/*! 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 63 (a) John has a larger mass than ... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) John has a larger mass than Barbara has. He is standing on the \(x\) axis at \(x_{\mathrm{J}}=+9.0 \mathrm{~m}\), while she is standing on the \(x\) axis at \(x_{\mathrm{B}}=+2.0 \mathrm{~m}\). Is their centerof-mass point closer to the \(9.0\) -m point or the \(2.0\) -m point? (b) They switch positions. Is their center-of-mass point now closer to the \(9.0\) -m point or the \(2.0\) -m point? (c) In which direction, toward or away from the origin, does their center of mass move as a result of the switch?

Short Answer

Expert verified
(a) Closer to 9.0 m. (b) Closer to 2.0 m. (c) Towards the origin.

Step by step solution

01

Understanding the Concept of Center of Mass

The center of mass of a two-object system along a line can be determined using the formula: \( x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \), where \( m_1 \) and \( m_2 \) are the masses and \( x_1 \) and \( x_2 \) are the positions of the objects. A point closer to the larger mass will result in a center of mass closer to that position.
02

Determining Initial Center of Mass

Let John have a mass \( m_J \) and Barbara a mass \( m_B \) with \( m_J > m_B \). Their positions are \( x_J = 9.0 \) m and \( x_B = 2.0 \) m, respectively. Since \( m_J > m_B \), the center of mass \( x_{cm} \) must be closer to John's position at 9 m.
03

Center of Mass After Switching Positions

After switching, Barbara is at \( x_J = 9.0 \) m and John at \( x_B = 2.0 \) m. Since John is heavier, the center of mass \( x_{cm} \) is now closer to his new position, 2 m.
04

Determining Directional Movement of Center of Mass

Originally, \( x_{cm} \) was closer to \( 9.0 \) m. After the switch, \( x_{cm} \) is closer to \( 2.0 \) m. Therefore, the center of mass has moved towards the origin.

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Key Concepts

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

Mass Distribution
When understanding the concept of center of mass, the mass distribution between two individuals is critical. Imagine a seesaw with different weights on each side; this is how mass distribution works in physics. In this exercise, John's mass is larger than Barbara's, which affects the overall balance. The position of the center of mass will shift towards the heavier individual's location.

To visualize this, consider the formula for center of mass:
  • \( x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \)
Here, \( m_1 \) and \( m_2 \) represent the masses, while \( x_1 \) and \( x_2 \) are the positions of the two masses. Intuitively, the more mass an object has, the more it "pulls" the center of mass towards itself, making mass distribution fundamental in determining where the center of mass falls.

Hence, initially, with John at 9 m and Barbara at 2 m, the center of mass leans towards John due to his larger mass.
Physics Problem Solving
Physics problems often involve multi-step solutions that require careful consideration of all variables. Here, we used the methodical application of physics principles to understand how mass and position affect the center of mass.

The strategy in this problem was straightforward:
  • Identify the mass of each individual and their respective positions.
  • Use the center of mass formula to determine where the center of mass lies.
  • Consider changes in position to re-evaluate the center of mass.
Every step brings us closer to understanding the outcome, showing the importance of a systematic approach in physics problem-solving.

Remember, sometimes visualization can help, so don't hesitate to draw diagrams or use analogies, like a seesaw, to better understand the problem dynamics.
Position Switching
Position switching can change the dynamics significantly in a physics problem. By switching places, Barbara and John alter the distribution of their masses on the x-axis.

Initially, because John is heavier, the center of mass was closer to his original position at 9 meters. However, when they switch:
  • Barbara takes John's position at 9 meters, and
  • John moves to her original position at 2 meters.

This switch results in a new center of mass closer to John's new position at 2 meters. This example illustrates how a simple position swap can notably affect the overall mass distribution and, as a result, shift the center of mass to a new location.
Step-by-Step Solution
Breaking down a physics problem into clear steps helps in grasping complex concepts. Here’s a recap of how this was handled:
  • Step 1: Understanding the center of mass formula: \( x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \).
  • Step 2: Determining the initial center of mass which is closer to John's position due to his larger mass.
  • Step 3: Identifying how the center of mass changes after they switch places. Now it points nearer to John's new 2-meter position.
  • Step 4: Observing the directional change in the center of mass, which has moved towards the origin.

This methodical approach not only solves the problem but also reinforces understanding of each concept involved, ensuring you build a solid foundation in physics problem-solving.

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Most popular questions from this chapter

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.

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. (a) Is it possible that the two-object system has a final total momentum of zero after the collision? (b) Roughly, what is the direction of the final total momentum of the two-object system after the collision? Problem Object A has a mass of \(m_{A}=17.0 \mathrm{~kg}\) and an initial velocity of \(\overrightarrow{\mathbf{v}}_{0 \mathrm{~A}}=8.00 \mathrm{~m} / \mathrm{s},\) due east. Object \(\mathrm{B},\) however, has a mass of \(m_{\mathrm{B}}=29.0 \mathrm{~kg}\) and \(\mathrm{an}\) initial velocity of \(\overrightarrow{\mathbf{v}}_{0 \mathrm{~B}}=5.00 \mathrm{~m} / \mathrm{s},\) due north. Find the magnitude and direction of the total momentum of the two-object system after the collision. Make sure that your answers are consistent with your answers to the Concept Questions.

Consult Interactive Solution \(\underline{7} .9\) at for a review of problem- solving skills that are involved in this problem. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of \(+16.0 \mathrm{~m} / \mathrm{s},\) while the exiting water stream has a velocity of \(-16.0 \mathrm{~m} / \mathrm{s}\). The mass of water per second that strikes the blade is \(30.0 \mathrm{~kg} / \mathrm{s}\). Find the magnitude of the average force exerted on the water by the blade.

A car (mass \(=1100 \mathrm{~kg}\) ) is traveling at \(32 \mathrm{~m} / \mathrm{s}\) when it collides head-on with a sport utility vehicle (mass \(=2500 \mathrm{~kg}\) ) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?

At provides a review of the concepts that are involved in this problem. A \(62.0-\mathrm{kg}\) person, standing on a diving board, dives straight down into the water. Just before striking the water, her speed is \(5.50 \mathrm{~m} / \mathrm{s}\). At a time of \(1.65 \mathrm{~s}\) after she enters the water, her speed is reduced to \(1.10 \mathrm{~m} / \mathrm{s}\). What is the net average force (magnitude and direction) that acts on her when she is in the water?

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