations:
Kf = (1/2) * M * |vsatellite,f|^2 + (1/2) * I * ?2^2 + (1/2) * m * |vjunk,f|^2
where M is the mass of the satellite, |vsatellite,f| is the magnitude of the final velocity of the satellite, I is the moment of inertia of the satellite, ?2 is the new angular velocity of the satellite, m is the mass of the space junk, and |vjunk,f| is the magnitude of the final velocity of the space junk.
12) The rise in thermal energy for the space junk and satellite combined can be calculated by subtracting the initial total kinetic energy (Ki) from the final total kinetic energy (Kf).
Follow these steps to solve the questions based on the given values and equations.