/*! 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 81 If \(E\) is the total energy of ... [FREE SOLUTION] | 91Ó°ÊÓ

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If \(E\) is the total energy of a particle with zero potential energy, show that \(d E / d p=v,\) where \(p\) and \(v\) are the momentum and velocity of the particle, respectively.

Short Answer

Expert verified
The derivative \(\frac{dE}{dp} = v\) shows that the rate of change of energy with respect to momentum is the velocity.

Step by step solution

01

Understanding Energy and Momentum

The total energy, \(E\), of a particle can be expressed as the sum of its kinetic and potential energies. Since the potential energy is zero, \(E\) is just the kinetic energy, given by \(E = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity of the particle.
02

Express Momentum

The momentum, \(p\), of a particle is defined as the product of its mass and velocity, hence \(p = mv\).
03

Differentiate Momentum with Respect to Velocity

Rearrange the expression for momentum to express velocity in terms of momentum: \(v = \frac{p}{m}\).
04

Differentiate Energy with Respect to Momentum

First, express \(E\) in terms of \(p\): substitute \(v = \frac{p}{m}\) into the kinetic energy equation to get \(E = \frac{1}{2}m \left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}\). Then find \(\frac{dE}{dp}\) by differentiating: \(\frac{dE}{dp} = \frac{d}{dp}\left(\frac{p^2}{2m}\right) = \frac{p}{m}\).
05

Relate Back to Velocity

Since \(\frac{dE}{dp} = \frac{p}{m}\), and we previously expressed \(v = \frac{p}{m}\), we conclude that \(\frac{dE}{dp} = v\).

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Key Concepts

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

Momentum
Momentum is an essential concept in understanding particle dynamics. A straightforward way to think about momentum is to consider it as the measure of the object's motion. The formula for momentum is given by \( p = mv \), where \( m \) is the mass of the particle, and \( v \) is its velocity.
This means that momentum depends not just on how fast something is moving but also on how much mass it has. Therefore:
  • If you double the mass but keep the velocity the same, the momentum doubles.
  • If you double the velocity but keep the mass the same, the momentum also doubles.
Momentum is a vector quantity, which means it has both a magnitude (how much) and a direction (where to). This gives it a unique place in physics, specifically in systems where direction matters. In our particular exercise, understanding momentum is crucial as it connects directly to energy calculations.
Kinetic Energy
Kinetic energy is the energy that an object possesses because of its motion. It's an important part of the total energy in dynamics problems and is defined by the formula \( E = \frac{1}{2}mv^2 \).
This tells us that kinetic energy:
  • Increases with the square of velocity. For example, if you double the velocity, the kinetic energy becomes four times larger.
  • Is directly proportional to mass. Doubling the mass will also double the kinetic energy for the same velocity.
Understanding kinetic energy helps us grasp how objects move and interact. It plays a vital role in this exercise, especially given the simplification that potential energy is zero. This highlights that all the particle's energy is kinetic. By looking at kinetic energy in terms of momentum \( \left(E = \frac{p^2}{2m}\right) \), we can better understand how the particle's speed and mass influence its energy.
Differentiation
Differentiation is a mathematical process used extensively in calculus to find how a quantity changes as another quantity changes. It's like zooming in on a curve to find the slope or steepness at any given point.
In our exercise, differentiation plays a crucial role as you need to find the rate of change of energy \( E \) with respect to momentum \( p \). This involves calculating \( \frac{dE}{dp} \). By substituting \( v = \frac{p}{m} \) back into our energy expression, we obtain \( E = \frac{p^2}{2m} \). Differentiating this expression with respect to \( p \) gives us \( \frac{dE}{dp} = \frac{p}{m} \).
This result shows that the change of kinetic energy with momentum equals velocity, which reflects a deeper understanding of the motion involved. By practicing differentiation in dynamics, you learn how particles behave under different conditions.

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Most popular questions from this chapter

A free neutron can decay into a proton, an electron, and a neutrino. Assume the neutrino's mass is zero; the other masses can be found in a Fundamental Constants Table. Determine the total kinetic energy shared among the three particles when a neutron decays at rest.

(II) What is the speed of an electron just before it hits a television screen after being accelerated from rest by the \(28,000 \mathrm{~V}\) of the picture tube?

(II) What magnetic field \(B\) is needed to keep 998-GeV protons revolving in a circle of radius 1.0 \(\mathrm{km}\) (at, say, the Fermilab synchrotron)? Use the relativistic mass. The proton's rest mass is 0.938 \(\mathrm{GeV} / c^{2} . \quad\left(1 \mathrm{GeV}=10^{9} \mathrm{eV} .\right)\) [Hint: In relativity, \(m_{\mathrm{rel}} v^{2} / r=q v B \quad\) is still valid in a magnetic field, where \(m_{\mathrm{rel}}=\gamma m . ]\)

(II) Reference frame \(\mathrm{S}^{\prime}\) moves at speed \(v=0.92 c\) in the \(+x\) direction with respect to reference frame \(\mathrm{S}\). The origins of \(\mathrm{S}\) and \(\mathrm{S}^{\prime}\) overlap at \(t=t^{\prime}=0 .\) An object is stationary in \(\mathrm{S}^{\prime}\) at position \(x^{\prime}=100 \mathrm{~m} .\) What is the position of the object in \(\mathrm{S}\) when the clock in \(\mathrm{S}\) reads \(1.00 \mu \mathrm{s}\) according to the (a) Galilean and (b) Lorentz transformation equations?

(III) A farm boy studying physics believes that he can fit a 12.0 -m long pole into a 10.0 -m long barn if he runs fast enough, carrying the pole. Can he do it? Explain in detail. How does this fit with the idea that when he is running the barn looks even shorter than 10.0 \(\mathrm{m} ?\)

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