Two manned satellites approaching one another at a relative speed of 0.200 m/s i
ID: 1459974 • Letter: T
Question
Two manned satellites approaching one another at a relative speed of 0.200 m/s intend to dock. The first has a mass of 5.00 103 kg, and the second a mass of 7.50 103 kg. Assume that the positive direction is directed from the second satellite towards the first satellite.
(a) Calculate the final velocity after docking, in the frame of reference in which the first satellite was originally at rest.
m/s
(b) What is the loss of kinetic energy in this inelastic collision?
J
(c) Repeat both parts, in the frame of reference in which the second satellite was originally at rest.
final velocity
m/s
loss of kinetic energy
55 J
The first three parts I got wrong, I'm not sure if its a calculation error. please help. Thank you
Explanation / Answer
Here ,
let the final velocity is v
a) Using conservation of momentum
(5 *10^3 + 7.5 *10^3) * v = 7.5 *10^3 * 0.2
v = 0.12 m/s
the final velocity after docking 0.12 m/s
b)
loss of kinetic energy = 0.5 * 7.5 *10^3 * 0.2^2 - 0.5 * (5 *10^3 + 7.5 *10^3) * 0.12^2
loss of kinetic energy = 60 J
the loss of kinetic energy is 60 J
c)
for the second satellite in rest
Using conservation of momentum
(5 *10^3 + 7.5 *10^3) * v = 5 *10^3 * 0.2
v = 0.08 m/s
the final velocity after docking 0.08 m/s
d)
loss of kinetic energy = 0.5 * 5 *10^3 * 0.2^2 - 0.5 * (5 *10^3 + 7.4 *10^3) * 0.08^2
loss of kinetic energy = 60 J
the loss of kinetic energy is 60 J
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