/*! 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 22 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 momenturn 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 \(K_{\mathrm{A}}\) 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) \( \frac{K_A}{K_B} = \frac{m_B}{m_A} \).

Step by step solution

01

Apply Conservation of Momentum

According to the conservation of momentum, the total momentum before and after the fission remains constant. Initially, the nucleus is at rest, so its total momentum is zero. After the fission, the momentum of piece A plus the momentum of piece B equals zero:\[ m_A v_A - m_B v_B = 0 \]Rearrange the equation to solve for \( v_B \):\[ v_B = \frac{m_A}{m_B} v_A \]
02

Derive Kinetic Energy Ratio Formula

The kinetic energies of pieces A and B are given by:\[ K_A = \frac{1}{2} m_A v_A^2 \] and \[ K_B = \frac{1}{2} m_B v_B^2 \]Substitute the expression for \( v_B \) from step 1 into the equation for \( K_B \):\[ K_B = \frac{1}{2} m_B \left( \frac{m_A}{m_B} v_A \right)^2 = \frac{1}{2} \frac{m_A^2}{m_B} v_A^2 \]Now find the ratio \( \frac{K_A}{K_B} \):\[ \frac{K_A}{K_B} = \frac{\frac{1}{2} m_A v_A^2}{\frac{1}{2} \frac{m_A^2}{m_B} v_A^2} = \frac{m_B}{m_A} \]
03

Conclude the Energy Ratio

After simplifying, we observe that the ratio of kinetic energies \( \frac{K_A}{K_B} \) is equivalent to \( \frac{m_B}{m_A} \), as required. This is consistent with the relationship derived from the conservation of momentum. The kinetic energy behaves inversely proportional to the mass when the velocity is expressed in terms of mass ratios.

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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 form of energy that an object possesses due to its motion. It depends on both the mass and the velocity of the object. We define kinetic energy with the formula \( K = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the speed of the object.
In the context of the fission of an atomic nucleus, kinetic energy becomes highly significant. When the nucleus splits, it releases pieces that move rapidly away from each other. Each piece has its own kinetic energy.
Using the conservation of momentum, we find the relationship between the velocities of the two pieces. From this, we can derive the relationship between their kinetic energies. For the nucleus breaking into two pieces, piece A and piece B, the exercise demonstrates that the ratio of their kinetic energies \( \frac{K_A}{K_B} \) is simply the inverse of their mass ratio \( \frac{m_B}{m_A} \). This shows how kinetic energy distributes when masses and velocities are related through momentum conservation.
Nuclear Fission
Nuclear fission is a reaction in which a heavy atomic nucleus splits into two or more smaller nuclei, along with the release of energy. This process is often spontaneous in nature for some heavy nuclei but can also be induced in reactors or weapons.
Nuclear fission is fundamental in nuclear physics and power generation. When a nucleus fissions, it releases a significant amount of energy, primarily in the form of kinetic energy of the fission fragments. This is because the mass of the initial nucleus is slightly larger than the sum of the masses of the products. This mass difference is converted into energy, as per Einstein's mass-energy equivalence principle \( E=mc^2 \).
In our example, when the nucleus disassembles into pieces A and B, conservation laws like momentum and energy help us predict the resulting speeds and how the energy disperses as kinetic energy.
  • The process is initiated by a neutron colliding with the nucleus.
  • The nucleus becomes unstable and splits into smaller nuclei.
  • Energy released can be used in energy applications or weapons.
Understanding these processes is key to harnessing nuclear energy efficiently and safely.
Physics Problem Solving
Problem solving in physics involves using fundamental principles to understand and predict physical situations. Here, we applied the conservation laws to solve a practical problem in nuclear physics.
To tackle such problems, you should follow certain steps:
  • Identify the physical principles at work: In this case, it was the conservation of momentum.
  • Set up equations: This involved equating the momentum before and after fission to find relationships between velocities.
  • Compute unknowns: Solve algebraically for unknowns such as the velocity of one of the pieces.
  • Apply results to find other quantities: Use known expressions to find the kinetic energy and its ratios.
Solving physics problems requires a structured approach, focusing on mathematical equations derived from laws of physics. This allows for understanding deep-seated principles governing the physical universe. By breaking down complex processes into manageable steps using math, we demystify how physical systems operate, just as with our fission example.

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Most popular questions from this chapter

In a shipping company distribution center, an open cart of mass 50.0 \(\mathrm{kg}\) is rolling to the left at a speed of 5.00 \(\mathrm{m} / \mathrm{s}(\text { Fig } .8 .46)\) . You can ignore friction between the cart and the floor. A 15.0 \(\mathrm{kg}\) package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . The package lands in the cart and they roll off together. If the lower end of the chute is a vertical distance of 4.00 \(\mathrm{m}\) above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

A \(0.150-\mathrm{kg}\) frame, when suspended from a coil spring, stretches the spring 0.050 \(\mathrm{m}\) . A \(0.200-\mathrm{kg}\) lump of putty is dropped from rest onto the frame from a height of 30.0 \(\mathrm{cm}\) (Fig. 8.42\()\) . Find the maximum distance the frame moves downward from its initial position.

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 80.0 \(\mathrm{kg}\) , is given a push and slides eastward. Abigail, with mass 50.0 \(\mathrm{kg}\) , is sent sliding northward. They collide, and after the collision Sam is moving at \(37.0^{\circ}\) north of east with a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) and Abigail is moving ar \(23.0^{\circ}\) south of east with a speed of 9.00 \(\mathrm{m} / \mathrm{s}\) . (a) What was the speed of each person before the collision? (b) By how much did the total kinetic energon of the two people decrease during the collision?

In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t}$$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k v\) . This gives, since \(F_{\text { ext }}=m g .\) $$m g=m \frac{d v}{d t}+v(k v)$$ Or, dividing by \(k\) $$x g=x \frac{d v}{d t}+v^{2}$$ This is a differential equation that has a solution of the form \(v=a t,\) where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v,\) find the acceleration \(a\) . (b) Find the distance the raindrop has fallen in \(t=3.00 \mathrm{s}\) . (c) Given that \(k=2.00 \mathrm{g} / \mathrm{m}\) , find the mass of the raindrop at \(t=3.00 \mathrm{s}\) . For many more intriguing aspects of this problem, see \(\mathbf{K} .\) S. Krane, Amer Jour. Phys, Vol. 49\((1981)\) pp. \(113-117\)

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