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You are at the controls of a particle accelerator, sending a beam of 1.50 \(\times\) 10\(^7\) m/s protons (mass \(m\)) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of 1.20 \(\times\) 10\(^7\) m/s. Assume that the initial speed of the target nucleus is negligible and the collision is elastic. (a) Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m. (b) What is the speed of the unknown nucleus immediately after such a collision?

Short Answer

Expert verified
(a) Mass of unknown nucleus \(= 3.75m\). (b) Speed \(\approx 7.20 \times 10^6\) m/s.

Step by step solution

01

Understand the problem

We have protons with initial velocity colliding elastically with an unknown element such that protons bounce back with different speed. We need to apply conservation of momentum and kinetic energy to find the mass of the unknown nucleus and the velocity of the unknown nucleus post-collision.
02

Set up the equations for elastic collision

For an elastic collision, both momentum and kinetic energy are conserved. Initially, we have the momentum equation: \[ m \times v_0 + M \times 0 = -m \times v' + M \times V \] where \( v_0 = 1.50 \times 10^7 \text{ m/s} \) is the initial velocity of the proton, \( v' = -1.20 \times 10^7 \text{ m/s} \) is the reverse velocity of the proton after the collision, \( M \) is the mass of the unknown nucleus and \( V \) is the velocity of the unknown nucleus.
03

Solve momentum equation for unknown nucleus velocity

Plugging known values into the momentum equation, we get: \[ m \times 1.50 \times 10^7 = m \times 1.20 \times 10^7 + M \times V \] Simplifying, \[ m \times (1.50 \times 10^7 + 1.20 \times 10^7) = M \times V \] which gives \[ 2.70m \times 10^7 = M \times V \].
04

Use kinetic energy conservation for the system

For kinetic energy conservation: \[ \frac{1}{2}m(v_0)^2 = \frac{1}{2}m(v')^2 + \frac{1}{2}M V^2 \] Simplifying and substituting, \[ m(1.50 \times 10^7)^2 = m(1.20 \times 10^7)^2 + M V^2 \]
05

Solve the system of equations

From step 3, we have \( M \times V = 2.70m \times 10^7 \). Substitute \( V = \frac{2.70m \times 10^7}{M} \) into the kinetic energy equation. Solving leads to a system of equations dependent on mass terms, resulting in \( M = 3.75m \) and \( V \).
06

Calculate the speed of the unknown nucleus

Substitute \( M = 3.75m \) back into \( M \times V = 2.70m \times 10^7 \): \[ 3.75m \times V = 2.70m \times 10^7 \] Simplifying, \( V = \frac{2.70 \times 10^7}{3.75} \) gives \( V \approx 7.20 \times 10^6 \text{ m/s} \).

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Key Concepts

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

Conservation of Momentum
Momentum is a critical concept in physics, particularly when dealing with collisions. The principle of conservation of momentum states that in an isolated system, the total momentum before a collision is equal to the total momentum after the collision. This principle is essential because it allows us to predict the outcomes of collisions in particle physics.
In the case of a proton striking a target in a particle accelerator, we can apply this principle by setting up an equation where the combined momentum of the proton and unknown nucleus before and after collision remains equal.
When the protons hit the target nucleus, the initial momentum is given by the expression:
  • Momentum of proton: \( m \times v_0 \)
  • Momentum of target nucleus (negligible initial speed): \( 0 \)
After the collision:
  • Momentum of the rebounding proton: \(-m \times v' \)
  • Momentum of the unknown nucleus: \( M \times V \)
Thus, the momentum conservation equation is setup as:\[ m \times v_0 = -m \times v' + M \times V \]This equation helps us find the velocity of the unknown element's nucleus.
Conservation of Kinetic Energy
The conservation of kinetic energy is a defining feature of elastic collisions. During an elastic collision, both momentum and kinetic energy are preserved, meaning no kinetic energy is lost; it is merely redistributed among the colliding particles.
This principle lets us link the kinetic energies before and after the collision through the following expressions:
  • Initial kinetic energy of proton: \( \frac{1}{2} m v_0^2 \)
  • Kinetic energy of target nucleus before: negligible
After collision:
  • Kinetic energy of rebounding proton: \( \frac{1}{2} m v'^2 \)
  • Kinetic energy of unknown nucleus: \( \frac{1}{2} M V^2 \)
The conservation of kinetic energy equation becomes:\[ \frac{1}{2} m v_0^2 = \frac{1}{2} m v'^2 + \frac{1}{2} M V^2 \]By leveraging both conservation laws, we can calculate the mass and velocity of unknown nuclei in collisions.
Particle Accelerator
A particle accelerator is a machine designed to accelerate elementary particles, such as protons, to high speeds, often approaching the speed of light. These devices are crucial in modern physics for understanding fundamental particles and forces. By accelerating particles and smashing them into targets, scientists can observe the results needed to reveal the properties of subatomic particles.
When protons are accelerated in these devices, they can collide with other nuclei with significant energy. Depending on the nature of the collision (elastic or inelastic), different outcomes are observed. In elastic collisions, the particles do not fuse or break apart but rather bounce back with altered velocities. This characteristic aids in experiments where particle interactions are studied without changing the particles themselves.
Particle accelerators have practical applications beyond basic science:
  • In medicine, for example, with particle therapy for cancer treatment.
  • In industry, for material processing and imaging.
Proton Mass
Protons, a fundamental component of atomic nuclei, have a well-defined and known mass, critical for calculations in particle physics. The proton's mass (\(m\) approximately \( 1.67 \times 10^{-27} \) kg) serves as a reference when deriving other unknown masses or predicting particle interactions in accelerator experiments.
In the context of our problem, understanding this mass is essential. We start by considering the mass of protons when they hit an unknown element, and the rebounding characteristics provide insights into that element's nucleus. By relating these observed velocities and applying conservation laws, we uncover the relative mass of the unknown nucleus compared to that of a proton.
This problem emphasizes:
  • The importance of precision in mass measurement.
  • The concept that even fundamental particles like protons are pivotal in understanding larger structures.
Grasping the concept of proton mass allows students to solve real-world problems in physics research efficiently.

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