Chapter 4, Problem 166P

(a)

To determine

The net force on rocket during the first $1.5\text{s}$ after the liftoff.

Answer to Problem 166P

Net force is $1.5\text{N}$ in upward direction.

Explanation of Solution

Acceleration of rocket in first $1.5\text{s}$ is $17.5\text{m}/{\text{s}}^{\text{2}}$, mass of rocket is $87\text{g}$, and the fuel is much less than $87\text{g}$.

Write the equation to find the net force on rocket.

$F_{net}=ma$

Here, the net force is $F_{net}$, mass of rocket is $m$, and the acceleration is $a$.

Conclusion:

Substitute $87\text{g}$ for $m$ and $17.5\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to find $F_{net}$.

$\begin{array}{c}{F}_{net}=\left(87\text{g}\frac{1\text{kg}}{{10}^{3}g}\right)\left(17.5\text{m}/{\text{s}}^{\text{2}}\right)\\ =1.5\text{N}\end{array}$

The direction of $F_{net}$ is the direction of motion of rocket itself.

Therefore, the net force is $1.5\text{N}$ in upward direction.

(b)

To determine

The force exerted by the burning fuel on the rocket.

Answer to Problem 166P

Exerted force is $2.4\text{N}$ in upward direction.

Explanation of Solution

Acceleration of rocket in first $1.5\text{s}$ is $17.5\text{m}/{\text{s}}^{\text{2}}$, mass of rocket is $87\text{g}$, and the fuel is much less than $87\text{g}$.

Write the equation to find the force exerted by the burning fuel on rocket.

$F_{fuel}=F_{net}+mg$

Here, the force exerted by the burning fuel is $F_{fuel}$ and the gravitational acceleration is $g$.

Conclusion:

Substitute $1.5\text{N}$ for $F_{net}$, $87\text{g}$ for $m$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $F_{net}$.

$\begin{array}{c}{F}_{fuel}=1.5\text{N}+\left(\left(87\text{g}\frac{1\text{kg}}{{10}^{3}g}\right)\left(17.5\text{m}/{\text{s}}^{\text{2}}\right)\right)\\ =1.5\text{N}+0.853\\ =2.4\text{N}\end{array}$

The direction of $F_{fuel}$ is the direction of motion of rocket itself.

Therefore, the exerted force is $2.4\text{N}$ in upward direction.

(c)

To determine

The distance travelled by the rocket.

Answer to Problem 166P

Distance travelled is $55\text{m}$.

Explanation of Solution

Acceleration of rocket in first $1.5\text{s}$ is $17.5\text{m}/{\text{s}}^{\text{2}}$, mass of rocket is $87\text{g}$, and the fuel is much less than $87\text{g}$.

Write the equation to calculate the distance travelled by rocket during the burning time.

$x_1=ut+\frac{1}{2}a{t}^2$ (I)

Here, the distance travelled by the rocket during the burning time is $x_1$, initial velocity is $u$, and the time taken is $t$.

Write the equation for the upward speed.

$v=\frac{a}{t}$ (II)

Here, the upward speed achieved by the rocket is $a$.

Write the equation to calculate the time taken to reduce the velocity from $v$ to zero in the free fall.

$t_{free}=\frac{v}{g}$ (III)

Write the equation to calculate the distance travelled by rocket during the free fall period.

$x_2=\frac{1}{2}\left(v_1+v\right)t_{free}$ (IV)

Here, the final speed of motion in free fall is $v_1$

Write the equation for the total distance traveled by the rocket.

$x=x_1+x_2$ (V)

Here, the total distance travelled is $x$.

Conclusion:

Substitute $0\text{m}/\text{s}$ for $u$, $1.5\text{s}$ for $t$, and $17.5\text{m}/{\text{s}}^{\text{2}}$ for $a$ in equation (I) to find $x_1$.

$\begin{array}{c}{x}_1=\left(0\text{m}/\text{s}\right)\left(1.5\text{s}\right)+\frac{1}{2}\left(17.5\text{m}/{\text{s}}^{\text{2}}\right){\left(1.5\text{s}\right)}^2\\ =0\text{m+19}\text{.7}\text{m}\\ \text{=19}\text{.7}\text{m}\end{array}$

Substitute $17.5\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $1.5\text{s}$ for $t$ in equation (II) to find $v$.

$\begin{array}{c}v=\frac{17.5\text{m}/{\text{s}}^{\text{2}}}{1.5\text{s}}\\ =26.25\text{m}/\text{s}\end{array}$

Substitute $26.25\text{m}/\text{s}$ for $v$ and $9.80\text{m}/\text{s}$ for $g$ in equation (III) to find $t_{free}$.

$\begin{array}{c}{t}_{free}=\frac{26.25\text{m}/\text{s}}{9.80\text{m}/\text{s}}\\ =2.68\text{s}\end{array}$

Substitute $0\text{m}/\text{s}$ for $v_1$, $26.25\text{m}/\text{s}$ for $v$, and $2.68\text{s}$ for $t_{free}$ in equation (IV) to find $x_2$.

$\begin{array}{c}{x}_2=\frac{1}{2}\left(0\text{m}/\text{s}+26.25\text{m}/\text{s}\right)\left(2.68\text{s}\right)\\ =35.2\text{m}\end{array}$

Substitute $\text{19}\text{.7}\text{m}$ for $x_1$ and $35.2\text{m}$ for $x_2$ in equation (V) to find $x$.

$\begin{array}{c}x=\text{19}\text{.7}\text{m}+35.2\text{m}\\ \text{=55}\text{m}\end{array}$

Therefore, the distance travelled is $55\text{m}$.

(d)

To determine

The time taken by the rocket to return to the ground after liftoff.

Answer to Problem 166P

Time taken is $3.35\text{s}$.

Explanation of Solution

Acceleration of rocket in first $1.5\text{s}$ is $17.5\text{m}/{\text{s}}^{\text{2}}$, mass of rocket is $87\text{g}$, and the fuel is much less than $87\text{g}$.

Write the equation to find the time taken by the rocket to fall from rest from height $x$.

$t_{ele}=\sqrt{\frac{2x}{g}}$ (VI)

Here, the time taken by the rocket to fall from rest from height $x$ is $t_{ele}$.

Write the equation to find the time taken by the rocket to return to the ground after liftoff.

$t_{total}=t+t_{free}+t_{ele}$ (VII)

Here, the total time taken by the rocket to return to the ground after liftoff is $t_{total}$.

Conclusion:

Substitute $55\text{m}$ for $x$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (VI) to find $t_{ele}$.

$\begin{array}{c}{t}_{ele}=\sqrt{\frac{2\left(55\text{m}\right)}{9.80\text{m}/{\text{s}}^{\text{2}}}}\\ =\sqrt{11.22}\text{s}\\ \text{=3}\text{.35}\text{s}\end{array}$

Substitute $1.5\text{s}$ for $t$, $2.68\text{s}$ for $t_{free}$, and $\text{3}\text{.35}\text{s}$ for $t_{ele}$ in equation (VII) to find $t_{total}$.

$\begin{array}{c}{t}_{total}=1.5\text{s}+2.68\text{s}+\text{3}\text{.35}\text{s}\\ \text{=7}\text{.5}\text{s}\end{array}$

Therefore, the time taken is $3.35\text{s}$.

(e)

To determine

Plot the velocity –time relation of rocket for the entire travel from the launch to the return back to the ground.

Explanation of Solution

Acceleration of rocket in first $1.5\text{s}$ is $17.5\text{m}/{\text{s}}^{\text{2}}$, mass of rocket is $87\text{g}$, and the fuel is much less than $87\text{g}$.

The velocity –time relation of rocket for the entire travel from the launch to the return back to the ground is shown below in figure 1.

In first $1.5\text{s}$, rocket travels upwards. After that, rocket starts to move towards ground. The point of maxima denotes the instant at which rocket changes its direction.

Therefore, the time taken is $3.35\text{s}$.

(f)

To determine

The net force on rocket after the complete usage of fuel.

Answer to Problem 166P

The net force is $0.85\text{N}$ in downward direction.

Explanation of Solution

Acceleration of rocket in first $1.5\text{s}$ is $17.5\text{m}/{\text{s}}^{\text{2}}$, mass of rocket is $87\text{g}$, and the fuel is much less than $87\text{g}$.

After the fuel is completely used, the rocket is in free fall. At that moment, the net force on the rocket is its weight itself.

Write the expression for the weight of rocket.

$F_{net}=mg$

Here, the net force is $F_{net}$.

Conclusion:

Substitute $87\text{g}$ for $m$ and $9.80\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $F_{net}$.

$\begin{array}{l}{F}_{net}=\left(87\text{g}\left(\frac{\text{1}\text{kg}}{{\text{10}}^{\text{3}}\text{g}}\right)\right)\left(9.80\text{m}/{\text{s}}^{\text{2}}\right)\\ =0.85\text{N}\end{array}$

Since weight is acting vertically downwards, $F_{net}$ is also in downward direction.

Therefore, the net force is $0.85\text{N}$ in downward direction.