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Q54 PE

College Physics (Urone)
Found in: Page 291

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Short Answer

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 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,

\[{\rm{Fdt = }}{{\rm{P}}_{\rm{f}}}{\rm{ - }}{{\rm{P}}_{\rm{i}}}\]


\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}

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