Tag Archives: momentum of an airplane taking off

College Physics by Openstax Chapter 8 Problem 3

(a) At what speed would a 2.00×104-kg airplane have to fly to have a momentum of 1.60×109 kg⋅m/s (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of 60.0 m/s? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.

Solution:

Linear momentum (momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity. In symbols, linear momentum \textbf{p} is defined as

\textbf{p}=m \textbf{v},

where m is the mass of the system and \textbf{v} is its velocity.

Part A. The Speed of an Airplane Given its Momentum

The airplane has the following quantities given:

\begin{align*}
m_{\text{airplane}} & = 2.00 \times 10^{4}\ \text{kg} \\
\textbf{p}_{\text{airplane}} & = 1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s}
\end{align*}

Using the formula of momentum, we can solve for the velocity in terms of the other variables. We can then substitute the given quantities to solve for the velocity of the airplane.

\begin{align*}
\textbf{p}_{\text{airplane}} & = m_{\text{airplane}} \textbf{v}_{\text{airplane}} \\
\textbf{v}_{\text{airplane}} & = \frac{\textbf{p}_{\text{airplane}}}{m_{\text{airplane}}} \\
\textbf{v}_{\text{airplane}} & = \frac{1.60 \times 10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{2.00 \times 10^{4}\ \text{kg}} \\
\textbf{v}_{\text{airplane}} & = 80000\ \text{m}/\text{s} \\
\textbf{v}_{\text{airplane}} & = 8.00 \times 10^{4}\ \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The Airplane’s Momentum When it is Taking Off

In this case, we are given the following properties of an airplane taking off:

\begin{align*}
m_{\text{airplane}} & = 2.00 \times 10^{4}\ \text{kg} \\
\textbf{v}_{\text{airplane}} & = 60.0\ \text{m}/\text{s} \\
\end{align*}

The momentum of the airplane at this instance is calculated as

\begin{align*}
\textbf{p}_{\text{airplane}} & = m_{\text{airplane}} \textbf{v}_{\text{airplane}} \\
\textbf{p}_{\text{airplane}} & = \left( 2.00 \times 10^{4}\ \text{kg} \right)\left( 60.0\ \text{m}/\text{s} \right) \\
\textbf{p}_{\text{airplane}} & = 1200000\ \text{kg} \cdot \text{m}/\text{s} \\
\textbf{p}_{\text{airplane}} & = 1.20 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

Since the momentum of the airplane is 3 orders of magnitude smaller than the ship, the ship will not recoil very much. The recoil would be -0.01 m/s, which is probably not noticeable.