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Q54 PE

Expert-verified
Found in: Page 291

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# What is the acceleration of a $5000\;kg$ rocket taking off from the Moon, where the acceleration due to gravity is only $1.6\;m/{s^2}$, if the rocket expels $8.00\;kg$ of gas per second at an exhaust velocity of ${\rm{2}}{\rm{.20 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}$?

The takeoff acceleration is ${\rm{1}}{\rm{.886}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.

See the step by step solution

## Step 1: Definition of Acceleration

The rate at which velocity changes is referred to as acceleration.

## Step 2: Given Data

The mass of rocket is ${{\rm{m}}_{\rm{1}}}{\rm{ = 5000}}\;{\rm{kg}}$.

The expulsion rate is $\frac{{{\rm{dm}}}}{{{\rm{dt}}}}{\rm{ = 8}}\;{\rm{kg/s}}$.

The exhaust velocity of rocket is ${{\rm{v}}_{\rm{g}}}{\rm{ = 2}}{\rm{.20 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}$.

The gravitational acceleration is ${\rm{g = 1}}{\rm{.6}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.

## Step 3: Calculating acceleration

Using the conservation of momentum along vertical direction we get,

${\rm{Fdt = }}{{\rm{P}}_{\rm{f}}}{\rm{ - }}{{\rm{P}}_{\rm{i}}}$

So,

\begin{align}{\rm{ - mg = m}}\frac{{{\rm{dv}}}}{{{\rm{dt}}}}{\rm{ - }}{{\rm{v}}_{\rm{g}}}\frac{{{\rm{dm}}}}{{{\rm{dt}}}}\end{align}\begin{align}\frac{{{\rm{dv}}}}{{{\rm{dt}}}}{\rm{ = }}\frac{{{{\rm{v}}_{\rm{g}}}\frac{{{\rm{dm}}}}{{{\rm{dt}}}}{\rm{ - mg}}}}{{\rm{m}}}\end{align}

Substituting the values we get,

\begin{align}\frac{{dv}}{{dt}} = \,a\, = \frac{{\left( {2.20 \times {{10}^3}\;m/s} \right) \times \left( {8\;kg/s} \right) - \left( {5000\,kg} \right) \times \left( {1.6\;m/{s^2}} \right)}}{{\left( {5000\ kg}\right)}}\end{align} \begin{align}= 1.886\;m/{s^2}\end{align}