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VE M2 where Therefore твув M₁ = MA+MB3 = M₂ - ME VB = MATMB VE 2MB 2VE

VE M2 where Therefore твув M₁ = MA+MB3 = M₂ - ME VB = MATMB VE 2MB 2VE

University Physics Volume 1
University Physics Volume 1
18th Edition
ISBN:9781938168277
Author:William Moebs, Samuel J. Ling, Jeff Sanny
Publisher:William Moebs, Samuel J. Ling, Jeff Sanny
Chapter9: Linear Momentum And Collisions
Section: Chapter Questions
Problem 103AP: Three skydivers are plummeting earthward. They are initially holding onto each other, but then push...
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Can you explain part b I attached the solution I don’t understand what they are doing with the formula of conservation of linear momentum why are there two types of Mf and MF
Problem 2: Rocket propulsion, two-stage rocket, escape velocity with (a) A fully fueled rocket has initial mass Mo, of which Mr is the mass of the fuel. The rocket launched a given constant velocity, Viel, and at vertically upward with initial velocity vo = 0. fuel is ejected burnout time, t-too, the rocket reaches velocity 4₂. = VE, where VE is the escape of planet Earth. Find a formula for the burnout time, too (no numbers needed) velocity from the surface bo 2 bo * Mo= mass of fully fueled rocket Mr = mass of fuel The rocket velocity a function of time is v(t) = vo - gt + vrel- as a At burnout time, M(tbo) = M₁-MF; set V(0) = 0 fully fueled rocket ; Vrel = fuel ejected at est velocity ln MCĐ Mo mass of fuel = vallo (1-1)-9 но зд g b Yoo = -gt bot vrel In (1- ME) ⇒ Vbo = VE ⇒ VE = - gtbo + vurel In (1-MF) <-divide by g bo 2 2 isolate bo (6) At burnout time, a prearranged explosion seperates the rocket into two parts, with part A stopped in midair by the prearranged explosion, while part B is propelled to a velocity VB7VE. Assuming MA/MB = 3, find the velocity of part B after the explosion. a mass ratio,
Transcribed Image Text:Problem 2: Rocket propulsion, two-stage rocket, escape velocity with (a) A fully fueled rocket has initial mass Mo, of which Mr is the mass of the fuel. The rocket launched a given constant velocity, Viel, and at vertically upward with initial velocity vo = 0. fuel is ejected burnout time, t-too, the rocket reaches velocity 4₂. = VE, where VE is the escape of planet Earth. Find a formula for the burnout time, too (no numbers needed) velocity from the surface bo 2 bo * Mo= mass of fully fueled rocket Mr = mass of fuel The rocket velocity a function of time is v(t) = vo - gt + vrel- as a At burnout time, M(tbo) = M₁-MF; set V(0) = 0 fully fueled rocket ; Vrel = fuel ejected at est velocity ln MCĐ Mo mass of fuel = vallo (1-1)-9 но зд g b Yoo = -gt bot vrel In (1- ME) ⇒ Vbo = VE ⇒ VE = - gtbo + vurel In (1-MF) <-divide by g bo 2 2 isolate bo (6) At burnout time, a prearranged explosion seperates the rocket into two parts, with part A stopped in midair by the prearranged explosion, while part B is propelled to a velocity VB7VE. Assuming MA/MB = 3, find the velocity of part B after the explosion. a mass ratio,
VE 29 (b) Use conservation of lineal momentum: tbo VE ME 2 where Therefore mBVB M₂ = MA+MB3 VB - = M₁-M= MATMA VE = 2VE 2MB -2- The
Transcribed Image Text:VE 29 (b) Use conservation of lineal momentum: tbo VE ME 2 where Therefore mBVB M₂ = MA+MB3 VB - = M₁-M= MATMA VE = 2VE 2MB -2- The
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