A two stage rocket is traveling at 1200 m/s with respect to the Earth when the f

Question

A two stage rocket is traveling at 1200 m/s with respect to the Earth when the first stage runs out of fuel.  Explosive bolts release the first stage and push it backward with a speed of 35 m/s relative to the second stage.  The first stage is three times as massive as the second stage.  What is the speed of the second stage after separation?

Explanation / Answer

The answer in the back of the book is 1226m/s . I believe there are a couple issues with the provided solution. First, the velocity of the second stage is RELATIVE to the first:so it is not -35m/s, but 1200 - 35 = 1165m/s relative to therelevant reference frame. So: pi = 1200 m/s * mr = pf pf = (1200 m/s - 35 m/s) * 3/4 mr +vdesired * 1/4 mr = 1200 m/s *mr    = 1165 m/s * 3/4 + vdesired* 1/4 = 1200 m/s    -->dividing out the mass mr     =873.75 m/s + vdesired* 1/4 = 1200 m/s    =>vdesired = 1305 m/s This answer still doesn't agree with the book... but passes the'reasonable' test. Any ideas?

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