Question

Far in space, where gravity is negligible, a $425kg$rocket traveling at $75m/s$fires its engines. FIGURE EX11.11 shows the thrust force as a function of time. The mass lost by the rocket during these $30s$is negligible.

a. What impulse does the engine impart to the rocket?

b. At what time does the rocket reach its maximum speed? What is the maximum speed?

Short answer

a. The engine get impulse of $J_x=(1.5\times {10}^{4}N)\times \left(s\right)$impart to the rocket.

b. ) The speed is maximum at time $30s$and this speed is$110m/s$.

Step-by-step solution

Part(a) Step 1: Given information    

We need to find that at What impulse does the engine impart to the rocket

Part(a) Step 2: Simplify    

For a short time a object receives a force is known as impulse. it is the area under the curve of the force-versus-time graph and same as momentum. The impulse is the quantity $J_x$and it is given by equation in the form

$$\begin{gathered} impulse=J_x={\int }_{t_i}^{t_f}{F}_x\left(t\right)dt \\ =areaunderthe{F}_x\left(t\right)curvebetween{t}_{i}and{t}_f\left(1\right) \end{gathered}$$

In a graph of force versus time the force is in the range of time $t_i=0sto{j}_f=30s$. the impulse takes the rectangle shape, where height is $h=1000N$and and the base is $b=30s-0s=30s$. The area of the rectangle is half the product of the heightand the base

$A=\frac{1}{2}\left(height\right)\left(base\right)$

Using equation $\left(1\right)$to get impulse

localid="1649766020389" $$\begin{gathered} J_x=\frac{1}{2}\left(height\right)\left(base\right) \\ =\frac{1}{2}(1000N)(30s) \\ =(1.5\times {10}^{4}N)\times \left(s\right) \end{gathered}$$

Part(b) step 1: Given information 

We need to find that At what time does the rocket reach its maximum speed and What is the maximum speed of rocket.

Part(b) Step 2: Simplify   

Based on part (a), the impulse is positive, so the rocket will accelerate faster. The force is given to the rocket for 30 seconds to increase its speed, and then it becomes zero. That means the rocket is at its maximum speed after 30 s.

$△t=30s$

The final speed after the impulse $v_{max}=v_f$must be the same as the maximum speed found. The impulse is the quantity $j_x$that is delivered when a force is applied to an object with an initial momentum $p_{ix}$and it changes to a new momentum $p_{fx}$, as shown in equation (11.9).

role="math" localid="1649767258766" $p_{fx}=p_{ix}+J_x\left(2\right)$

The object has mass $m$and moves with speed $v$that has momentum $p$, Vector is a product of object's mass and its velocity.The momentum is given by equation in the form

$p=mv$

To get final speed using expression of momentum into equation $\left(2\right)$

$$\begin{gathered} p_{fx}=p_{ix}+J_x \\ m{v}_f=m{v}_i+J_x \\ {v}_f=\frac{m{v}_i+J_x}{m}\left(3\right) \end{gathered}$$

The values for $m,v_{i}and{j}_x$putting into equation $\left(3\right)$to get $v_f$

$$\begin{gathered} v_f=\frac{m{v}_i+J_x}{m} \\ =\frac{\left(425kg\right)\left(75m/s\right)+\left(1.5\times {10}^{4}N\right)\left(s\right)}{425kg} \\ =110m/s \end{gathered}$$

The maximum speed in$110m/s$.