Problem 50 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

(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 1: 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.

Step 2: 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:\[ m_p v_{p1} = m_p v_{p2} + m_n v_{n2} \]

Step 3: 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 \(m_n\).

Step 4: 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:\[ m_n v_{n2} = m_p(1.50 + 1.20) \times 10^7 \]

Step 5: 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 \(m_n\).

Step 6: Calculate Mass of Unknown Nucleus

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

Step 7: 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:\[ v_{n2} = \frac{m_p (1.50 + 1.20) \times 10^7}{m_n} \]

Key Concepts

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.