/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 An atomic nucleus suddenly burst... [FREE SOLUTION] | 91Ó°ÊÓ

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An atomic nucleus suddenly bursts apart (fissions) into two pieces. Piece \(A,\) of mass \(m_{A},\) travels off to the left with speed \(v_{A} .\) Piece \(B,\) of mass \(m_{B},\) travels off to the right with speed \(v_{B}\). (a) Use conservation of momentum to solve for \(v_{B}\) in terms of \(m_{A}\) , \(m_{B},\) and \(v_{A}\) . (b) Use the results of part (a) to show that \(K_{A} / K_{B}=m_{B} / m_{A},\) where \(K_{A}\) and \(K_{B}\) are the kinetic energies of the two pieces.

Short Answer

Expert verified
(a) \(v_B = -\frac{m_A}{m_B}v_A\), (b) \(K_A / K_B = m_B / m_A\).

Step by step solution

01

Understand Conservation of Momentum

Momentum is conserved during the fission of the nucleus, meaning that the total momentum before and after the fission is the same. Initially, the system is at rest, so the total initial momentum is zero. After fission, the sum of the momenta of pieces \(A\) and \(B\) must also equal zero.
02

Set Up Momentum Conservation Equation

Write the conservation of momentum equation as \(m_A v_A + m_B v_B = 0\). This equation follows from the principle discussed in Step 1, where piece A's momentum is to the left and piece B's momentum is to the right.
03

Solve for Speed \(v_B\)

Rearrange the momentum equation to solve for \(v_B\):\[ v_B = -\frac{m_A}{m_B}v_A \]This formula expresses \(v_B\) in terms of \(m_A, m_B,\) and \(v_A\). The negative sign indicates that \(v_B\) is in the opposite direction to \(v_A\).
04

Understand Kinetic Energy Ratio

We need to show that the ratio of the kinetic energies \(K_A / K_B\) equals \(m_B / m_A\). Kinetic energy \(K\) is given by \(\frac{1}{2}mv^2\). So, find expressions for \(K_A\) and \(K_B\) in terms of mass and velocity.
05

Calculate Kinetic Energies

Use \(K_A = \frac{1}{2}m_Av_A^2\) and \(K_B = \frac{1}{2}m_B(-\frac{m_A}{m_B}v_A)^2\). Simplifying \(K_B\) gives:\[K_B = \frac{1}{2}m_B\left(\frac{m_A}{m_B}v_A\right)^2 = \frac{1}{2}\frac{m_A^2}{m_B}v_A^2\]
06

Simplify the Kinetic Energy Ratio

Find the ratio \(\frac{K_A}{K_B}\):\[\frac{K_A}{K_B} = \frac{\frac{1}{2}m_Av_A^2}{\frac{1}{2}\frac{m_A^2}{m_B}v_A^2} = \frac{m_A}{\frac{m_A^2}{m_B}} = \frac{m_B}{m_A}\]This shows that the ratio of kinetic energies \(K_A / K_B = m_B / m_A\), proving the required result.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a measure of the energy an object possesses due to its motion. When an object moves, it has a kinetic energy calculated using the expression \( K = \frac{1}{2}mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.
In the context of atomic nucleus fission, each fragment of the nucleus gains kinetic energy as it departs in opposite directions.
The kinetic energy depends not only on the velocity of each fragment but also on their mass.
  • For Piece A, its kinetic energy \( K_A \) can be expressed as \( K_A = \frac{1}{2}m_Av_A^2 \).
  • For Piece B, the kinetic energy \( K_B \) follows a similar formula but takes into account the derived velocity \( v_B \).
Understanding the relationship between the kinetic energies and masses of the fragments allows you to derive important results, such as the ratio \( K_A / K_B \), which demonstrates the inverse proportionality between kinetic energy and mass when momentum is conserved.
Atomic Nucleus Fission
Atomic nucleus fission is a process where a large atomic nucleus splits into smaller fragments, often releasing a significant amount of energy.
This process is at the heart of nuclear energy and atomic bombs, as it releases energy due to the strong forces within the nucleus.
  • During fission, the original nucleus, which was initially at rest, splits and the fragments move away in opposite directions.
  • Each fragment carries away a portion of the energy in the form of kinetic energy.
In nuclear fission, the principle of conservation of momentum ensures that as the nucleus disintegrates, the total momentum of all fragments remains equal to the initial state, which is zero if the original nucleus was stationary.
This is a key point of the exercise, where even the infinitely small rearrangement of tiny particles inside an atomic nucleus adheres strictly to the laws of physics.
Momentum Equation
The momentum equation is crucial in understanding physical interactions, especially in nuclear fission events.
Momentum is defined as the product of mass and velocity, \( p = mv \), and for a system at rest initially, like an atom before fission, the total initial momentum is zero.
  • Upon fission, fragments move in opposite directions, ensuring that the sum of their momenta remains zero.
  • The momentum equation after fissioned can be represented as \( m_A v_A + m_B v_B = 0 \).
Rearranging this equation allows us to find the velocity of one fragment in terms of the others: \( v_B = -\frac{m_A}{m_B}v_A \).
This result tells us how the momentum of one piece relates to the other, maintaining the balance dictated by the conservation principle.
Understanding this concept and its application provides a powerful tool for solving problems in physics that involve collisions and separations.

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Most popular questions from this chapter

A \(20.00-\) -kg lead sphere is hanging from a hook by a thin wire 3.50 \(\mathrm{m}\) long and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00 -kg steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

To warm up for a match, a tennis player hits the \(57.0-\) g ball vertically with her racket. If the ball is stationary just before it is hit and goes 5.50 \(\mathrm{m}\) high, what impulse did she impart to it?

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Just before it is struck by a racket, a tennis ball weighing 0.560 \(\mathrm{N}\) has a velocity of \((20.0 \mathrm{m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{m} / \mathrm{s}) \hat{\boldsymbol{J}}\) . During the 3.00 \(\mathrm{ms}\) that the racket and ball are in contact, the net force on the ball is constant and equal to \(-(380 \mathrm{N}) \hat{\boldsymbol{\imath}}+(110 \mathrm{N}) \hat{\boldsymbol{J}}\) . (a) What are the \(x\) - and \(y\) -components of the impulse of the net force applied to the ball? (b) What are the \(x\) - and \(y\) -components of the final velocity of the ball?

A bat strikes a 0.145 -kg baseball. Just before impact, the ball is traveling horizontally to the right at \(50.0 \mathrm{m} / \mathrm{s},\) and it leaves the bat traveling to the left at an angle of \(30^{\circ}\) above horizontal with a speed of 65.0 \(\mathrm{m} / \mathrm{s}\) . If the ball and bat are in contact for 1.75 \(\mathrm{ms}\) find the horizontal and vertical components of the average force on the ball.

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