The space shuttle launches an 900-kg satellite by ejecting it from the cargo bay

Question

The space shuttle launches an 900-kg satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.0 s to give it a velocity of 0.25 m/s in the z-direction relative to the shuttle. The mass of the shuttle is 9.2×104 kg .

Part A

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

Express your answer using two significant figures.

Part B

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

Express your answer using two significant figures.

Explanation / Answer

Mass of satellite m = 900-kg

Time t = 4 s

velocity v = 0.25 m/s

The mass of the shuttle M= 9.2×104 kg

From law of conservation of momentum ,MV+mv = 0    Since initial velocity is zero

          MV = - mv

From this the component of velocity of the shuttle in the minus z-direction resulting from the ejection is

           V = -mv / M

              = -(900x0.25) /( 9.2×104 )

              = -2.4456 x10 -3 m/s

Here negative sign indicates that it is in negative Z direction.

(B). the average force that the shuttle exerts on the satellite during the ejection F= mv / t

         F = (900 x 0.25 ) / 4    where t = time = 4 s

           = 56.25 N

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