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(a) A particle of mass \(m\) moves with momentum \(p .\) Show that the kinetic energy of the particle is \(K=p^{2} / 2 m\) (b) Express the magnitude of the particle's momentum in terms of its kinetic energy and mass.

Short Answer

Expert verified
The kinetic energy of a particle can be represented as \( K = \frac{p^{2}}{2m}\) and the magnitude of the particle's momentum can be expressed as \( p = \sqrt{2mK}\)

Step by step solution

01

Understand and establish the given equations

The first thing to do is to write down the known formulas. The formula for kinetic energy (K) is \( K = \frac{1}{2}m v^{2}\), where \(v\) is the velocity of the particle, and the formula for momentum (p) is \( p = m v \). These concepts are from basic high-school physics and are important to familiarize with.
02

Deriving the given energy equation

In order to prove that \( K = \frac{p^{2}}{2m}\), you can substitute the given momentum equation into the kinetic energy equation. So if \( p = m v \), then \( v = \frac{p}{m}\). Substituting this into the kinetic energy equation gives \( K = \frac{1}{2}m(\frac{p}{m})^{2}\), which leads to \( K = \frac{p^{2}}{2m}\). This proves the required equation.
03

Expressing momentum in terms of kinetic energy and mass

The second part of the problem asks to express the magnitude of the particle's momentum in terms of kinetic energy and mass. To do this, you can rearrange the previously derived equation \( K = \frac{p^{2}}{2m}\) to find \( p \). This leads to \( p = \sqrt{2mK}\). Now, momentum is expressed in terms of kinetic energy and mass.

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Key Concepts

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

Momentum and Kinetic Energy
Understanding the relationship between momentum and kinetic energy is essential in physics. Momentum, denoted by the symbol \( p \), is a measure of the quantity of motion an object has and is calculated by the product of an object's mass \( m \) and its velocity \( v \), given by the equation \( p = mv \).

Kinetic energy, on the other hand, is the energy an object possesses due to its motion. It's given by the formula \( K = \frac{1}{2}mv^2 \). The connection between these two quantities might not be immediately apparent, but by manipulating these equations, one can show that kinetic energy can also be expressed as \( K = \frac{p^2}{2m} \), thus connecting kinetic energy directly to momentum and mass.

This relationship signifies that if an object’s momentum increases, its kinetic energy will also increase as long as the mass remains constant. It's also an excellent example of how different physical concepts are interconnected and why understanding these relationships can solve various physics problems.
Deriving Equations in Physics
Deriving equations is a critical skill in physics, helping to reveal the underlying principles governing physical scenarios. It involves manipulating known formulas to express a particular quantity in terms of others. For instance, when we derive kinetic energy from momentum, we start with the known formulas for momentum \( p = mv \) and kinetic energy \( K = \frac{1}{2}mv^2 \), and through substitution, we derive a new expression for kinetic energy in terms of momentum and mass, \( K = \frac{p^2}{2m} \).

To derive effectively, one must understand the variables involved and the basic algebraic operations, such as substitution and square roots. Following these steps methodically can unveil new ways to understand and solve physics problems, like how one quantity, such as momentum, can define another, like kinetic energy.
Mass-Velocity Relation
In physics, the relationship between an object's mass, its velocity, and resulting quantities like momentum or kinetic energy is foundational. The mass-velocity relation can be seen directly in the formulas for momentum \( p = mv \) and kinetic energy \( K = \frac{1}{2}mv^2 \), indicating that both quantities grow with either increasing mass or velocity.

However, these relationships are not linear. While momentum is directly proportional to both mass and velocity, kinetic energy is proportional to the square of the velocity, indicating that a small increase in velocity will result in a much larger increase in kinetic energy than a corresponding increase in mass. This has important implications not only in academic problems but also in real-world scenarios such as vehicle safety and energy conservation.

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Most popular questions from this chapter

A \(12.0-\mathrm{g}\) wad of sticky clay is hurled horizontally at a \(100-\mathrm{g}\) wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides \(7.50 \mathrm{m}\) before coming to rest. If the coefficient of friction between the block and the surface is \(0.650,\) what was the speed of the clay immediately before impact?

A neutron in a nuclear reactor makes an elastic head-on collision with the nucleus of a carbon atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (b) If the initial kinetic energy of the neutron is \(1.60 \times 10^{-13} \mathrm{J},\) find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The mass of the carbon nucleus is nearly 12.0 times the mass of the neutron.)

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Consider as a system the Sun with the Earth in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system over a period of 6 months. Neglect the influence of other celestial objects. You may obtain the necessary astronomical data from the end papers of the book.

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