Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q55 PE

College Physics (Urone)
Found in: Page 291

Answers without the blur.

Just sign up for free and you're in.


Short Answer

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

Recommended explanations on Physics Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.