Chapter 4, Problem 102P

(a)

To determine

Net force on the rocket during the first $1.5\text{s}$ after the lift off.

Answer to Problem 102P

Net force is $\underset{\_}{\text{1}\text{.5}\text{N}}$ in upward direction.

Explanation of Solution

Write the equation to find the net force on rocket.

    $F_{\text{net}}=ma$        (I)

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

Conclusion

Substitute $87\text{g}$ for $m$ and $17.5\text{m}/{\text{s}}^{\text{2}}$ for $a$ in (I)

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

Net force is $\underset{\_}{\text{1}\text{.5}\text{N}}$ in upward direction.

(b)

To determine

The force exerted by the burning fuel on the rocket.

Answer to Problem 102P

The force is $\underset{\_}{2.4\text{N}}$ in upward direction.

Explanation of Solution

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

    $F_{\text{fuel}}=F_{\text{net}}+mg$        (II)

Here $F_{\text{fuel}}$ is the force exerted by the burning fuel and $g$ is the acceleration due to gravity.

Conclusion

Substitute $1.5\text{N}$ for $F_{\text{net}}$, $87\text{g}$ for $m$ and $\text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}$ for $g$.

    $\begin{array}{c}{F}_{\text{fuel}}=1.5\text{N}+\left(87\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\left(\text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}\right)\right)\\ =2.4\text{N}\end{array}$

The force is $\underset{\_}{2.4\text{N}}$ in upward direction.

(c)

To determine

Distance travelled by the rocket.

Answer to Problem 102P

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

Explanation of Solution

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

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

Here $x_1$ is the distance travelled, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Write the equation for the upward speed.

    $v=\frac{a}{t}$        (IV)

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

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

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

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

Here $v_1$ the final speed of motion in free fall.

Write the equation for total distance travelled by the rocket.

    $x=x_1+x_2$        (VII)

Conclusion

Substitute $0\text{m}/\text{s}$ for $u$, $\text{1}\text{.5}\text{s}$ for $t$, $17.5\text{m}/{\text{s}}^{\text{2}}$ for $a$ in (III)

    $\begin{array}{c}{x}_1=\left(0\text{m}/\text{s}\right)\left(\text{1}\text{.5}\text{s}\right)+\frac{1}{2}\left(17.5\text{m}/{\text{s}}^{\text{2}}\right){\left(\text{1}\text{.5}\text{s}\right)}^2\\ =19.7\text{m}\end{array}$

Substitute $\text{1}\text{.5}\text{s}$ for $t$, $17.5\text{m}/{\text{s}}^{\text{2}}$ for $a$ in (IV)

    $\begin{array}{c}v=\frac{17.5\text{m}/{\text{s}}^{\text{2}}}{\text{1}\text{.5}\text{s}}\\ =26.25\text{m}/\text{s}\end{array}$

Substitute $26.25\text{m}/\text{s}$ for $v$ and $9.8{\text{m}/\text{s}}^2$ for $g$ in (V)

    $\begin{array}{c}{t}_{\text{free}}=\frac{26.25\text{m}/\text{s}}{9.8{\text{m}/\text{s}}^2}\\ =2.68\text{s}\end{array}$

Substitute $0\text{m}/\text{s}$ for $v_1$, $26.25\text{m}/\text{s}$ for $v$ and $2.68\text{s}$ for $t_{\text{free}}$ in (VI)

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

Substitute $19.7\text{m}$ for $x_1$ and $35.2\text{m}$ for $x_2$

    $\begin{array}{c}x=19.7\text{m}+35.2\text{m}\\ =55\text{m}\end{array}$

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

(d)

To determine

Time taken by the rocket to return to the ground after lift-off.

Answer to Problem 102P

Time taken is $\underset{\_}{3.35\text{m}}$.

Explanation of Solution

Write the equation for time taken by the by the rocket to return to the ground to fall from rest from a height $x$.

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

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 lift-off.

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

Here $t_{total}$ the total time taken by the rocket to return to the ground after lift-off.

Conclusion

Substitute $55\text{m}$ for $x$ and $9.8{\text{m}/\text{s}}^2$ for $g$ in (VIII)

    $\begin{array}{c}{t}_{ele}=\sqrt{\frac{2\left(55\text{m}\right)}{9.8{\text{m}/\text{s}}^2}}\\ =3.35\text{s}\end{array}$

Substitute $3.35\text{s}$ for $t_{ele}$, $2.68\text{s}$ for $t_{free}$ and $1.5\text{s}$ for $t$ in (IX)

    $\begin{array}{c}{t}_{total}=1.5\text{s}+2.68\text{s}+3.35\text{s}\\ \text{=7}\text{.5}\text{s}\end{array}$

(e)

To determine

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

Answer to Problem 102P

The graph has been plotted.

Explanation of Solution

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

Conclusion

The graph has been plotted.

(f)

To determine

Net force on rocket after the complete usage of fuel.

Answer to Problem 102P

The net force is $\underset{\_}{0.85\text{N}}$.

Explanation of Solution

Write the expression for the weight of the rocket.

    $F_{net}=mg$        (X)

Conclusion

Substitute $87\text{g}$ for $m$ and $\text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}$ for $g$ in (X)

    $\begin{array}{c}{F}_{net}=\left(87\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(\text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}\right)\\ =0.85\text{N}\end{array}$

The net force is $\underset{\_}{0.85\text{N}}$.