Chapter 26, Problem 95P

(a)

To determine

The magnitude of the force.

Answer to Problem 95P

The magnitude of the force is $1.8\times {10}^7\text{N}$ .

Explanation of Solution

In problem, time for which force applied is $3.6\times {10}^4\text{s}$, mass of spaceship is $2200\text{kg}$ and speed of the spaceship is $0.70c$.

Write impulse-momentum equation.

  $F\Delta t=\Delta p$                                                                                                                  (I)

Here, $F$ is the constant force applied to spaceship, $\Delta t$ is the time during which force is applied and $\Delta p$ is the change in momentum.

Write the expression for $\Delta p$.

  $\Delta p=p_{\text{f}}-p_{\text{i}}$

Here, $p_{\text{i}}$ is the initial momentum and $p_{\text{f}}$ is the final momentum.

Since the spaceship starts from rest, $p_{\text{i}}$ is zero.

Substitute $0$ for $p_{\text{i}}$ in above equation to get $\Delta p$.

  $\Delta p=p_{\text{f}}$

Substitute $p_{\text{f}}$ for $\Delta p$ in equation (I) to get $F\Delta t$.

  $F\Delta t=p_{\text{f}}$                                                                                                                 (II)

In this problem speed of spaceship is a large fraction of $c$, so relativistic form of the momentum must be used.

Write the equation for the relativistic momentum of spaceship.

  $p_{\text{f}}=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $m$ is the mass of spaceship and $v$ is the speed of spaceship.

Put the above equation in equation (II) to get the final expression for $F$.

  $\begin{array}{c}F\Delta t=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\\ F=\frac{mv}{\Delta t\sqrt{1-\frac{v^2}{c^2}}}\end{array}$                                                                                                   (III)

Conclusion:

Substitute $2200\text{kg}$ for $m$, $0.70c$ for $v$ and $3.6\times {10}^4\text{s}$ in equation (III) to get $F$.

  $\begin{array}{c}F=\frac{\left(2200\text{kg}\right)\left(0.70c\right)}{\left(3.6\times {10}^4\text{s}\right)\sqrt{1-\frac{{\left(0.70c\right)}^2}{c^2}}}\\ =\frac{\left(2200\text{kg}\right)\left(0.70c\right)}{\left(3.6\times {10}^4\text{s}\right)\sqrt{1-{\left(0.70\right)}^2}}\end{array}$

Substitute $3.00\times {10}^8\text{m}/\text{s}$ for $c$ in above equation to get $F$.

  $\begin{array}{c}F=\frac{\left(2200\text{kg}\right)\left(0.70\left(3.00\times {10}^8\text{m}/\text{s}\right)\right)}{\left(3.6\times {10}^4\text{s}\right)\sqrt{1-{\left(0.70\right)}^2}}\\ =1.8\times {10}^7\text{N}\end{array}$

Therefore, the magnitude of the force is $1.8\times {10}^7\text{N}$.

(b)

To determine

The initial acceleration of the spaceship and comment on the magnitude of the acceleration.

Answer to Problem 95P

The initial acceleration of the spaceship is equal to $8200\text{m}/\text{s}$ in the direction of the force and this much larger than any human could survive.

Explanation of Solution

Write the expression for the acceleration.

  $a=\frac{F}{m}$                                                                                                                     (IV)

Here, $a$ is the initial acceleration of the spaceship.

Conclusion:

Substitute $1.8\times {10}^7\text{N}$ for $F$ and $2200\text{kg}$ for $m$ in (IV) to get $a$.

  $\begin{array}{c}a=\frac{1.8\times {10}^7\text{N}}{2200\text{kg}}\\ =8200\text{m}/{\text{s}}^2\times \frac{g}{9.8\text{m}/{\text{s}}^2}\\ =830g\end{array}$

Since acceleration is positive, direction of acceleration is in the direction of force. Since this acceleration is $830$ times the acceleration due to gravity, human could not survive in this acceleration.

Therefore, the initial acceleration of the spaceship is equal to $8200\text{m}/\text{s}$ in the direction of force and the magnitude of the initial acceleration of the spaceship is much larger than any human could survive