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You are at the controls of a particle accelerator, sending a beam of \(1.50 \times 10^{7} \mathrm{m} / \mathrm{s}\) 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 \(1.20 \times 10^{7} \mathrm{m} / \mathrm{s}\) . 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 the nucleus is \(\frac{m_p (2.7)^2}{(1.5)^2-(1.2)^2}\). (b) Speed of nucleus is found by plugging into velocity equation.

Step by step solution

01

Understand the Problem

We need to find the mass of the unknown nucleus using the given information about elastic collisions and velocities of protons before and after the collision. The conservation laws of momentum and kinetic energy will be used.
02

Use Conservation of Momentum

In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. For the proton and the nucleus:\[ m_p v_{p1} + m_n v_{n1} = m_p v_{p2} + m_n v_{n2} \]Given that the initial speed of the nucleus \(v_{n1} = 0\) and \(v_{p1} = 1.50 \times 10^7\, \text{m/s}\) and \(v_{p2} = -1.20 \times 10^7\, \text{m/s}\), the equation simplifies to:\[ m_p v_{p1} = m_p v_{p2} + m_n v_{n2} \]
03

Use Conservation of Kinetic Energy

Since the collision is elastic, the total kinetic energy before and after the collision is conserved:\[ \frac{1}{2} m_p v_{p1}^2 = \frac{1}{2} m_p v_{p2}^2 + \frac{1}{2} m_n v_{n2}^2 \]This equation will be used along with the momentum to find the unknown mass \(m_n\).
04

Solve for Unknown Mass (a)

From the momentum conservation equation:\[ m_p(1.50 \times 10^7) = m_p(-1.20 \times 10^7) + m_n v_{n2} \]Rearranging gives:\[ m_n v_{n2} = m_p(1.50 + 1.20) \times 10^7 \]
05

Relate Speeds Using Kinetic Energy (b)

From kinetic energy conservation:\[ m_p(1.50^2) = m_p(1.20^2) + m_n v_{n2}^2 \]Solve for \(v_{n2}\) in terms of \(m_p\) and substitute into previous expressions to solve for \(m_n\).
06

Calculate Mass of Unknown Nucleus

Combine equations from Steps 4 and 5 to find:\[ m_n = m_p \frac{(1.50 + 1.20)^2}{(1.50^2) - (1.20^2)} \]
07

Calculate Speed of Unknown Nucleus After Collision

After finding \(m_n\), use the relation found in Step 5 to solve for \(v_{n2}\). This is the speed of the unknown nucleus after the collision:\[ v_{n2} = \frac{m_p (1.50 + 1.20) \times 10^7}{m_n} \]

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Key Concepts

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

Momentum Conservation
Understanding momentum conservation is crucial for solving problems involving elastic collisions, like the ones occurring in particle accelerators. In the context of our exercise, momentum conservation dictates that the total momentum before the collision must equal the total momentum after the collision.
In simple terms, if you think of momentum as a way to measure the "quantity of motion," this principle tells us that the "quantity of motion" is conserved in an isolated system. Mathematically, for two objects—the proton and the unknown nucleus—the momentum before the collision can be expressed as:
  • The momentum of the proton: \(m_p v_{p1}\)

  • The initial momentum of the nucleus: \(m_n v_{n1}\)
After the collision, the momentum equations consider:
  • The momentum of the rebounding proton: \(m_p v_{p2}\)

  • The momentum of the nucleus: \(m_n v_{n2}\)
With the given scenario, simplifying this results in accounting primarily for the velocity changes and masses of the particles. Understanding this principle allows scientists to predict particle behavior in accelerators.
Kinetic Energy Conservation
Kinetic energy conservation in elastic collisions is another powerful concept that's especially relevant when dealing with particle physics. Elastic collisions are unique because they conserve not just momentum but also kinetic energy. This means that the total kinetic energy before and after the collision remains constant.
For the collision between a proton and an unknown nucleus, the kinetic energy concept can be mathematically represented as:
  • Initial kinetic energy (from the moving proton): \(\frac{1}{2} m_p v_{p1}^2\)

  • Final kinetic energy: the sum of individual energies of both moving particles after the collision.
To solve for unknowns, like the mass of the unknown nucleus, you equate the sum of kinetic energies before and after the event:
  • \(\frac{1}{2} m_p v_{p1}^2 = \frac{1}{2} m_p v_{p2}^2 + \frac{1}{2} m_n v_{n2}^2\)
This balance of kinetic energy is what confirms the event as elastic and offers clues towards solving for unknown variables.
Particle Accelerator
Particle accelerators are extraordinary scientific instruments utilized to propel charged particles, like protons, to high speeds—often near the speed of light. They are pivotal in studying the fundamental components of matter. In our scenario, an accelerator helps analyze collisions by speeding up protons and directing them at a target.
A better understanding of how an accelerator works can illuminate why and how these experiments happen. Inside, powerful electromagnetic fields swirl the particles rapidly in either circular or linear paths. The energy furnished by this acceleration defines the possible interactions these particles can undergo.
These devices enable experiments where particles collide at high speeds, replicating conditions only found in extreme cosmic scenarios, allowing us to reveal elements not visible otherwise and potentially discover unknown particles.
Unknown Element Identification
In particle physics experiments, identifying an unknown element can be likened to unraveling a cosmic puzzle. The rebounding protons in our experiment hold significant information. When protons collide with an element's nucleus and rebound, the manner of rebounding tells us about the nucleus they engaged with.
The unknown element's mass influences how particles rebounded in experiments. By using conservation laws, scientists deduce various properties of the elements. For example, the change in speed and the resulting kinetic energy provide clues to calculate the element's mass relative to known particles, helping in its identification.
This methodical approach showcases particle accelerators' importance in modern science, permitting the discovery of element masses and behaviors not easily identifiable by conventional means. Therefore, each collision and rebounding event is an opportunity for new discoveries.

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Most popular questions from this chapter

A small wooden block with mass 0.800 \(\mathrm{kg}\) is suspended from the lower end of a light cord that is 1.60 \(\mathrm{m}\) long. The block is initially at rest. A bullet with mass 12.0 \(\mathrm{g}\) is fired at the block with a horizontal velocity \(v_{0} .\) The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of \(0.800 \mathrm{m},\) the tension in the cord is 4.80 \(\mathrm{N} .\) What was the initial speed \(v_{0}\) of the bullet?

A ball with mass \(M\) , moving horizontally at \(4.00 \mathrm{m} / \mathrm{s},\) collides elastically with a block with mass 3\(M\) that is initially hanging at rest from the ceiling on the end of a \(50.0-\) cm wire. Find the maximum angle through which the block swings after it is hit.

Two friends, Burt and Ernie, are standing at opposite ends of a uniform log that is floating in a lake. The log is 3.0 \(\mathrm{m}\) long and has mass 20.0 \(\mathrm{kg} .\) Burt has mass 30.0 \(\mathrm{kg}\) and Ernie has mass 40.0 \(\mathrm{kg} .\) Initially the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Neglect any horizontal force that the water exerts on the log and assume that neither Burt nor Ernie falls off the log.

An engine of the orbital maneuvering system (OMS) on a space shuttle exerts a force of \((26,700 \mathrm{N}) \hat{J}\) for 3.90 \(\mathrm{s}\) , exhausting a negligible mass of fuel relative to the \(95,000-\mathrm{kg}\) mass of the shuttle. (a) What is the impulse of the force for this 3.90 s? (b) What is the shuttle's change in momentum from this impulse? (c) What is the shuttle's change in velocity from this impulse? (d) Why can't we find the resulting change in the kinetic energy of the shuttle?

A movie stuntman (mass 80.0 \(\mathrm{kg}\) ) stands on a window ledge 5.0 \(\mathrm{m}\) above the floor (Fig. P8.85). Grabbing a rope attached to a chandelier, he swings down to grapple with the movie's villain (mass 70.0 kg), who is standing directly under the chandelier. (Assume that the stuntman's center of mass moves downward 5.0 \(\mathrm{m} .\) He releases the rope just as he reaches the villain.) (a) With what speed do the entwined foes start to slide across the floor? (b) If the coefficient of kinetic friction of their bodies with the floor is \(\mu_{k}=0.250,\) how far do they slide?

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