Tag Archives: Momentum of a tranquilizer dart

College Physics by Openstax Chapter 8 Problem 1

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7.50 m/s. (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s. (c) What is the momentum of the 90.0-kg hunter running at 7.40 m/s after missing the elephant?

Solution:

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

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

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

Part A. The Momentum of the Elephant

We are given the mass and the velocity of the elephant, so we can just directly substitute these values in the formula for momentum.

pelephant=melephantvelephantpelephant=(2000 kg)(7.50 m/s)pelephant=15000 kg⋅m/spelephant=1.50×104 kg⋅m/s  (Answer)\begin{align*} \textbf{p}_{\text{elephant}} & = m_{\text{elephant}} \textbf{v}_{\text{elephant}} \\ \textbf{p}_{\text{elephant}} & = \left( 2000\ \text{kg} \right)\left( 7.50\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{elephant}} & = 15000\ \text{kg} \cdot \text{m}/\text{s} \\ \textbf{p}_{\text{elephant}} & = 1.50 \times 10 ^{4} \ \text{kg} \cdot \text{m}/\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. Comparing the momentum of the elephant in part A with the momentum of a tranquilizer

First, we need to calculate the momentum of the tranquilizer.

ptranquilizer=mtranquilizervtranquilizerptranquilizer=(0.0400 kg)(600 m/s)ptranquilizer=24 kg⋅m/s\begin{align*} \textbf{p}_{\text{tranquilizer}} & = m_{\text{tranquilizer}} \textbf{v}_{\text{tranquilizer}} \\ \textbf{p}_{\text{tranquilizer}} & = \left( 0.0400\ \text{kg} \right)\left( 600\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{tranquilizer}} & = 24\ \text{kg} \cdot \text{m}/\text{s} \end{align*}

Now, we can compare their momentums.

pelephantptranquilizer=15000 kg⋅m/s24 kg⋅m/spelephantptranquilizer=625pelephant=625 ptranquilizer  (Answer)\begin{align*} \frac{\textbf{p}_{\text{elephant}}}{\textbf{p}_{\text{tranquilizer}}} & = \frac{15000\ \text{kg} \cdot \text{m}/\text{s}}{24\ \text{kg} \cdot \text{m}/\text{s} } \\ \frac{\textbf{p}_{\text{elephant}}}{\textbf{p}_{\text{tranquilizer}}} & = 625 \\ \textbf{p}_{\text{elephant}} & = 625\ \textbf{p}_{\text{tranquilizer}} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The momentum of the elephant is 625 times larger than the momentum of the tranquilizer.

Part C. The Momentum of a Hunter Running after missing the elephant

phunter=mhuntervhunterphunter=(90.0 kg)(7.40 m/s)phunter=666 kg⋅m/s  (Answer)\begin{align*} \textbf{p}_{\text{hunter}} & = m_{\text{hunter}} \textbf{v}_{\text{hunter}} \\ \textbf{p}_{\text{hunter}} & = \left( 90.0\ \text{kg} \right)\left( 7.40\ \text{m}/\text{s} \right) \\ \textbf{p}_{\text{hunter}} & = 666\ \text{kg} \cdot \text{m}/\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}