Chapter 7, Problem 108P

(a)

To determine

The spaceship having greater kinetic energy and that having greater momentum if both the engines are fired for same time.

Answer to Problem 108P

The Vulcan spaceship will have greater kinetic energy and both the ships will have same momentum if both the engines are fired for same time.

Explanation of Solution

Given that the mass of Vulcan spaceship is $65000\text{kg}$, the mass of Romulan spaceship is $2\times 65000\text{kg}$, the total force of engines of both spaceships is $9.5\times {10}^6\text{N}$, and the engines are fired for $100\text{s}$.

Write the expression for the distance travelled by the spaceship.

$\Delta x=\frac{1}{2}a{\left(\Delta t\right)}^2$ (I)

Here, $\Delta x$ is the distance travelled, $a$ is the acceleration, and $\Delta t$ is the time duration.

Modify the equation (I) using Newton’s second law.

$\Delta x=\frac{1}{2}\frac{F}{m}{\left(\Delta t\right)}^2$ (II)

Here, $F$ is the force, and $m$ is the mass.

Write the expression for the work done on the spaceship by the engine.

$W=F\Delta x$ (III)

Here, $W$ is the work done.

According to work energy theorem, the work done is equal to change in kinetic energy $\Delta K$. Hence, equation (III) can be written as,

$\Delta K=F\Delta x$ (IV)

Use equation (II) in (III).

$\begin{array}{c}\Delta K=F\left(\frac{1}{2}\frac{F}{m}\right)\\ =\frac{F^2}{2m}{\left(\Delta t\right)}^2\end{array}$ (V)

Write the expression for the change in momentum of the spaceships.

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

Here, $\Delta p$ is the change in momentum.

Conclusion:

Equation (V) indicate that, the change in kinetic energy of the spaceships is inversely proportional to its mass. Since both the spaceships are starting from rest, the spaceship with lesser mass will have greater change in kinetic energy and hence the Vulcan ship will have greater kinetic energy.

Since both the spaceships are provided with same force and same duration of engine firing, according to equation (VI) both the ships will have same change in momentum, and hence both the sips will have same momentum.

Therefore, the Vulcan spaceship will have greater kinetic energy and both the ships will have same momentum if both the engines are fired for same time.

(b)

To determine

The spaceship having greater kinetic energy and that having greater momentum if both the engines are fired for same distance.

Answer to Problem 108P

Both the ships will have same momentum, and the Romulan spaceship will have greater kinetic energy if both the engines are fired for same distance.

Explanation of Solution

Given that the mass of Vulcan spaceship is $65000\text{kg}$, the mass of Romulan spaceship is $2\times 65000\text{kg}$, the total force of engines of both spaceships is $9.5\times {10}^6\text{N}$, and the engines are fired for $100\text{m}$.

Equation (IV) gives the change in kinetic energy of the spaceships.

$\Delta K=F\Delta x$

Equation (VI) indicates that the change in momentum of the spaceships is directly proportional to the time for which the engine is fired.

$\Delta p=F\Delta t$

Conclusion:

The force and the distance for which the engine fired are same for both the spaceships, This results the change in kinetic energy of the spaceships to be the same according to equation (IV). Since both ships are starting from rest, both will have same kinetic energy.

The more massive ship needs to fire its engine for long time to cover a particular distance. Thus, according to equation (VI), the change in momentum will be greater for the Romulan ship. Hence, Romulan ship will have greater momentum.

Therefore, both the ships will have same momentum, and the Romulan spaceship will have greater kinetic energy if both the engines are fired for same distance.

(c)

To determine

The kinetic energy and momentum of the spaceships when they are fired for same time and when they are fired for same distance.

Answer to Problem 108P

When the ships are fired for same time, the kinetic energy of Vulcan ship is $\underset{\_}{6.9\times {10}^{12}\text{J}}$, and that of Romulan ship is $\underset{\_}{3.5\times {10}^{12}\text{J}}$, and the momentum of both the ships is $\underset{\_}{9.5\times {10}^8\text{kg}\cdot \text{m/s}}$. When the ships are fired for same distance, the momentum of Vulcan ship is $\underset{\_}{1.1\times {10}^7\text{kg}\cdot \text{m/s}}$, and that of Romulan ship is $\underset{\_}{1.6\times {10}^7\text{kg}\cdot \text{m/s}}$, and the kinetic energy of both the ships is $\underset{\_}{9.5\times {10}^8\text{J}}$.

Explanation of Solution

Given that in part (a), the mass of Vulcan spaceship is $65000\text{kg}$, the mass of Romulan spaceship is $2\times 65000\text{kg}$, the total force of engines of both spaceships is $9.5\times {10}^6\text{N}$, and the engines are fired for $100\text{s}$.

Equation (V) gives the kinetic energy of the spaceships.

$\Delta K=\frac{F^2}{2m}{\left(\Delta t\right)}^2$

Equation (VI) gives the momentum of both the ships.

$\Delta p=F\Delta t$

Given that in part (b), the engines are fired for $100\text{m}$.

Equation (IV) gives the kinetic energy of both the spaceships when the engines are fired for same distance.

$\Delta K=F\Delta x$

Write the expression for the momentum of the spaceship in terms of its kinetic energy.

$p=\sqrt{2mK}$ (VII)

Conclusion:

Consider the condition in part (a), the engines are fired for same time.

Substitute $9.5\times {10}^6\text{N}$ for $F$, $65000\text{kg}$ for $m$, and $100\text{s}$ for $\Delta t$ in equation (V) to find the kinetic energy of the Vulcan ship.

$\begin{array}{c}\Delta K=\frac{{\left(9.5\times {10}^6\text{N}\right)}^2}{2\left(65000\text{kg}\right)}{\left(100\text{s}\right)}^2\\ =6.9\times {10}^{12}\text{J}\end{array}$

Substitute $9.5\times {10}^6\text{N}$ for $F$, $2\times 65000\text{kg}$ for $m$, and $100\text{s}$ for $\Delta t$ in equation (V) to find the kinetic energy of the Romulan ship.

$\begin{array}{c}\Delta K=\frac{{\left(9.5\times {10}^6\text{N}\right)}^2}{2\left(2\times 65000\text{kg}\right)}{\left(100\text{s}\right)}^2\\ =3.5\times {10}^{12}\text{J}\end{array}$

Substitute $9.5\times {10}^6\text{N}$ for $F$, and $100\text{s}$ for $\Delta t$ in equation (VI) to find the momentum of both the ships.

$\begin{array}{c}\Delta p=\left(9.5\times {10}^6\text{N}\right)\left(100\text{s}\right)\\ =9.5\times {10}^8\text{kg}\cdot \text{m/s}\end{array}$

Consider the condition in part (b), the engines are fired for same distance.

Substitute $9.5\times {10}^6\text{N}$ for $F$, and $100\text{m}$ for $\Delta x$ in equation (IV) to find the kinetic energy of both the ships.

$\begin{array}{c}\Delta K=\left(9.5\times {10}^6\text{N}\right)\left(100\text{m}\right)\\ =9.5\times {10}^8\text{J}\end{array}$

Substitute $9.5\times {10}^8\text{J}$ for $K$, and $65000\text{kg}$ for $m$ in equation (VII) to find the momentum of the Vulcan ship.

$\begin{array}{c}p=\sqrt{2\left(65000\text{kg}\right)\left(9.5\times {10}^8\text{J}\right)}\\ =1.1\times {10}^7\text{kg}\cdot \text{m/s}\end{array}$

Substitute $9.5\times {10}^8\text{J}$ for $K$, and $2\times 65000\text{kg}$ for $m$ in equation (VII) to find the momentum of the Romulan ship.

$\begin{array}{c}p=\sqrt{2\left(2\times 65000\text{kg}\right)\left(9.5\times {10}^8\text{J}\right)}\\ =1.6\times {10}^7\text{kg}\cdot \text{m/s}\end{array}$

Therefore, when the ships are fired for same time, the kinetic energy of Vulcan ship is $\underset{\_}{6.9\times {10}^{12}\text{J}}$, and that of Romulan ship is $\underset{\_}{3.5\times {10}^{12}\text{J}}$, and the momentum of both the ships is $\underset{\_}{9.5\times {10}^8\text{kg}\cdot \text{m/s}}$. When the ships are fired for same distance, the momentum of Vulcan ship is $\underset{\_}{1.1\times {10}^7\text{kg}\cdot \text{m/s}}$, and that of Romulan ship is $\underset{\_}{1.6\times {10}^7\text{kg}\cdot \text{m/s}}$, and the kinetic energy of both the ships is $\underset{\_}{9.5\times {10}^8\text{J}}$.