/*! 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 26 As indicated by the derivation, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

As indicated by the derivation, the principle of impulse and momentum is valid for observers in any inertial reference frame. Show that this is so, by considering the 10 -kg block which slides along the smooth surface and is subjected to a horizontal force of \(6 \mathrm{N}\). If observer \(A\) is in a fixed frame \(x\), determine the final speed of the block in 4 s if it has an initial speed of \(5 \mathrm{m} / \mathrm{s}\) measured from the fixed frame. Compare the result with that obtained by an observer \(B\), attached to the \(x^{\prime}\) axis that moves at a constant velocity of \(2 \mathrm{m} / \mathrm{s}\) relative to \(A\).

Short Answer

Expert verified
Observer A observes the block moving at a speed of 7.4 m/s, while Observer B observes it to be moving at a speed of 5.4 m/s.

Step by step solution

01

Calculate final speed for Observer A

The final speed of the block as observed by \(A\) can be calculated using the equation of motion \(v = u + a*t\), where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration and \(t\) is time. The acceleration \(a\) is given by \(a = F/m\), where \(F\) is the force acting on the object and \(m\) is its mass. In this case, \(a = 6N / 10kg = 0.6 m/s^2\). So, \(v = 5 m/s + 0.6 m/s^2 * 4s = 7.4 m/s\).
02

Calculate relative motion for Observer B

For observer \(B\), the velocity of the block is the velocity observed by \(A\) minus the velocity of \(B\) relative to \(A\). Therefore, the final speed of the block according to Observer B can be calculated as \(v' = v - v_B = 7.4 m/s - 2 m/s = 5.4 m/s\).
03

Compare the results

Comparing the results, we see that the observed final speed of the block differs in the two reference frames, which confirms that the principle of impulse and momentum holds for any inertial reference frame. For observer A, the speed is 7.4 m/s, which is higher than the 5.4 m/s observed by observer B, due to B moving along with the block.

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.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Inertial Reference Frame
An inertial reference frame is a frame of reference in which Newton's laws of motion are valid. This means that an object either remains at rest or continues to move at a constant velocity, except when acted upon by a force. For example, a stationary observer or one moving with constant velocity can be considered to be in an inertial reference frame.
In the original exercise, Observer A is in a fixed frame, which is an ideal example of an inertial reference frame. Because there are no accelerating forces acting on the reference frame itself, the principles of impulse and momentum can be directly applied to determine the motion of the block as seen by Observer A.
It's crucial to note that inertial frames are relatively moving with respect to each other and not accelerating. In our exercise, Observer B's frame of reference moves with a constant velocity, hence it is still considered an inertial frame.
Relative Motion
Relative motion is a straightforward but important concept. It explains how the movement of one object is perceived differently from different reference points or frames. In essence, motion can appear different depending on where you are looking from.
In the context of our exercise, Observer B experiences relative motion. While the block accelerates due to an external force, Observer B is moving with a constant velocity. Because Observer B's frame is moving relative to Observer A, the block's speed seems to be different when viewed from B's perspective. This effect is calculated by subtracting the velocity of the moving frame (which is 2 m/s in this case) from the velocity observed in the fixed frame.
Understanding relative motion helps clarify why the same event or motion can give different results when viewed from different points of references. It underscores the importance of clearly specifying the frame of reference when discussing motion.
Equation of Motion
The equation of motion is fundamental in analyzing the movement of objects. The basic equation used here is: \[ v = u + at \] where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time period.
This formula results from integrating the acceleration over time and is necessary to find how fast something is moving after a period of acceleration. In our exercise, acceleration is derived from the force acting (using \( a = F/m \)), and we plug this into the equation along with given initial values to solve for the final speed of the block.
The equation of motion is powerful in determining future velocities in a precise manner and is one of the building blocks in classical mechanics. It shows how velocity changes in response to acceleration over a given span of time, showing its invaluable role in momentum and impulse calculations.

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

Ball \(A\) has a mass of \(3 \mathrm{kg}\) and is moving with a velocity of \(8 \mathrm{m} / \mathrm{s}\) when it makes a direct collision with ball \(B,\) which has a mass of \(2 \mathrm{kg}\) and is moving with a velocity of \(4 \mathrm{m} / \mathrm{s} .\) If \(e=0.7,\) determine the velocity of each ball just after the collision. Neglect the size of the balls.

An earth satellite of mass \(700 \mathrm{kg}\) is launched into a free-flight trajectory about the earth with an initial speed of \(v_{A}=10 \mathrm{km} / \mathrm{s}\) when the distance from the center of the earth is \(r_{A}=15 \mathrm{Mm}\). If the launch angle at this position is \(\phi_{A}=70^{\circ},\) determine the speed \(v_{B}\) of the satellite and its closest distance \(r_{B}\) from the center of the earth. The earth has a mass \(M_{e}=5.976\left(10^{24}\right)\) kg. Hint: Under these conditions, the satellite is subjected only to the earth's gravitational force, \(F=G M_{e} m_{s} / r^{2},\) Eq. \(13-1 .\) For part of the solution, use the conservation of energy.

Water is discharged from a nozzle with a velocity of \(12 \mathrm{m} / \mathrm{s}\) and strikes the blade mounted on the 20 -kg cart. Determine the tension developed in the cord, needed to hold the cart stationary, and the normal reaction of the wheels on the cart. The nozzle has a diameter of \(50 \mathrm{mm}\) and the density of water is \(\rho_{w}=1000 \mathrm{kg} / \mathrm{m}^{3}\)

The jet is traveling at a speed of \(720 \mathrm{km} / \mathrm{h}\). If the fuel is being spent at \(0.8 \mathrm{kg} / \mathrm{s}\), and the engine takes in air at \(200 \mathrm{kg} / \mathrm{s},\) whereas the exhaust gas (air and fuel) has a relative speed of \(12000 \mathrm{m} / \mathrm{s}\), determine the acceleration of the plane at this instant. The drag resistance of the air is \(F_{D}=\left(55 v^{2}\right),\) where the speed is measured in \(\mathrm{m} / \mathrm{s}\). The jet has a mass of \(7 \mathrm{Mg}\)

A \(4-16\) ball \(B\) is traveling around in a circle of radius \(r_{1}=3 \mathrm{ft}\) with a speed \(\left(v_{B}\right)_{1}=6 \mathrm{ft} / \mathrm{s}\). If the attached cord is pulled down through the hole with a constant speed \(v_{r}=2 \mathrm{ft} / \mathrm{s},\) determine how much time is required for the ball to reach a speed of \(12 \mathrm{ft} / \mathrm{s}\). How far \(r_{2}\) is the ball from the hole when this occurs? Neglect friction and the size of the ball.

See all solutions

Recommended explanations on Physics Textbooks

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.