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An oxygen atom is four times as massive as a helium atom. In an experiment, a helium atom and an oxygen atom have the same kinetic energy. What is the ratio \(v_{\mathrm{H}} / v_{\mathrm{O}}\) of their speeds?

Short Answer

Expert verified
The ratio of the speed of the helium atom to the oxygen atom is 2:1.

Step by step solution

01

Understanding kinetic energy

The kinetic energy (K.E) of a moving object is given by the equation K.E = \(0.5 * m * v^2\), where \(m\) is the mass of the object and \(v\) is its velocity.
02

Formulate equations for the kinetic energy of helium and oxygen atoms

The kinetic energy of the helium atom is given as K.E = \(0.5 * m_{\mathrm{H}} * v_{\mathrm{H}}^2\). The oxygen atom, being four times as heavy as helium, has kinetic energy given as K.E = \(0.5 * 4 * m_{\mathrm{H}} * v_{\mathrm{O}}^2\). Since it's given that the atoms have the same kinetic energy, we equate the two equations: \(0.5 * m_{\mathrm{H}} * v_{\mathrm{H}}^2 = 0.5 * 4 * m_{\mathrm{H}} * v_{\mathrm{O}}^2\).
03

Solve for the velocity ratio

Solving the equation from Step 2 for the velocity ratio \(v_{\mathrm{H}} / v_{\mathrm{O}}\), we get: \(v_{\mathrm{H}} / v_{\mathrm{O}} = \sqrt{4} = 2\).

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Key Concepts

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

Kinetic Energy Equation
Kinetic energy is the energy possessed by an object in motion. The kinetic energy equation plays a crucial role in understanding how objects exchange energy through motion. The formula for kinetic energy (K.E) is given by the expression \( K.E = \frac{1}{2} m v^2 \), where \( m \) represents mass and \( v \) stands for velocity. This equation illustrates that the kinetic energy of an object is directly proportional to its mass and the square of its velocity. Hence, if you double the velocity, the kinetic energy increases by four times due to the square relationship.

Understanding the kinetic energy equation empowers students to solve problems involving motion and the conversion of energy forms, which is pivotal in many physics applications. For instance, in the case of the oxygen and helium atoms from the textbook exercise, despite the significant mass difference, the kinetic energy is constant for both, leading to insightful observations about their velocities.
Mass-Velocity Relationship
The kinetic energy equation unveils a fundamental principle in physics: the mass-velocity relationship. This relationship signifies that for a constant kinetic energy, the velocity of an object is inversely proportional to the square root of its mass. Simply put, more massive objects move slower when compared to less massive ones at the same kinetic energy, which is demonstrated in the provided textbook exercise.

In the exercise, the kinetic energies of a helium atom and an oxygen atom, which is four times heavier, are equal. According to the mass-velocity relation \(v = \sqrt{\frac{2K.E}{m}}\), heavier mass leads to a lower velocity. This intrinsic relationship is key in many physical phenomena and technologies, like particle accelerators where controlling speeds is vital. A solid grasp of this concept is crucial for students to predict motion outcomes in physics problems.
Comparing Atomic Speeds
In physics, comparing atomic speeds becomes interesting when analyzing gases or studying molecular behavior. Atomic speeds are influenced by factors like temperature and mass as seen in the kinetic energy equation. When comparing the speeds of different atoms, like helium and oxygen from our example, we use the principle that equal kinetic energies will result in different speeds based on their mass.

The exercise demonstrated that the lighter helium atom travels at a higher speed compared to the heavier oxygen atom, given the same amount of kinetic energy. This insight is part of the basis for the kinetic theory of gases, which describes how gas pressure is due to collisions of moving atoms with the walls of a container. Therefore, understanding how to compare atomic speeds based on their mass and kinetic energy is essential for students studying thermodynamics, gas laws, and molecular dynamics.

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Most popular questions from this chapter

The desperate contestants on a TV survival show are very hungry. The only food they can see is some fruit hanging on a branch high in a tree. Fortunately, they have a spring they can use to launch a rock. The spring constant is \(1000 \mathrm{N} / \mathrm{m},\) and they can compress the spring a maximum of \(30 \mathrm{cm} .\) All the rocks on the island seem to have a mass of \(400 \mathrm{g}\). a. With what speed does the rock leave the spring? b. If the fruit hangs 15 m above the ground, will they feast or go hungry?

You have been asked to design a "ballistic spring system" to measure the speed of bullets. A bullet of mass \(m\) is fired into a block of mass \(M .\) The block, with the embedded bullet, then slides across a frictionless table and collides with a horizontal spring whose spring constant is \(k .\) The opposite end of the spring is anchored to a wall. The spring's maximum compression \(d\) is measured. a. Find an expression for the bullet's speed \(v_{\mathrm{B}}\) in terms of \(m, M\) \(k,\) and \(d\) b. What was the speed of a \(5.0 \mathrm{g}\) bullet if the block's mass is \(2.0 \mathrm{kg}\) and if the spring, with \(k=50 \mathrm{N} / \mathrm{m},\) was compressed by \(10 \mathrm{cm} ?\) c. What fraction of the bullet's energy is "lost"? Where did it go?

A 500 g rubber ball is dropped from a height of \(10 \mathrm{m}\) and undergoes a perfectly elastic collision with the earth. a. What is the earth's velocity after the collision? Assume the earth was at rest just before the collision. b. How many years would it take the earth to move 1.0 mm at this speed?

A very slippery ice cube slides in a vertical plane around the inside of a smooth, \(20-\) cm-diameter horizontal pipe. The ice cube's speed at the bottom of the circle is \(3.0 \mathrm{m} / \mathrm{s}\) a. What is the ice cube's speed at the top? b. Find an algebraic expression for the ice cube's speed when it is at angle \(\theta\), where the angle is measured counterclockwise from the bottom of the circle. Your expression should give \(3.0 \mathrm{m} / \mathrm{s}\) for \(\theta=0^{\circ}\) and your answer to part a for \(\theta=180^{\circ}\) c. Make a graph of \(v\) versus \(\theta\) for one complete revolution.

A block sliding along a horizontal frictionless surface with speed \(v\) collides with a spring and compresses it by \(2.0 \mathrm{cm} .\) What will be the compression if the same block collides with the spring at a speed of \(2 v ?\)

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