A rocket is moving away from the solar system at a speed of $\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. It fires its engine, which ejects exhaust with a speed of$\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$relative to the rocket. The mass of the rocket at this time is$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathbf{kg}$ , and its acceleration is$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. (a) What is the thrust of the engine? (b) At what rate, in kilograms per second, is exhaust ejected during the firing?

1. Speed of the rocket,$v_r$is$6.0\times {10}^3\mathrm{m}/\mathrm{s}$.
2. Mass of the rocket,$m_r$is$4.0\times {10}^4\mathrm{kg}$.
3. Speed of the exhaust relative to the speed of the rocket,$v_{rel}$is$3.0\times {10}^3\mathrm{m}/\mathrm{s}$.
4. Acceleration of the rocket, $a_r$is$2.0\mathrm{m}/{\mathrm{s}}^2$.

Using the formula of thrust, you can find the thrust oftheengine usingthegiven mass and acceleration of the engine. Using this thrust and speed of the exhaust relative to the speed of the rocket, you can find the rate of exhaust during the firing.

The trust of the engine can be calculated as,

$T=m_r{a}_r$

Substitute the values in the above expression, and we get,

$$\begin{gathered} T=\left(4.0\times {10}^4\mathrm{kg}\right)\left(2.0\frac{\mathrm{m}}{{\mathrm{s}}^2}\right) \\ =8.0\times {10}^4·\left(1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}\right) \\ =8.0\times {10}^4\mathrm{N} \end{gathered}$$

Therefore, the thrust of the engine is $8.0\times {10}^4\mathrm{N}$.

For the rate of exhaust relative to the speed of the rocket, we can write the formula as,

$R=\frac{T}{v_{rel}}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} R=\frac{8.0\times {10}^4\mathrm{N}}{3.0\times {10}^3\mathrm{m}/\mathrm{s}} \\ =26.66·\left(1\mathrm{N}\times \frac{1}{1\mathrm{m}/\mathrm{s}}\times \frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right) \\ =26.66\frac{\mathrm{kg}}{\mathrm{s}} \end{gathered}$$

Therefore, the rate of the exhaust (in kg/s) during the firing is 27 kg/s .