Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 3: Problem 414

An impulsive force of \(100 \mathrm{~N}\) acts on a body for \(1 \mathrm{sec}\) What is the change in its linear momentum ? (A) \(10 \mathrm{~N}-\mathrm{S}\) (B) \(100 \mathrm{~N}-\mathrm{S}\) (C) \(1000 \mathrm{~N}-\mathrm{S}\) (D) \(1 \mathrm{~N}-\mathrm{S}\)

Short Answer

Expert verified
The change in linear momentum is \(100 \mathrm{~N}-\mathrm{S}\).

Step by step solution

01

Recall the impulse-momentum theorem formula

The formula for the impulse-momentum theorem is given as: Impulse = Change in Momentum Impulse = Force × Time Change in Momentum = Force × Time
02

Plug in the given values and calculate the change in momentum

From the problem, the force is 100 N and the time is 1 sec. Plug these values into the formula: Change in Momentum = 100 N × 1 s Change in Momentum = 100 N·s
03

Compare our result with the options

Our calculation gave us a change in momentum of 100 N·s. Compare this result to the options given. The correct answer is: (B) \(100 \mathrm{~N}-\mathrm{S}\)

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

The velocity of a body of mass \(20 \mathrm{~kg}\) decrease from $20 \mathrm{~ms}^{-1}\( to \)5 \mathrm{~ms}^{-1}\( in a distance of \)100 \mathrm{~m}$. Force on the body is (A) \(-27.5 \mathrm{~N}\) (B) \(-47.5 \mathrm{~N}\) (C) \(-37.5 \mathrm{~N}\) (D) \(-67.5 \mathrm{~N}\)

The average force necessary to stop a hammer with 25 NS momentum in $0.04 \mathrm{sec}\( is \)\quad \mathrm{N}$ (A) 625 (B) 125 (C) 50 (D) 25

Formula for true force is (A) \(\mathrm{F}=\mathrm{ma}\) (B) \(\mathrm{F}=[\\{\mathrm{d}(\mathrm{mv})\\} / \mathrm{dt}]\) (C) \(\mathrm{F}=\mathrm{m}(\mathrm{dv} / \mathrm{dt})\) (D) \(F=m\left(d^{2} x / d t^{2}\right)\)

Three Forces \(F_{1}, F_{2}\), and \(F_{3}\) together keep a body in equilibrium. If \(F_{1}=3 \mathrm{~N}\) along the positive \(\mathrm{X}\) - axis, \(\mathrm{F}_{2}=4 \mathrm{~N}\) along the positive Y-axis then the third force \(F_{3}\) is (A) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (B) \(5 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis (C) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(3 / 4)\) with negative \(\mathrm{y}\) -axis (D) \(7 \mathrm{~N}\) -making an angle \(\theta=\tan ^{-1}(4 / 3)\) with negative \(\mathrm{y}\) -axis

A player caught a cricket ball of mass \(150 \mathrm{~g}\) moving at the rate of \(20 \mathrm{~ms}^{-1}\). If the catching process be completed in $0.1 \mathrm{~s}$ the force of the blow exerted by the ball on the hands of player is (A) \(0.3 \mathrm{~N}\) (B) \(30 \mathrm{~N}\) (C) \(300 \mathrm{~N}\) (D) \(3000 \mathrm{~N}\)
See all solutions

Recommended explanations on English Textbooks

Essay Plan

Read Explanation

Linguistic Terms

Read Explanation

Rhetorical Analysis Essay

Read Explanation

Rhetoric

Read Explanation

Pragmatics

Read Explanation

English Grammar Summary

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