Chapter 10, Problem 48PQ

(a)

To determine

The average exhaust speed of the rocket engine.

Answer to Problem 48PQ

The average exhaust speed of the rocket engine is $\underset{\_}{770\text{m}/\text{s}}$.

Explanation of Solution

Write the equation to find the force of thrust of the rocket.

  ${\overrightarrow{F}}_{thrust}={\left(\overrightarrow{v_E}\right)}_R\frac{d{M}_R}{dt}$                                                                                                   (I)

Here, ${\overrightarrow{F}}_{thrust}$ is the force of thrust of the rocket, ${\left(\overrightarrow{v_E}\right)}_R$ is the exhaust speed of the rocket, and $\frac{d{M}_R}{dt}$ is the rate of burning of fuel.

The engine uses $13.5\text{g}$ of fuel at the rate of $2.10\text{s}$. Therefore rate of burning of fuel , $\frac{d{M}_R}{dt}=\frac{13.5\text{g}}{2.10\text{s}}$.

Convert $\frac{d{M}_R}{dt}$ into SI units.

  $\begin{array}{c}\frac{d{M}_R}{dt}=\frac{13.5\text{g}}{2.10\text{s}}\left(\frac{1\text{kg}}{{10}^3\text{g}}\right)\\ =6.43\times {10}^{-3}\text{kg}/\text{s}\end{array}$

Rewrite equation (I) in terms of ${\overrightarrow{F}}_{thrust}$ to find $\left|{\left(\overrightarrow{v_E}\right)}_R\right|$.

  $\left|{\left(\overrightarrow{v_E}\right)}_R\right|=\frac{{\overrightarrow{F}}_{thrust}}{\frac{d{M}_R}{dt}}$                                                                                                    (II)

Conclusion:

Substitute $4.95\text{N}$ for ${\overrightarrow{F}}_{thrust}$ and $6.43\times {10}^{-3}\text{kg}/\text{s}$ for $\frac{d{M}_R}{dt}$ in equation (II) to get $\left|{\left(\overrightarrow{v_E}\right)}_R\right|$.

  $\begin{array}{c}\left|{\left(\overrightarrow{v_E}\right)}_R\right|=\frac{4.95\text{N}}{6.43\times {10}^{-3}\text{kg}/\text{s}}\left(\frac{1\text{kg}/\text{s}}{1\text{kgm}/{\text{s}}^2}\right)\\ =770\text{m}/\text{s}\end{array}$

Therefore, the average exhaust speed of the rocket engine is $\underset{\_}{770\text{m}/\text{s}}$.

(b)

To determine

The final velocity of the rocket.

Answer to Problem 48PQ

The final velocity of the rocket is $\underset{\_}{123\widehat{i}\text{m}/\text{s}}$.

Explanation of Solution

Write the equation to find the total mass of the rocket.

  $M_{Rf}=M_C+M_E+M_{fuel}$                                                                                        (III)

Here, $M_{Rf}$ is the mass of rocket, $M_C$ is the mass of chassis, $M_E$ is the mass of engine, and $M_{fuel}$ is the mass of fuel exhausted.

Since the fuel is decreasing the mass of fuel possess a negative sign. Hence rewrite equation (III).

  $M_{Rf}=M_C+M_E-M_{fuel}$                                                                                        (IV)

Write the expression to calculate the mass of the rocket before propulsion.

  $M_{Ri}=M_C+M_E$                                                                                                   (V)

Here, $M_{Ri}$ is the mass of rocket before propulsion.

Write the equation to find the final velocity of rocket.

  $\overrightarrow{v_f}={\left(\overrightarrow{v_E}\right)}_R\ln \left(\frac{M_{Rf}}{M_{Ri}}\right)$                                                                                             (VI)

Here, $\overrightarrow{v_f}$ is the final velocity of rocket.

Conclusion:

Substitute $60.7\text{g}$ for $M_C$, $30.5\text{g}$ for $M_E$, $13.5\text{g}$ for $M_{fuel}$ in equation (IV) to get $M_{Rf}$.

  $\begin{array}{c}{M}_{Rf}=60.7\text{g+30}\text{.5}\text{g}-13.5\text{g}\\ =\text{77}\text{.7}\text{g}\end{array}$

Substitute $60.7\text{g}$ for $M_C$ and $30.5\text{g}$ for $M_E$ in (V) to calculate $M_{Ri}$

  $\begin{array}{c}{M}_{Ri}=60.7\text{g+30}\text{.5}\text{g}\\ =9\text{1}\text{.2}\text{g}\end{array}$

Substitute $-770\widehat{i}\text{m}/\text{s}$ for ${\left(v_E\right)}_R$, $77.7\text{g}$ for $M_{Rf}$ and $91.2\text{g}$ for $M_{Ri}$ in equation (VI) to get $\overrightarrow{v_f}$    $\begin{array}{c}\overrightarrow{v_f}=\left(-770\widehat{i}\text{m}/\text{s}\right)\ln \left(\frac{77.7\text{g}}{91.2\text{g}}\right)\\ =123\widehat{i}\text{m}/\text{s}\end{array}$

Therefore, the final velocity of the rocket is $\underset{\_}{123\widehat{i}\text{m}/\text{s}}$.