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Q55 PE

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Found in: Page 291

### College Physics (Urone)

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

# Calculate the increase in velocity of a $4000\;kg$ space probe that expels$3500\;kg$ of its mass at an exhaust velocity of$2.00\times{10^3}\;m/s$. . You may assume the gravitational force is negligible at the probe’s location.

The increase in velocity is${\rm{4}}{\rm{.159 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}$.

See the step by step solution

## Step 1: Definition of Force

A force is a factor that can impact an object's motion. A force can cause a mass item to accelerate (e.g., from a standstill).

## Step 2: Given Data

The mass of space probe is ${\rm{m = 4000}}\;{\rm{kg}}$.

Original mass of the probe now is ${{\rm{m}}_{\rm{o}}}{\rm{ = 4000 - 3500 = 500}}\;{\rm{kg}}$.

The exhaust velocity is ${{\rm{v}}_{\rm{e}}}{\rm{ = 2}}{\rm{.00 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}$.

## Step 3: Calculation of change in velocity

The relationship between velocity and mass is as follows:

\begin{align}{\rm{v = }}{{\rm{v}}_{\rm{o}}}{\rm{ + }}{{\rm{v}}_{\rm{e}}}{\rm{ln}}\frac{{\rm{m}}}{{{{\rm{m}}_{\rm{o}}}}}\end{align}

Substituting the obtained values,

${\rm{v - }}{{\rm{v}}_{\rm{o}}}{\rm{ = }}{{\rm{v}}_{\rm{e}}}{\rm{ln}}\frac{{\rm{m}}}{{{{\rm{m}}_{\rm{o}}}}}$

${\rm{ = 2 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{ln}}\frac{{{\rm{4000 kg}}}}{{{\rm{500 kg}}}}$

${\rm{ = 4}}{\rm{.159 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}$

Hence, the answer is ${\rm{4}}{\rm{.159 \times 1}}{{\rm{0}}^{\rm{3}}}\;{\rm{m/s}}$.