/*! 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 36 A \(0.25 \mathrm{~kg}\) puck is ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 36

A \(0.25 \mathrm{~kg}\) puck is initially stationary on an ice surface with negligible friction. At time \(t=0,\) a horizontal force begins to move the puck. The force is given by \(\vec{F}=\left(12.0-3.00 t^{2}\right) \mathrm{i},\) with \(\vec{F}\) in newtons and \(t\) in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between \(t=0.500 \mathrm{~s}\) and \(t=1.25 \mathrm{~s} ?\) (b) What is the change in momentum of the puck between \(t=0\) and the instant at which \(F=0 ?\)

Short Answer

Expert verified
(a) Impulse is 7.171875 Ns. (b) Change in momentum is 16 kg路m/s.

Step by step solution

01

Understanding the Impulse Equation

Impulse is defined as the integral of force over a time interval. Mathematically, it is expressed as \( J = \int F(t) \, dt \), where \( F(t) \) is the force as a function of time.
02

Compute the Impulse (t=0.5s to t=1.25s)

Given \( F(t) = 12.0 - 3.00 t^2 \), the impulse from \( t = 0.5 \) s to \( t = 1.25 \) s is calculated as: \( J = \int_{0.5}^{1.25} (12.0 - 3.00 t^2) \, dt \). First, find the antiderivative: \( J = \left[ 12.0t - \frac{3.00}{3}t^3 \right]_{0.5}^{1.25} \).Calculate: \( J = (12.0 \times 1.25 - 1.00 \times 1.25^3) - (12.0 \times 0.5 - 1.00 \times 0.5^3) \). Evaluate the expressions: \( J = (15 - 1.953125) - (6 - 0.125) \).\( J = 15 - 1.953125 - 6 + 0.125 \) which results in \( J = 7.171875 \) Ns.
03

Determine When Force F(t) becomes Zero

The force given is \( F(t) = 12.0 - 3.00 t^2 \). Set this equal to zero to find when \( F(t) \) becomes zero: \( 12.0 - 3.00 t^2 = 0 \). Solve for \( t \): \( 3.00 t^2 = 12.0 \)\( t^2 = 4 \)\( t = 2 \) s.
04

Calculate Change in Momentum from t=0 to t=2s

Impulse is also equal to the change in momentum, \( \Delta p = J \). To find the change in momentum from \( t = 0 \) to \( t = 2 \) s, calculate: \( \Delta p = \int_{0}^{2} (12.0 - 3.00 t^2) \, dt \). Use a similar process as in Step 2: \( \Delta p = \left[ 12.0t - \frac{3.00}{3}t^3 \right]_{0}^{2} \).Evaluate at bounds: \( \Delta p = (12.0 \times 2 - 1.00 \times 2^3) \) which results in \( \Delta p = 24 - 8 \), giving \( \Delta p = 16 \) kg路m/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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Calculus in Physics
In physics, calculus is a powerful tool that helps to understand and solve real-world problems involving changing quantities. It provides a way to quantify the relationships between physical variables.
For instance, in the problem about the puck, the force applied to it changes over time, which requires calculus to evaluate the overall effect on the puck. By integrating a function that describes the force over time, you determine the impulse on the puck.
  • Integration determines areas under curves, which in this context means the accumulated effect of the force over a specified interval.
  • The force function given as a polynomial, \(F(t) = 12.0 - 3.00 t^2\), can be integrated to calculate quantities like impulse.
Calculus allows precise calculations of these quantities, turning complex motion-related and force-related problems into solvable equations.
Newton's Laws of Motion
Newton's laws are fundamental to understanding motion in physics. Specifically, his second law of motion is relevant to the problem at hand.
  • This law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The relationship is given by \(F = ma\).
  • It implies that when a force is applied, objects will change their velocity over time, depending on their mass.
Applying Newton's second law clarifies how the force applied to the puck results in its acceleration and subsequent change in momentum.
This is the core mechanism behind calculating changes in motion, such as how impulses and forces affect velocities over time.
Force-Time Integral
The force-time integral is a vital concept in physics as it directly relates impulse and momentum. This integral helps you compute the impulse given a time-dependent force.
Impulse, defined as \(J = \int F(t) \, dt\), is the total change in momentum produced by the force over a time interval.
  • For the puck, the impulse was computed over intervals, showing how integrals represent the accumulated effects of force.
  • The integral is evaluated by finding the antiderivative and using bounds, essential for finding precise moments when forces have tangible impacts.
Through understanding this concept, you grasp that integration of force over time explains how gradual or time-varying applications of force lead to changes in motion and momentum.

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

Basilisk lizards can run across the top of a water surface (Fig. \(9-52\) ). With each step, a lizard first slaps its foot against the water and then pushes it down into the water rapidly enough to form an air cavity around the top of the foot. To avoid having to pull the foot back up against water drag in order to complete the step, the lizard withdraws the foot before water can flow into the air cavity. If the lizard is not to sink, the average upward impulse on the lizard during this full action of slap, downward push, and withdrawal must match the downward impulse due to the gravitational force. Suppose the mass of a basilisk lizard is \(90.0 \mathrm{~g}\), the mass of each foot is \(3.00 \mathrm{~g}\), the speed of a foot as it slaps the water is \(1.50 \mathrm{~m} / \mathrm{s},\) and the time for a single step is \(0.600 \mathrm{~s}\). (a) What is the magnitude of the impulse on the lizard during the slap? (Assume this impulse is directly upward.) (b) During the \(0.600 \mathrm{~s}\) duration of a step, what is the downward impulse on the lizard due to the gravitational force? (c) Which action, the slap or the push, provides the primary support for the lizard, or are they approximately equal in their support?

A railroad freight car of mass \(3.18 \times 10^{4} \mathrm{~kg}\) collides with a stationary caboose car. They couple together, and \(27.0 \%\) of the initial kinetic energy is transferred to thermal cnergy, sound, vibrations, and so on. Find the mass of the caboose.

a \(10 \mathrm{~g}\) bullet moving directly upward at \(1000 \mathrm{~m} / \mathrm{s}\) strikes and passes through the center of mass of a \(5.0 \mathrm{~kg}\) block initially at rest. The bullet emerges from the block moving directly upward at \(400 \mathrm{~m} / \mathrm{s} .\) To what maximum height does the block then rise above its initial position?

A \(1000 \mathrm{~kg}\) automobile is at rest at a traffic signal. At the instant the light turns green, the automobile starts to move with a constant acceleration of \(4.0 \mathrm{~m} / \mathrm{s}^{2}\). At the same instant a \(2000 \mathrm{~kg}\) truck, travcling at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s},\) overtakes and passcs the automobile. (a) How far is the com of the automobile-truck system from the traffic light at \(t=3.0 \mathrm{~s} ?\) (b) What is the speed of the com then?

An object is tracked by a radar station and determined to have a position vector given by \(\vec{r}=(3500-160 t) \hat{i}+2700 \mathrm{j}+300 \hat{\mathrm{k}},\) with \(\vec{r}\) in meters and \(t\) in scconds. The radar station's \(x\) axis points cast. its \(y\) axis north, and its \(z\) axis vertically up. If the object is a \(250 \mathrm{~kg}\) meteorological missile, what are (a) its linear momentum, (b) its direction of motion, and (c) the net force on it?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Translational Dynamics

Read Explanation

Turning Points in Physics

Read Explanation

Space Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Fluids

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.