/*! 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 124 A \(0.550 \mathrm{~kg}\) ball fa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 124

A \(0.550 \mathrm{~kg}\) ball falls directly down onto concrete, hitting it with a speed of \(12.0 \mathrm{~m} / \mathrm{s}\) and rebounding directly upward with a speed of \(3.00 \mathrm{~m} / \mathrm{s} .\) Extend a \(y\) axis upward. In unit-vector notation, what are (a) the change in the ball's momentum, (b) the impulse on the ball, and (c) the impulse on the concrete?

Short Answer

Expert verified
(a) \(8.25\hat{j} \text{ kg} \cdot \text{m/s}\), (b) \(8.25\hat{j} \text{ kg} \cdot \text{m/s}\), (c) \(-8.25\hat{j} \text{ kg} \cdot \text{m/s}\)

Step by step solution

01

Define the initial momentum

First, calculate the initial momentum when the ball hits the concrete. Since the motion is downwards, the initial velocity is negative in the y-axis direction. Use the formula for momentum: \( p_i = m \cdot v_i \), where \( m = 0.550 \) kg and \( v_i = -12.0 \) m/s. \[p_i = 0.550 \times (-12.0) = -6.60 \text{ kg} \cdot \text{m/s}\]
02

Define the final momentum

Next, calculate the final momentum after the ball rebounds. Since the ball moves upwards, the final velocity is positive in the y-axis direction. The momentum is given by \( p_f = m \cdot v_f \), where \( v_f = 3.00 \) m/s.\[p_f = 0.550 \times 3.00 = 1.65 \text{ kg} \cdot \text{m/s}\]
03

Calculate change in momentum

The change in the ball's momentum is the difference between the final and initial momentum: \( \Delta p = p_f - p_i \).\[\Delta p = 1.65 - (-6.60) = 1.65 + 6.60 = 8.25 \text{ kg} \cdot \text{m/s}\]Therefore, the change in momentum \( \Delta \vec{p} \) in unit vector notation is:\[\Delta \vec{p} = 8.25\hat{j} \text{ kg} \cdot \text{m/s}\]
04

Determine the impulse on the ball

The impulse on an object is equal to the change in momentum of the object. Therefore, the impulse \( \vec{I} \) on the ball is:\[\vec{I} = \Delta \vec{p} = 8.25\hat{j} \text{ kg} \cdot \text{m/s}\]
05

Calculate the impulse on the concrete

By Newton's third law, the impulse exerted by the ball on the concrete is equal in magnitude but opposite in direction to the impulse on the ball. Thus, the impulse \( \vec{I}_c \) on the concrete is:\[\vec{I}_c = -8.25\hat{j} \text{ kg} \cdot \text{m/s}\]

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.

Impulse Explained
Impulse is a concept that helps us understand how the force applied over a period changes an object's motion. Impulse acts as a bridge between force and momentum.
Think about impulse as the kick that changes how fast something is moving and in what direction. The formula for impulse is: \[ \vec{I} = \Delta \vec{p} = \vec{F} \cdot \Delta t \]where:
  • \( \vec{I} \) represents impulse
  • \( \Delta \vec{p} \) is the momentum change
  • \( \vec{F} \) is the average force exerted
  • \( \Delta t \) is the time duration for which the force is applied
When a 0.550 kg ball hits and rebounds with differing velocities, impulse tells us how much its momentum has changed.
It shows us which way the force acted and how strong it was.
Newton's Third Law in Action
Newton's third law is simple yet powerful: For every action, there's an equal and opposite reaction. This applies everywhere, even when a ball hits concrete.
When the ball smashes into the concrete, it exerts a force downward. According to Newton鈥檚 third law, the concrete pushes back with equal force in the opposite direction. This force happens over the time the ball and concrete are in contact, creating impulse.
Simply put, if the impulse on the ball is 8.25 kg鈰卪/s upward, then the impulse on the concrete is the same in magnitude but downward.
  • Action and Reaction: Equal forces, opposite sides
  • Instant interaction: Force acts while items are in contact
  • Impulse: Measures this interaction as a whole
Understanding Momentum Calculation
Momentum combines an object鈥檚 mass and velocity into one quantity. In simple terms, it鈥檚 a measure of motion: how hard it is to stop a moving object or how much punch it delivers upon impact. The formula to find momentum is straightforward:\[ p = m \cdot v\]where:
  • \( p \) is momentum
  • \( m \) is mass
  • \( v \) is velocity
The momentum change \(\Delta p\) is calculated by the difference between final and initial momentum:
\[\Delta p = p_f - p_i \]This concept helps us see the effect of forces like gravity or collisions on moving objects.
Applying it to our ball, the initial downward momentum is -6.60 kg鈰卪/s and the final upward momentum is 1.65 kg鈰卪/s. The difference, 8.25 kg鈰卪/s, shows how much the initial impact changed the ball鈥檚 motion. This isn't just a number; it's a look at the physical push the ball experienced.

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

(a) How far is the center of mass of the Farth-Moon system from the center of Farth? (Appendix C gives the masses of Earth and the Moon and the distance between the two.) (b) What percentage of Earth's radius is that distance?

Particle 1 of mass \(200 \mathrm{~g}\) and speed \(3.00 \mathrm{~m} / \mathrm{s}\) undergoes a onedimensional collision with stationary particle 2 of mass \(400 \mathrm{~g}\). What is the magnitude of the impulse on particle 1 if the collision is (a) elastic and (b) completely inelastic?

Two particles are launched from the origin of the coordinate system at time \(t=0 .\) Particle 1 of mass \(m_{1}=5.00 \mathrm{~g}\) is shot directly along the \(x\) axis on a frictionless floor, with constant speed \(10.0 \mathrm{~m} / \mathrm{s} .\) Particle 2 of mass \(m_{2}=3.00 \mathrm{~g}\) is shot with a velocity of magnitude \(20.0 \mathrm{~m} / \mathrm{s},\) at an upward angle such that it always stays directly above particle \(1 .\) (a) What is the maximum height \(H_{\max }\) reached by the com of the two-particle system? In unit- vector notation, what are the (b) velocity and (c) acceleration of the com when the com reaches \(H_{\max } ?\)

A \(4.0 \mathrm{~kg}\) mess kit sliding on a frictionless surface explodes into two \(2.0 \mathrm{~kg}\) parts: \(3.0 \mathrm{~m} / \mathrm{s},\) due north, and \(5.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) north of cast. What is the original speed of the mess kit?

block 1 (mass \(2,0 \mathrm{~kg}\) ) is moving rightward at \(10 \mathrm{~m} / \mathrm{s}\) and block 2 (mass \(5.0 \mathrm{~kg}\) ) is moving rightward at \(3.0 \mathrm{~m} / \mathrm{s}\) The surface is frictionless, and a spring with a spring constant of \(1120 \mathrm{~N} / \mathrm{m}\) is fixed to block 2 . When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Turning Points in Physics

Read Explanation

Engineering Physics

Read Explanation

Waves Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Physical Quantities And Units

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.