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a) the momentum of the gas is equayl but opposite in direction to the momentum of the rocket
b) gas pushes on the air at the back of rocket
c) change in momentum of the gas gives rise to a force on rocket
d) ejected gas creates a region of high pressure behind rocket

2006-12-17 23:03:36 · 4 answers · asked by Diana K 1 in Science & Mathematics Physics

4 answers

c) change in momentum of the gas gives rise to a force on the rocket.

a is not the correct answer, because the momentum of the gas and the rocket only sum to zero if you choose a reference frame in which the rocket was stationary before it activated its engines. In all other reference frames this is false, and in all reference frames it is irrelevant - only changes in momentum are important when discussing acceleration, not the momentum itself.

b is not the correct answer. First of all, what air? and second, even if the rocket IS in an atmosphere, this describes an interaction between the exhaust gas and the air, neither of which is the rocket, so this interaction by itself is irrelevant to hte rocket's motion.

d is not the correct answer. Insofar as we can describe the forces acting on the rocket in terms of pressure differentials, it is the high pressure inside the reaction chamber that moves the rocket forward, not the high pressure behind the rocket (to see why this would cause the rocket to move forward, consider that this pressure is being exerted on the front wall of the chamber, but not the back wall, because there is no back wall).

c is the correct answer: changing the momentum of the gas requires exerting a force on the gas, which by Newton's third law causes an equal and opposite force on the rocket, accelerating it forward.

2006-12-18 00:20:08 · answer #1 · answered by Pascal 7 · 2 0

a) can't be because both momentum cancel each other out resulting in no net movement


b) or d) most plausible

d)the force created by the ejection of the air propels the rocket

2006-12-18 07:48:40 · answer #2 · answered by Zidane 3 · 0 2

For every action there is an equal but opposite reaction, so I'd circle a.

2006-12-19 03:29:16 · answer #3 · answered by Anonymous · 0 0

a), c), & d) seem most plausible answers.

2006-12-18 07:33:38 · answer #4 · answered by Paleologus 3 · 0 1

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