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 Fnd that the rocket has drifted o± course and that it must change direction by 12 if it is to hit the moon. By radio control the rocket’s engines are Fred 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 . ²or 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 . ~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- ~p r - f y = ~p f y ( M - m ) v = m v g . Therefore v = m M - m v g . (5) ²rom 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
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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

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