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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 \(\mathrm{m}\) ) 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
The mass of the unknown nucleus in terms of the proton mass \(m\) is equal to \(M=m\). The speed of the unknown nucleus immediately after the collision is \(3.0 \times 10^6 m/s\).

Step by step solution

01

Identify the known variables

The initial speed of the proton, \(u_p=1.50 \times 10^{7} m/s\); the speed of the proton after rebounding, \(v_p = 1.20 \times 10^{7} m/s\); and there is no initial speed of the target nucleus, \(u_n =0 m/s\). The proton's mass is given as \(m\). Let's represent the unknown nucleus mass as \(M\), the speed of the unknown nucleus after collision as \(v_n\) and assume that all the motions are along the same line.
02

Apply conservation of momentum principle

Based on the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. This implies that \(m u_p +(M*u_n)= m*v_p + M*v_n\). Substitute \(u_n=0\) and rearrange the equation to solve for \(v_n\), which gives \(v_n= \frac{m(u_p-v_p)}{M}\).
03

Apply conservation of kinetic energy principle

According to the conservation of kinetic energy, the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision. This implies that \(\frac{1}{2} m u_p^2 + 0 = \frac{1}{2} m v_p^2+ \frac{1}{2} M v_n^2\). Substitute the value of \(v_n\) from step 2 and expand and rearrange the equation, which gives \(M= m\frac{u_p^2-v_p^2}{u_p^2-v_p^2}\), or \(M=m\).
04

Solve for the speed of the unknown nucleus

Substitute \(M=m\) into the equation from step 2 and solve for \(v_n\), which gives \(v_n=u_p-v_p\).
05

Calculate the values

Substitute the given values into the equations found for \(M\) and \(v_n\) to calculate the numerical values.

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Key Concepts

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

Conservation of Momentum
In physics, momentum is a key concept that helps us understand how objects move and interact. Momentum is defined as the product of an object's mass and velocity and is a vector quantity, meaning it has both magnitude and direction. The principle of conservation of momentum is one of the fundamental laws of physics, stating that the total momentum of a closed system remains constant before and after a collision, provided no external forces act on it. In simpler terms, what goes in must come out.

During the exercise, we see this principle in action as a proton collides with an unknown nucleus in an elastic collision. Prior to the collision, only the proton has significant speed since the nucleus is stationary. The mathematical expression for this conservation in the problem is:
  • Before collision: Total momentum is given by the momentum of the proton, which is \( m u_p \).
  • After collision: Both the proton and the nucleus move, so total momentum is \( m v_p + M v_n \).
The equation balances because the initial momentum of the system equals its final momentum. By rearranging this equation, we can solve for unknown variables such as the speed of the nucleus after the collision.
Conservation of Kinetic Energy
Closely related to momentum, kinetic energy is another crucial concept studied during particle collisions. Kinetic energy is the energy an object Kinetic energy is the energy an object possesses due to its motion, given by the equation \( KE = \frac{1}{2} mv^2 \). For an elastic collision, the kinetic energy of the entire system is conserved.

In our exercise, we are dealing with an elastic collision between a proton and an unknown nucleus. Before the collision, all of the system's kinetic energy is within the moving proton because the nucleus is stationary. After the collision, both particles have some kinetic energy. The principle of conservation of kinetic energy for the collision process is expressed as:
  • Initial kinetic energy: \( \frac{1}{2} m u_p^2 \)
  • Final kinetic energy: \( \frac{1}{2} m v_p^2 + \frac{1}{2} M v_n^2 \)
By equating the initial and final kinetic energies, and using known variables, we can deduce properties like the mass of the unknown nucleus. In this instance, the calculations show that the mass of the unknown nucleus is equal to the mass of the proton, demonstrating how these principles guide us to quantitative conclusions.
Elastic Collision
Elastic collisions are a special type of collision where both momentum and kinetic energy are conserved. This makes them particularly useful when solving problems in physics because they provide two mathematical relationships we can use to find unknown quantities. Elastic collisions are often perfectly reversible and occur without the generation of heat or sound.

In the given problem, the proton and the unknown nucleus undergo such an elastic collision. After rebounding, the proton’s speed is reduced, transferring some of its kinetic energy to the nucleus. Both the conservation laws applicable to elastic collisions help us solve for unknowns, such as the mass of the nucleus \( M \) and its speed \( v_n \) after collision. Analyzing elastic collisions can thus reveal detailed information about systems in motion, leveraging both momentum and energy conservation to uncover hidden properties of objects involved.

By understanding elastic collisions, students can appreciate the predictable nature of motion at a fundamental level and accurately predict outcomes in various physical systems.

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Most popular questions from this chapter

Object \(B\) is at rest when object \(A\) collides with it. The collision is one- dimensional and elastic. After the collision object \(B\) has half the velocity that object \(A\) had before the collision. (a) Which object has the greater mass? (b) How much greater? (c) If the velocity of object \(A\) before the collision was \(6.0 \mathrm{~m} / \mathrm{s}\) to the right. what is its velocity after the collision?

A uniform cube with mass \(0.500 \mathrm{~kg}\) and volume \(0.0270 \mathrm{~m}^{3}\) is sitting on the floor. A uniform sphere with radius \(0.400 \mathrm{~m}\) and mass \(0.800 \mathrm{~kg}\) sits on top of the cubc. How far is the center of mass of the two-object system above the floor?

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Two carts of cqual mass are on a horizontal, frictionless air track. Initially cart \(A\) is moving toward stationary cart \(B\) with a speed of \(v_{A}\). The carts undergo an inelastic collision, and after the collision the total kinetic energy of the two carts is one-half their initial total kinctic energy before the collision. What is the speed of cach cart after the collision?

Animal Propulsion. Squids and octopuses propel themsclves by expelling water, They do this by kecping water in a cavity and then suddenly contracting the cavity to force out the water through an opening. A \(6.50 \mathrm{~kg}\) squid (including the water in the cavity) at rest suddenly sees a dangerous predator. (a) If the squid has \(1.75 \mathrm{~kg}\) of watcr in its cavity, at what speed must it expel this water suddenly to achieve a speed of \(2.50 \mathrm{~m} / \mathrm{s}\) to cscape the predator? Ignore any drag cffects of the surrounding water. (b) How much kinetic energy does the squid create by this mancuver?

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