The distance between a carbon atom \((m=12 \mathrm{u})\) and an oxygen atom \((m=16 \mathrm{u})\) in a carbon monoxide \((\mathrm{CO})\) molecule is \(1.13 \cdot 10^{-10} \mathrm{~m} .\) How far from the carbon atom is the center of mass of the molecule? \((1 \mathrm{u}=1\) atomic mass unit. \()\)

Short Answer

Expert verified
Answer: The center of mass is located at approximately \(6.483\times10^{-11} m\) away from the carbon atom.

Step by step solution

01

Write down given information

We have the following information given: - Mass of carbon atom (C): \(m_C=12u\) - Mass of oxygen atom (O): \(m_O=16u\) - Distance between C and O: \(d=1.13\times10^{-10}m\) - Conversion factor of Atomic mass unit (u) to kilogram (kg): \(1u=1.66\times10^{-27}kg\)
02

Convert the atomic mass units to kilograms

We need to convert the masses of the carbon and oxygen atoms to kilograms using the conversion factor. \(m_C = 12u\times (1.66\times10^{-27}kg/u) = 1.992\times10^{-26}kg\) \(m_O = 16u\times (1.66\times10^{-27}kg/u) = 2.656\times10^{-26}kg\)
03

Apply the center of mass formula for a two-point system

For a two-point mass system like this, the formula for the center of mass is: \(x_{cm} = \frac{m_Cx_C + m_Ox_O}{m_C + m_O}\) Let's assume that the carbon atom is at the origin \(x_C=0\), and the oxygen atom is at position \(x_O=d\). So, the formula becomes: \(x_{cm} = \frac{0 + 2.656\times10^{-26}kg\times (1.13\times10^{-10}m)}{1.992\times10^{-26}kg + 2.656\times10^{-26}kg}\)
04

Calculate the position of the center of mass

Now, use the equation derived in step 3 to calculate \(x_{cm}\): \(x_{cm} = \frac{2.656\times10^{-26}kg\times (1.13\times10^{-10}m)}{4.648\times10^{-26}kg}\) \(x_{cm} = 6.483\times10^{-11} m\) So, the center of mass is located at approximately \(6.483\times10^{-11} m\) away from the carbon atom.

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Key Concepts

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

Atomic Mass Unit
An atomic mass unit (u) is a standard unit of mass that quantifies mass on an atomic or molecular scale. It is defined as one twelfth the mass of a carbon-12 atom in its ground state and is approximately equal to 1.66053906660 x 10^-27 kilograms. This seemingly small number is incredibly significant in the world of chemistry and physics, as it allows scientists to compare masses of different atoms and molecules on a relative scale.

Using atomic mass units simplifies the calculations involving the masses of particles that are too small to weigh individually. For example, in the carbon monoxide molecule exercise, the atoms of carbon and oxygen were given in atomic mass units, which were later converted to kilograms to calculate the center of mass. The atomic mass unit provides a bridge between the macroscopic and microscopic worlds, enabling calculations that involve fundamental particles of matter.
Two-Point Mass System
A two-point mass system refers to a simplified physical model where only two distinct masses are considered for the purpose of calculation. This model is particularly useful in the study of the center of mass in a system composed of two particles.

The center of mass, or centroid, is a crucial concept in physics as it represents the mean position of all the mass in a system. It is the point where the system can be balanced or the point that moves as if all of the system's mass were concentrated there and all external forces were applied there. For a two-point mass system, the calculation of the center of mass involves weighting the position of each mass by its magnitude and summing these products.

By assuming that the carbon atom in the exercise is at the origin and the oxygen atom is at a defined distance away, we can apply the center of mass equation to this two-point system, which results in a simple and elegant solution.
Physics Problem Solving
Physics problem solving is a structured approach to understanding and finding solutions to physical phenomena. It often involves several steps, including identifying the relevant principles, analyzing given information, formulating equations, and performing calculations to arrive at an answer. This methodical approach breaks down complex physics problems into manageable pieces.

In the context of the center of mass problem presented, problem-solving begins by clearly stating the given information and ends up with the application of the center of mass formula. This step-by-step process not only makes the solution comprehensible but also ensures that the underlying physical concepts are well understood. Embracing this structured form of problem-solving is essential for mastering physics and succeeding in this scientific field. Slight improvements in understanding each step can make a significant difference in solving similar problems in the future.

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Most popular questions from this chapter

Two objects with masses \(m_{1}\) and \(m_{2}\) are moving along the \(x\) -axis in the positive direction with speeds \(v_{1}\) and \(v_{2}\), respectively, where \(v_{1}\) is less than \(v_{2}\). The speed of the center of mass of this system of two bodies is a) less than \(v_{1}\). b) equal to \(v_{1}\). c) equal to the average of \(v_{1}\) and \(v_{2}\). d) greater than \(v_{1}\) and less than \(v_{2}\). e) greater than \(v_{2}\).

Starting at rest, two students stand on 10.0 -kg sleds, which point away from each other on ice, and they pass a 5.00 -kg medicine ball back and forth. The student on the left has a mass of \(50.0 \mathrm{~kg}\) and can throw the ball with a relative speed of \(10.0 \mathrm{~m} / \mathrm{s}\). The student on the right has a mass of \(45.0 \mathrm{~kg}\) and can throw the ball with a relative speed of \(12.0 \mathrm{~m} / \mathrm{s} .\) (Assume there is no friction between the ice and the sleds and no air resistance.) a) If the student on the left throws the ball horizontally to the student on the right, how fast is the student on the left moving right after the throw? b) How fast is the student on the right moving right after catching the ball? c) If the student on the right passes the ball back, how fast will the student on the left be moving after catching the pass from the student on the right? (d) How fast is the student on the right moving after the pass?

Can the center of mass of an object be located at a point outside the object, that is, at a point in space where no part of the object is located? Explain.

Many nuclear collisions studied in laboratories are analyzed in a frame of reference relative to the laboratory. A proton, with a mass of \(1.6605 \cdot 10^{-27} \mathrm{~kg}\) and traveling at a speed of \(70.0 \%\) of the speed of light, \(c\), collides with a tin\(116\left({ }^{116} \mathrm{Sn}\right)\) nucleus with a mass of \(1.9096 \cdot 10^{-25} \mathrm{~kg} .\) What is the speed of the center of mass with respect to the laboratory frame? Answer in terms of \(c\), the speed of light.

A uniform chain with a mass of \(1.32 \mathrm{~kg}\) per meter of length is coiled on a table. One end is pulled upward at a constant rate of \(0.47 \mathrm{~m} / \mathrm{s}\). a) Calculate the net force acting on the chain. b) At the instant when \(0.15 \mathrm{~m}\) of the chain has been lifted off the table, how much force must be applied to the end being raised?

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