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An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision?

Short Answer

Expert verified
The ratio is \( \frac{4}{1837 \times 1838} \).

Step by step solution

01

Identify Variables and Initial Conditions

Let the mass of the electron be \(m\) and the mass of the hydrogen atom be \(M = 1837m\). Let's denote the initial velocity of the electron as \(v_{e}\) and the initial velocity of the hydrogen atom as \(0\), since it is stationary.
02

Apply Conservation of Momentum

For an elastic collision, momentum before collision equals momentum after collision. The equation is:\[ m v_{e} = m v_{e'} + M v_{H'} \]where \(v_{e'}\) and \(v_{H'}\) are velocities of the electron and hydrogen atom after collision, respectively.
03

Apply Conservation of Kinetic Energy

In elastic collisions, kinetic energy is also conserved. Thus:\[ \frac{1}{2}m v_{e}^2 = \frac{1}{2}m v_{e'}^2 + \frac{1}{2}M v_{H'}^2 \]
04

Solve for Velocity of Hydrogen Atom

Substitute \(v_{e'}\) from momentum equation into kinetic energy equation and solve for \(v_{H'}\):\[ m v_{e} = m v_{e'} + 1837m v_{H'} \]Rearrange to find:\[ v_{H'} = \frac{v_{e} - v_{e'}}{1837} \]
05

Express Kinetic Energy Ratio

The kinetic energy of hydrogen atom after collision is \(\frac{1}{2}M v_{H'}^2\) and the kinetic energy of the electron before collision is \(\frac{1}{2}m v_{e}^2\). The ratio is:\[ KE_{ratio} = \frac{\frac{1}{2}M v_{H'}^2}{\frac{1}{2}m v_{e}^2} = \frac{M v_{H'}^2}{m v_{e}^2} \]Substitute \(M = 1837m\) and \(v_{H'} = \frac{v_{e} - v_{e'}}{1837} \):\[ KE_{ratio} = \frac{1837m \left(\frac{v_{e} - v_{e'}}{1837}\right)^2}{m v_{e}^2} \]Simplify to find:\[ KE_{ratio} = \frac{(v_{e} - v_{e'})^2}{1837 v_{e}^2} \]
06

Relationship Between Velocities

Assuming equations are set correctly from conservation laws, the relationship between velocities gives us:\[ v_{e'} = \frac{m - M}{m + M} v_{e} = \frac{1 - 1837}{1 + 1837} v_{e} = -\frac{1836}{1838}v_{e} \]
07

Simplify Kinetic Energy Ratio

Substitute \(v_{e'}\) into the kinetic energy ratio:\[ KE_{ratio} = \frac{\left(v_{e} - \left(-\frac{1836}{1838}v_{e}\right)\right)^2}{1837 v_{e}^2} = \frac{\left(\frac{2}{1838}v_{e}\right)^2}{1837 v_{e}^2} \]Simplify:\[ KE_{ratio} = \frac{4}{1837 \times 1838} \]
08

Result

The ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision is \( \frac{4}{1837 \times 1838} \).

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Key Concepts

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

Conservation of Momentum
In the realm of physics, the principle of conservation of momentum is a fundamental concept. It states that within a closed system, where no external forces act, the total momentum remains constant throughout the interaction. This concept is notably crucial in understanding elastic collisions.

In an elastic collision, like the one between an electron and a stationary hydrogen atom, momentum is conserved before and after the collision. This is due to the absence of external forces. Mathematically, this can be expressed as:
  • Initial momentum = Final momentum
  • For the electron-hydrogen scenario, the equation is: \( m v_{e} = m v_{e'} + M v_{H'} \)
Here, \( m \) is the mass of the electron, \( v_{e} \) is its initial velocity, and \( v_{e'} \) is its velocity after collision.

The mass of the hydrogen atom is represented as \( M \) (or \( 1837m \)), and \( v_{H'} \) is its velocity post-collision. By conserving momentum, we can understand how the particles move post-collision.
Conservation of Kinetic Energy
Another significant principle in physics is the conservation of kinetic energy. This concept is particularly important in elastic collisions, where not only momentum, but also kinetic energy is conserved.

In an elastic collision, such as that between our electron and hydrogen atom, the kinetic energy before the collision equals the sum of the kinetic energies after. This is because no energy is lost to factors like heat or sound. Expressed mathematically:
  • Initial kinetic energy = Final kinetic energy
  • This can be shown as: \( \frac{1}{2}m v_{e}^2 = \frac{1}{2}m v_{e'}^2 + \frac{1}{2}M v_{H'}^2 \)
In this equation, the left side represents the initial kinetic energy of the electron. The right side shows the sum of kinetic energies of both the electron and hydrogen atom after the collision.

The conservation of kinetic energy allows us to further explore the relations between velocities and energy transformations in elastic collisions.
Kinetic Energy Ratio
Understanding the kinetic energy ratio in an elastic collision provides insight into energy distributions between colliding bodies. In this particular scenario, we are looking to find the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision.

This ratio encompasses the idea of how energy transitions from one body to another during their interaction. To capture this, we use the formula:
  • \( KE_{ratio} = \frac{\frac{1}{2}M v_{H'}^2}{\frac{1}{2}m v_{e}^2} \)
  • By simplifying, \( KE_{ratio} = \frac{1837m \left(\frac{v_{e} - v_{e'}}{1837}\right)^2}{m v_{e}^2} \)
Subsequently, the final kinetic energy ratio calculation provides a precise comparison:
  • \( KE_{ratio} = \frac{4}{1837 \times 1838} \)
This ratio highlights how much of the electron's initial kinetic energy is transferred to the hydrogen atom.
Electron-Hydrogen Collision
The collision between an electron and a hydrogen atom provides a classic example of studying elastic collisions. When an electron, which is much lighter, collides with the stationary hydrogen atom, the interactions involve both momentum and kinetic energy conservation.

Initially, the electron moves with a velocity \( v_{e} \) while the hydrogen is stationary. Upon collision, these principles guide us to calculate the resulting velocities and energy transformation. From our computations, the electron’s post-collision velocity \( v_{e'} \) can be found using the relationship:
  • \( v_{e'} = \frac{m - M}{m + M} v_{e} \)
This makes the electron rebound while the hydrogen atom gains speed. The mass disparity plays a crucial role in dictating the outcome of such a collision due to the vast difference in masses, where the hydrogen atom's mass is 1837 times that of the electron.

This scenario efficiently demonstrates the elegance of fundamental physics laws governing motion, energy, and interactions.

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Most popular questions from this chapter

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.

At presents a method for modeling this problem. The carbon monoxide molecule (CO) consists of a carbon atom and an oxygen atom separated by a distance of \(1.13 \times 10^{-10} \mathrm{~m}\). The mass \(m_{\mathrm{C}}\) of the carbon atom is 0.750 times the mass \(m_{\mathrm{O}}\) of the oxygen atom, or \(\mathrm{m}_{\mathrm{c}}=0.750 \mathrm{~m}_{0} .\) Determine the location of the center of mass of this molecule relative to the carbon atom.

(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?

ssm 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.

When jumping straight down, you can be seriously injured if you land stiff- legged One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 75 -kg man just before contact with the ground has a speed of \(6.4 \mathrm{~m} / \mathrm{s}\). (a) In a stiff-legged landing he comes to a halt in \(2.0 \mathrm{~ms}\). Find the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in \(0.10 \mathrm{~s}\). Find the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the force of the ground on the man in parts (a) and (b).

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