/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 An 85 -kg jogger is heading due ... [FREE SOLUTION] | 91Ó°ÊÓ

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An 85 -kg jogger is heading due east at a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). A \(55-\mathrm{kg}\) jogger is heading \(32^{\circ}\) north of east at a speed of \(3.0 \mathrm{~m} / \mathrm{s}\). Find the magnitude and direction of the sum of the momenta of the two joggers.

Short Answer

Expert verified
The total momentum is approximately 322.8 kg·m/s at 15.7° north of east.

Step by step solution

01

Calculate the momentum of the 85-kg jogger

The momentum of an object is given by the formula \( p = mv \), where \( m \) is mass and \( v \) is velocity. For the 85-kg jogger, the velocity is \( 2.0 \ \mathrm{m/s} \) east. Thus, the momentum \( \mathbf{p}_1 \) is:\[ \mathbf{p}_1 = 85 \, \mathrm{kg} \times 2.0 \, \mathrm{m/s} = 170 \, \mathrm{kg \cdot m/s} \] This momentum is directed due east (along the positive x-axis).
02

Resolve the velocity of the 55-kg jogger into components

The velocity of the 55-kg jogger is \( 3.0 \, \mathrm{m/s} \) at an angle of \( 32^{\circ} \) north of east. We resolve this velocity into x (eastward) and y (northward) components using trigonometry:\[v_{x,2} = 3.0 \, \mathrm{m/s} \times \cos(32^{\circ}) \approx 2.54 \, \mathrm{m/s} \]\[v_{y,2} = 3.0 \, \mathrm{m/s} \times \sin(32^{\circ}) \approx 1.59 \, \mathrm{m/s} \]
03

Compute the momentum of the 55-kg jogger

Using the resolved components, compute the momentum \( \mathbf{p}_2 \) of the 55-kg jogger for each axis:\[\mathbf{p}_{x,2} = 55 \, \mathrm{kg} \times 2.54 \, \mathrm{m/s} \approx 139.7 \, \mathrm{kg \cdot m/s} \] (eastward)\[\mathbf{p}_{y,2} = 55 \, \mathrm{kg} \times 1.59 \, \mathrm{m/s} \approx 87.5 \, \mathrm{kg \cdot m/s} \] (northward)
04

Sum the x-components of momentum

Add the x-components of the momenta from both joggers to find the total x-component:\[ \mathbf{p}_x = \mathbf{p}_1 + \mathbf{p}_{x,2} = 170 \, \mathrm{kg \cdot m/s} + 139.7 \, \mathrm{kg \cdot m/s} = 309.7 \, \mathrm{kg \cdot m/s} \]
05

Sum the y-components of momentum

Since only the 55-kg jogger has a y-component, the total y-momentum is:\[ \mathbf{p}_y = 87.5 \, \mathrm{kg \cdot m/s} \]
06

Find the magnitude of the total momentum

Use the Pythagorean theorem to calculate the magnitude of the total momentum:\[ p_{total} = \sqrt{\mathbf{p}_x^2 + \mathbf{p}_y^2} \]\[ p_{total} = \sqrt{(309.7 \, \mathrm{kg \cdot m/s})^2 + (87.5 \, \mathrm{kg \cdot m/s})^2} \approx 322.8 \, \mathrm{kg \cdot m/s} \]
07

Determine the direction of the total momentum

Calculate the angle \( \theta \) using the inverse tangent function:\[ \theta = \tan^{-1}\left(\frac{\mathbf{p}_y}{\mathbf{p}_x}\right) = \tan^{-1}\left(\frac{87.5}{309.7}\right) \]\[ \theta \approx 15.7^{\circ} \]Thus, the total momentum is approximately \( 15.7^{\circ} \) north of east.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Components
In physics, understanding vector components is essential when dealing with quantities that have both direction and magnitude, like momentum or velocity. A vector, such as the velocity of a jogger, is often not aligned perfectly along a single axis. This is where breaking the vector into components becomes useful. By doing so, we can analyze the horizontal and vertical influences separately.

To break a vector into its components:
  • Use trigonometric functions – cosine and sine – to resolve it along the x and y axes.
  • For example, if a velocity vector is 3.0 m/s at an angle of 32° north of east, the x-component (eastward) can be found using cosine, and the y-component (northward) using sine.
The equations would look like:
  • \(v_x = v \cdot \cos(\theta)\)
  • \(v_y = v \cdot \sin(\theta)\)
Here, \(v_x\) is the eastward component, and \(v_y\) is the northward component, helping simplify complex movement into more manageable parts.
Trigonometry in Physics
Trigonometry plays a vital role in resolving forces and understanding motions that aren't aligned along the standard x or y axes. By using trigonometric functions such as sine, cosine, and tangent, we can find angles and distances, which are key to analyzing physical phenomena.

Key trigonometric functions:
  • **Sine (sin)** - Used to find the opposite side in a right triangle. In our example, it's used to find the northward component of velocity.
  • **Cosine (cos)** - Helps find the adjacent side. In the jogger's case, it calculates the eastward component.
  • **Tangent (tan)** - Used for finding angles based on opposite and adjacent sides.
This can apply to a variety of physics problems, helping convert angles and distances into useful data for calculating other physical quantities, like momentum. It is particularly important whenever a direction is indicated, such as 'north of east', requiring careful resolution of components.
Pythagorean Theorem
The Pythagorean theorem is a cornerstone in physics when dealing with components and overall vectors. It states that in a right triangle, the square of the hypotenuse (longest side) is equal to the sum of the squares of the other two sides.

Mathematically, the theorem is expressed as:\[a^2 + b^2 = c^2\]where \(a\) and \(b\) are the two shorter sides, and \(c\) is the hypotenuse.
Applying this to vectors:
  • Calculate the total magnitude of a vector, such as the combined momentum of two joggers, by knowing the x and y components.
For instance, if \(p_x\) and \(p_y\) are the components of the momentum, then the magnitude of the total momentum \(p_{total}\) can be obtained by:\[p_{total} = \sqrt{p_x^2 + p_y^2}\]This gives a single value representing the entire vector's magnitude. It's an essential method in physics for combining perpendicular components into a single coherent measurement.

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