Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 1: Problem 102

Kinetic energy \(\mathrm{K}\) and linear momentum \(\mathrm{P}\) are related as \(\mathrm{K}=\left(\mathrm{P}^{2} / 2 \mathrm{~m}\right) .\) What is the equation of the relative error \(\Delta \mathrm{k} / \mathrm{k}\) in measurement of the \(\mathrm{K} ?\) (mass in constant) (a) \((\mathrm{P} / \Delta \mathrm{P})\) (b) \(2(\Delta \mathrm{P} / \mathrm{P})\) (c) \((\mathrm{P} / 2 \Delta \mathrm{P})\) (d) \(4(\Delta \mathrm{P} / \mathrm{P})\)

Short Answer

Expert verified
The equation for the relative error in the measurement of K is \(2(\Delta P / P)\), which corresponds to option (b).

Step by step solution

01

We are given the equation: \(K = \frac{P^2}{2m}\) #Step 2: Differentiate K with respect to P#

To differentiate K with respect to P, we can use the power rule: \(\frac{d}{dx} x^n = nx^{n-1}\) \(\frac{dK}{dP} = \frac{d}{dP} \frac{P^2}{2m} = \frac{2P}{2m} = \frac{P}{m}\) #Step 3: Find the equation for ΔK#
02

Now we need to find the equation for ΔK, the change in the kinetic energy. At P = P₀, K = K₀. If P changes by ΔP, K changes by ΔK. Knowing that, we can write: \(\Delta K = \frac{dK}{dP} \Delta P = \frac{P}{m} \Delta P\) #Step 4: Divide ΔK by K to find the relative error#

The relative error is given by \(\frac{\Delta K}{K}\). We can now plug in the equations we derived in the previous steps: \(\frac{\Delta K}{K} = \frac{\frac{P}{m}\Delta P}{\frac{P^2}{2m}} = \frac{2 \Delta P}{P}\) Answer: The equation for the relative error in the measurement of K is \(\boxed{2(\Delta P / P)}\), which corresponds to option (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In the experiment of simple pendulum error in length of pendulum \((\ell)\) is \(5 \%\) and that of \(g\) is \(3 \%\) then find percentage error in measurement of periodic time for pendulum (a) \(4.2 \%\) (b) \(1.2 \%\) (c) \(2 \%\) (d) \(4 \%\)

Equation of force \(F=a t+b t^{2}\) where \(F\) is force in Newton \(t\) is time in second, then write unit of \(\mathrm{b}\). (a) \(\mathrm{Ns}^{-1}\) (b) \(\mathrm{Ns}^{2}\) (c) \(\mathrm{Ns}\) (d) \(\mathrm{Ns}^{-2}\)

Equation of \(\ell_{1}=\ell_{0}\left[1+\alpha\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)\right]\) find out the dimensions of the coefficient of linear expansion \(\alpha\) suffix. (a) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1} \mathrm{~K}^{1}\) (b) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{1} \mathrm{~K}^{1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0} \mathrm{~K}^{1}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \mathrm{~K}^{-1}\)

Write the dimensional formula of the ratio of linear momentum to angular momentum. (a) \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{1}\)

\(F=A_{0}\left(1-e^{-(B x t) 2}\right)\) where \(F\) is force and \(x\) is displacement. write the dimension formula of \(B\) (a) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{-2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-1}\)
See all solutions

Recommended explanations on English Textbooks

Linguistic Terms

Read Explanation

Email

Read Explanation

Text Comparison

Read Explanation

Synthesis Essay

Read Explanation

Single Paragraph Essay

Read Explanation

Sociolinguistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free