Homework 18

# Homework 18 - homework 18 BAUTISTA ALDO Due Mar 6 2006 4:00...

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homework 18 – BAUTISTA, ALDO – Due: Mar 6 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 2. 10 points. A(n) 6100 kg rocket traveling at 3200 m / s is moving freely through space on a journey to the moon. The ground controllers find that the rocket has drifted off course and that it must change direction by 12 if it is to hit the moon. By radio control the rocket’s engines are fired instantaneously ( i.e., as a single pel- let) in a direction perpendicular to that of the rocket’s motion. The gases are expelled ( i.e., the pellet) at a speed of 4800 m / s (relative to the rocket). What mass of gas must be expelled to make the needed course correction? Correct answer: 757 . 111 kg (tolerance ± 1 %). Explanation: Note: The gases are expelled in a single burst, so we do not consider the change in mass of the rocket as the gas is expelled. Let : m = mass of expelled gas , M = 6100 kg = initial mass of rocket , v i = 3200 m / s = initial velocity of rocket , v g = 4800 m / s = velocity of gas , and θ = 12 = course correction angle . The thrust of a rocket engine is T = v g m Δ t and the impulse due to this force is simply I T = v g m . The result of this impulse is the change of the rocket’s momentum vector ~p f - ~p i . For the problem at hand, ~ T and hence ~ I T is perpendicular to the rocket’s initial momen- tum M ~v i . The momentum diagrams for the “rocket minus fuel” system and for the “fuel” system are shown below. p f θ rocket minus fuel ( M - m ) v i ( M - m ) v where rocket - fuel : ~p r - f x = ( M - m ) v i , (1) rocket - fuel : ~p r - f y = ( M - m ) v . (2) p g m v i m v g fuel where fuel : ~p f x = m v i , (3) fuel : ~p f y = m v g . (4) Note: The initial momentum of the rocket (including fuel) is the sum of ~p r - f x and ~p f x . Using Eqs. (1) & (3) gives ~p i = ~p r - f x + ~p f x = ( M - m ) v i + m v i = M v i . Conservation of momentum in the direc- tion, using Eqs. (2) & (4) gives ~p r - f y = ~p f y ( M - m ) v = m v g . Therefore v = m M - m v g . (5) From the “rocket minus fuel” momentum diagram we have tan θ = ( M - m ) v ( M - m ) v i = v v i . (6) Substituting v from Eq. (5) into Eq. (6) tan θ = m ( M - m ) v g v i . (7) Solving Eq. (7) for m gives M v i tan θ - m v i tan θ = m v g m v g + m v i tan θ = M v i tan θ

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homework 18 – BAUTISTA, ALDO – Due: Mar 6 2006, 4:00 am 2 m ( v g + v i tan θ ) = M v i tan θ Since v i tan θ = (3200 m / s) tan 12 = 680 . 181 m / s , we have m = M v i tan θ v g + v i tan θ (8) = (6100 kg) (680 . 181 m / s) 4800 m / s + 680 . 181 m / s = 757 . 111 kg . Question 2 Part 2 of 2. 0 points. In the realistic world, the gas is not ex- hausted in one short burst as a single gas pellet. The gas is released over a time span in molecular form.
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