/*! 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 13 An \(85-\mathrm{kg}\) jogger is ... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 13

An \(85-\mathrm{kg}\) 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 magnitude of the total momentum is 322.73 kg路m/s, directed 15.77掳 north of east.

Step by step solution

01

Calculate Individual Momenta

The momentum of an object is given by \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. First, calculate the momentum of the 85-kg jogger: \( p_1 = 85 \, \text{kg} \times 2.0 \, \text{m/s} = 170 \, \text{kg} \cdot \text{m/s} \) going due east. Next, calculate the momentum of the 55-kg jogger. The speed is given as \(3.0\, \text{m/s}\) and they are moving 32掳 north of east: \( p_2 = 55 \, \text{kg} \times 3.0 \, \text{m/s} = 165 \, \text{kg} \cdot \text{m/s} \).
02

Resolve the 55-kg Jogger's Momentum Into Components

The 55-kg jogger's momentum needs to be broken into eastward and northward components using trigonometry. The eastward component \( p_{2x} = 165 \cdot \cos(32^\circ) \). The northward component \( p_{2y} = 165 \cdot \sin(32^\circ) \). Calculate these using a calculator:\( p_{2x} \approx 139.71 \, \text{kg} \cdot \text{m/s} \) and \( p_{2y} \approx 87.29 \, \text{kg} \cdot \text{m/s}. \)
03

Sum the Momentum Components in Each Direction

Now, sum the momentum components for the east direction:\( p_{\text{east}} = 170 + 139.71 = 309.71 \, \text{kg} \cdot \text{m/s} \).For the north direction:\( p_{\text{north}} = 0 + 87.29 = 87.29 \, \text{kg} \cdot \text{m/s} \).
04

Calculate the Magnitude of the Total Momentum

The magnitude of the total momentum \( p \) can be found using the Pythagorean Theorem:\[p = \sqrt{(p_{\text{east}})^2 + (p_{\text{north}})^2} = \sqrt{(309.71)^2 + (87.29)^2} \approx 322.73 \, \text{kg} \cdot \text{m/s} \].
05

Determine the Direction of the Total Momentum

The direction \( \theta \) with respect to east can be found using the inverse tangent:\[\theta = \tan^{-1}\left( \frac{p_{\text{north}}}{p_{\text{east}}} \right) = \tan^{-1}\left( \frac{87.29}{309.71} \right) \approx 15.77^\circ.\]Thus, the momentum is directed 15.77掳 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
To understand momentum's direction and magnitude, we first break vectors into components. This is like finding the shorter legs of a right triangle when you know the hypotenuse. For our jogger heading 32掳 north of east, we use trigonometric functions:
  • The eastward component is obtained using cosine: \( p_{2x} = p \cos(\theta) \).
  • The northward component is found with sine: \( p_{2y} = p \sin(\theta) \).
Both components are like building blocks for the jogger鈥檚 momentum. Calculating these allows us to analyze movement along both directions separately.
Trigonometry in Physics
Trigonometry helps us understand angles and distances in physics. It鈥檚 vital for breaking vectors into components. For the jogger moving 32掳 north of east:
  • Use \( \cos(32^\circ) \) to find the eastward movement.
  • Use \( \sin(32^\circ) \) for the northward movement.
This process is crucial in physics to decompose different forces or movements into understandable parts, especially when they don't align purely along conventional axes.
Resultant Vector
A resultant vector combines individual vectors into a single vector that has the same effect. When two joggers have different momentum directions, we calculate each one's effects and sum them:
  • Add eastward components to find the total eastward momentum.
  • Add northward components for total northward momentum.
The resulting vector is like drawing a diagonal from the origin to the endpoint formed by these total components. This gives a clear image of both the magnitude and direction.
Pythagorean Theorem
The Pythagorean Theorem, \( a^2 + b^2 = c^2 \), is key when combining momentum components. It allows us to calculate the overall magnitude of combined momentum:
  • Here, \( a = p_{\text{east}} \) and \( b = p_{\text{north}} \).
  • The resultant magnitude \( c = \sqrt{a^2 + b^2} \).
This formula helps bridge components to reveal the total effect as a single vector, giving insight into the joggers' combined momentum.
Inverse Tangent
The inverse tangent function, \( \tan^{-1} \), helps find directions of vectors from their components. In our example:
  • Use \( \theta = \tan^{-1} \left( \frac{p_{\text{north}}}{p_{\text{east}}} \right) \) to find the angle from the east axis.
  • This gives a direction relative to the east, creating a clear representation of motion.
By applying inverse tangent, we can precisely determine how the momentum vector points, completing the picture of both direction and magnitude.

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Most popular questions from this chapter

A 40.0 -kg boy, riding a 2.50 -kg skateboard at a velocity of \(+5.30 \mathrm{m} / \mathrm{s}\) across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is \(6.00 \mathrm{m} / \mathrm{s}, 9.50^{\circ}\) above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard"s velocity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

Two arrows are fired horizontally with the same speed of \(30.0 \mathrm{m} / \mathrm{s}\) Each arrow has a mass of \(0.100 \mathrm{kg} .\) One is fired due east and the other due south. Find the magnitude and direction of the total momentum of this two-arrow system. Specify the direction with respect to due east.

In a performance test, each of two cars takes \(9.0 \mathrm{s}\) to accelerate from rest to \(27 \mathrm{m} / \mathrm{s}\). Car A has a mass of \(1400 \mathrm{kg},\) and car \(\mathrm{B}\) has a mass of \(1900 \mathrm{kg} .\) Find the net average force that acts on each car during the test.

You and your crew must dock your \(25000 \mathrm{kg}\) spaceship at Spaceport Alpha, which is orbiting Mars. In the process, Alpha's control tower has requested that you ram another vessel, a freight ship of mass \(16500 \mathrm{kg},\) latch onto it, and use your combined momentum to bring it into dock. The freight ship is not moving with respect to the colossal Spaceport Alpha, which has a mass of \(1.85 \times 10^{7} \mathrm{kg} .\) Alpha's automated system that guides incoming spacecraft into dock requires that the incoming speed is less than \(2.0 \mathrm{m} / \mathrm{s}\). (a) Assuming a perfectly linear alignment of your ship's velocity vector with the freight ship (which is stationary with respect to Alpha) and Alpha's docking port, what must be your ship's speed (before colliding with the freight ship) so that the combination of the freight ship and your ship arrives at Alpha's docking port with a speed of \(1.50 \mathrm{m} / \mathrm{s} ?\) (b) How does the velocity of Spaceport Alpha change when the combination of your vessel and the freight ship successfully docks with it? (c) Suppose you made a mistake while maneuvering your vessel in an attempt to ram the freight ship and, rather than latching on to it and making a perfectly inelastic collision, you strike it and knock it in the direction of the spaceport with a perfectly elastic collision. What is the speed of the freight ship in that case (assuming your ship had the same initial velocity as that calculated in part (a))?

A golf ball bounces down a flight of steel stairs, striking several steps on the way down, but never hitting the edge of a step. The ball starts at the top step with a vertical velocity component of zero. If all the collisions with the stairs are elastic, and if the vertical height of the staircase is \(3.00 \mathrm{m},\) determine the bounce height when the ball reaches the bottom of the stairs. Neglect air resistance.
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