Tag Archives: momentum of an artillery shell fired

College Physics by Openstax Chapter 8 Problem 2

(a) What is the mass of a large ship that has a momentum of 1.60×109 kg⋅m/s, when the ship is moving at a speed of 48.0 km/h? (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.

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 Mass of a Large Ship Given its Momentum

For this part, we are given the following quantities of the ship:

\begin{align*}
\textbf{p}_{\text{ship}} & = 1.60 \times  10^{9}\ \text{kg} \cdot \text{m}/\text{s} \\
\textbf{v}_{\text{ship}} & = 48.0\ \text{km}/\text{h} = 13.3333\ \text{m}/\text{s} 
\end{align*}

From the formula of momentum, we can solve for m in terms of \textbf{p} and \textbf{v}. Then we can substitute the given values to solve for the mass of the ship. 

\begin{align*}
\textbf{p}_{\text{ship}} & = m_{\text{ship}} \textbf{v}_{\text{ship}} \\
m_{\text{ship}}  & = \frac{\textbf{p}_{\text{ship}}}{\textbf{v}_{\text{ship}}} \\
m_{\text{ship}}  & = \frac{1.60 \times  10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{13.3333\ \text{m}/\text{s}} \\
m_{\text{ship}}  & = 120000000\ \text{kg} \\
m_{\text{ship}}  & = 1.20 \times 10^{8} \ \text{kg}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. Comparing the Momentum of a Large Ship with an Artillery Shell

The artillery shell has a mass of 1100 kilograms and a speed of 1200 m/s. Therefore, its momentum is

\begin{align*}
\textbf{p}_{\text{shell}} & = m_{\text{shell}} \textbf{v}_{\text{shell}} \\
\textbf{p}_{\text{shell}} & = \left( 1100\ \text{kg} \right)\left( 1200\ \text{m}/\text{s} \right) \\
\textbf{p}_{\text{shell}} & = 1320000\ \text{kg} \cdot \text{m}/\text{s} \\
\textbf{p}_{\text{shell}} & = 1.32 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s}
\end{align*}

Comparing the momentum of the large ship and the shell, we have.

\begin{align*}
\frac{\textbf{p}_{\text{ship}}}{\textbf{p}_{\text{shell}}} & = \frac{1.60 \times  10^{9}\ \text{kg} \cdot \text{m}/\text{s}}{1.32 \times 10^{6}\ \text{kg} \cdot \text{m}/\text{s}} \\
\frac{\textbf{p}_{\text{ship}}}{\textbf{p}_{\text{shell}}} & = 1212.1212 \\
\textbf{p}_{\text{ship}} & = 1212.1212\ \textbf{p}_{\text{shell}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The ship has a momentum that is about 1200 times the momentum of the artillery shell.