Problem 4 - Applied Physics

Question

Two vehicles of equal mass collide at a \(90^{\circ}\) intersection. If the momentum of vehicle \(A\) is \(1.20 \times 10^{5} \mathrm{~kg} \mathrm{~km} / \mathrm{h}\) east and the momentum of vehicle \(\mathrm{B}\) is \(8.50 \times 10^{4} \mathrm{~kg} \mathrm{~km} / \mathrm{h}\) north, what is the resulting momentum of the final mass?

Step-by-Step Solution

Answer

The resulting momentum is approximately \(1.47 \times 10^5\, \text{kg km/h}\) at \(35.54^\circ\) north of east.

Step 1: Identify Given Values

You are given the momentum of vehicle A as \(1.20 \times 10^{5} \, \text{kg km/h}\) east and the momentum of vehicle B as \(8.50 \times 10^{4} \, \text{kg km/h}\) north.

Step 2: Understanding Momentum Vector Addition

Momentum is a vector quantity. To find the resultant momentum of the vehicles, we need to add these momentum vectors using vector addition principles.

Step 3: Represent Momentum as Vectors

Represent the momentum of vehicle A as a vector \( \vec{p}_A = 1.20 \times 10^5 \, \hat{i} \) (east direction) and the momentum of vehicle B as \( \vec{p}_B = 8.50 \times 10^4 \, \hat{j} \) (north direction).

Step 4: Calculate Resultant Momentum Magnitude

To find the magnitude of the resultant momentum \( \vec{p}_R \), use the Pythagorean theorem: \[ |\vec{p}_R| = \sqrt{(\vec{p}_A)^2 + (\vec{p}_B)^2} \] Substituting the given values: \[ |\vec{p}_R| = \sqrt{(1.20 \times 10^5)^2 + (8.50 \times 10^4)^2} \] Calculate this to get the magnitude.

Step 5: Calculate Resultant Momentum Magnitude

Calculate the magnitude: \[ |\vec{p}_R| = \sqrt{(1.20 \times 10^5)^2 + (8.50 \times 10^4)^2} = \sqrt{1.44 \times 10^{10} + 7.225 \times 10^9} = \sqrt{2.165 \times 10^{10}} \] This gives \( |\vec{p}_R| \approx 1.47 \times 10^5 \, \text{kg km/h}\).

Step 6: Determine Direction

To find the direction \( \theta \) of the resultant momentum, use the tangent function: \( \tan \theta = \frac{\text{momentum north (vehicle B)}}{\text{momentum east (vehicle A)}} = \frac{8.50 \times 10^4}{1.20 \times 10^5} \). Calculate \( \theta \) using the inverse tangent function.

Step 7: Calculate Direction

Calculate the angle: \( \theta = \tan^{-1} \left( \frac{8.50 \times 10^4}{1.20 \times 10^5} \right) \approx \tan^{-1}(0.7083) \approx 35.54^\circ \). This means the direction is \(35.54^\circ\) north of east.

Key Concepts

Vector Addition

Vector addition is a method used to combine quantities that have both magnitude and direction. In our exercise, momentum is represented as vectors. Each vehicle has its own momentum vector, pointed in the respective directions of east and north. To add vectors, you treat each component (in this case, the momentum in each direction) separately, combining them as you would numbers in algebra. This results in a new vector which represents the overall effect of both original vectors.
Understanding vector addition is crucial because it's how we calculate the resultant momentum of two colliding vehicles. By breaking down the momentum into their x (east) and y (north) components, we can easily find how they interact. The combined effect of these perpendicular vectors can be calculated using the Pythagorean theorem, solving for the hypotenuse which represents the resultant vector.

Resultant Momentum

The resultant momentum is simply the combined effect of the two separate momentum vectors after they collide. Think of resultant momentum as the final summary of the momentum considering both vectors, showing the direction and magnitude of the total motion after the collision.
In order to find it, you need to use vector addition, as explained before. The resultant momentum is essential because it gives you a complete picture of what happens post-collision. It helps predict the future motion of the combined mass of the vehicles, telling us where they're headed and with what force. This is critical in understanding collisions in physics, particularly in managing traffic incident simulations.
Calculating the resultant momentum involves mathematical formulas that compute both its direction and magnitude based on the given data for each involved party.

Momentum Magnitude

Momentum magnitude refers to the size or length of the momentum vector. After adding the vectors of the two vehicles we calculate how intense the resultant motion is. In the collision exercise, this is achieved by using the Pythagorean theorem.
The theorem \[ | \vec{p}_R | = \sqrt{( \vec{p}_A )^2 + ( \vec{p}_B )^2} \]allows us to find the magnitude. Here, \( \vec{p}_A \) and \( \vec{p}_B \) are the eastward and northward momentum vectors respectively. This results in \( | \vec{p}_R | = 1.47 \times 10^5 \; kg \; km/h \) showing the collision strength.
The magnitude is important because it quantifies the energy and momentum involved, which is crucial for understanding how intense the resulting impact will be, significant in both physics and safety assessments!

Momentum Direction

Momentum direction tells us where the combined mass will head after the collision. In physics, direction matters since momentum is a vector, not just a scalar. Calculating this involves using trigonometry. Specifically, you use the tangent function with the momentum components.
To find the angle \( \theta \), we utilize:\[ \tan \theta = \frac{\text{momentum north (vehicle B)}}{\text{momentum east (vehicle A)}} \]Here, the angle \( \theta \) represents the arc tangent of the ratio, calculating where the resultant direction leans towards. Solving this, \( \theta \approx 35.54^\circ \).This angle is measured from a specific direction (east, in this context), pointing north of east, showing the precise direction resulting from the crash. This calculation is crucial for predicting exactly how both vehicles will move after colliding, an essential knowledge in designing responsive safety measures.