The space shuttle launches an 885-kg satellite by ejecting it from the cargo bay

Question

The space shuttle launches an 885-kg satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.8 s to give it a velocity of 0.30 m/s in the x direction relative to the shuttle. The mass of the shuttle is 92700 kg .

Part A

Determine the component of velocity vf of the shuttle in the minus x direction resulting from the ejection.

Express your answer to two significant figures and include the appropriate units.

Part B

Find the average force that the shuttle exerts on the satellite during the ejection.

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Part-A : The component of velocity of the shuttle in minus x direction resulting from the ejection which is given as -

using conservation of momentum, we have

pinitial = pfinal

m1 v0,1 + m2 vo,2 = m1 v1 + m2 v2

where, v0,1 = vo,2 = v0 (initial speed is zero)

m1 = mass of satellite = 885 kg

v1 = final speed of satellite = 0.3 m/s

m2 = mass of space shuttle = 92700 kg

v2 = final speed of space shuttle = ?

inserting the values in above eq.

0 = (885 kg) (0.3 m/s) + (92700 kg) v2

(265.5 kg.m/s) = - (92700 kg) v2

v2 = - (265.5 kg.m/s) / (92700 kg)

v2 = - 0.00286 m/s

v2 = - 2.8 x 10-3 m/s

(the negative sign indicates direction which is opposite to the velocity of satellite)

(b) The average force that the shuttle exerts on the satellite during the ejection which will be given as :

using an equation,    Favg = m1 a                                           

where, a = acceleration = v / t = (0.3 m/s) / (4.8 s)

Favg = (885 kg) [(0.3 m/s) / (4.8 s)]

Favg = 55.3 N

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