/*! 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 47 A ball is dropped from rest from... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A ball is dropped from rest from the top of a \(6.10-\mathrm{m}\) -tall building, falls straight downward, collides inelastically with the ground, and bounces back. The ball loses \(10.0 \%\) of its kinetic energy every time it collides with the ground. How many bounces can the ball make and still reach a windowsill that is \(2.44 \mathrm{m}\) above the ground?

Short Answer

Expert verified
The ball can bounce 11 times before it cannot reach the windowsill.

Step by step solution

01

Calculate Initial Potential Energy

The ball starts at the roof with gravitational potential energy. The formula to calculate potential energy is:\[ PE = mgh \]where \( m \) is mass, \( g \) is the acceleration due to gravity \(9.81 \text{ m/s}^2\), and \( h \) is the height. Since time and speed are not dependent on mass and we won't need to compare energy between masses, we will handle without \( m \). Thus, the initial energy at 6.10 meters is calculated as:\[ E_{initial} = gh = 9.81 \times 6.10 \].
02

Calculate Energy After Each Bounce

When the ball hits the ground, it loses 10% of its energy. Therefore, the energy retained after each bounce can be expressed as:\[ E' = 0.90 imes E \]where \( E' \) is the energy after the bounce. Use this formula iteratively to track the energy after each bounce.
03

Determine Bounce Height

After every bounce, the energy the ball has dictates how high it can reach. To reach a height \( h' \), the relation is given by:\[ E' = mgh' \]Solve for \( h' \):\[ h' = \frac{E'}{mg} \].
04

Calculate Maximum Number of Bounces

Continue to decrease the ball's energy using the equation from Step 2, and find its corresponding bounce height using the equation from Step 3, until \( h' \) becomes less than the windowsill height (2.44 m). Count the number of bounces possible before this occurs.

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.

Potential Energy
When discussing the concept of potential energy, particularly in physics problems involving heights, it's important to visualize energy as a stored capability. For a ball held at the top of a building, potential energy originates from its position above ground. The formula used to calculate potential energy is \( PE = mgh \). Here, \( m \) denotes mass, \( g \) is gravitational acceleration which is a constant \( 9.81 \text{ m/s}^2 \), and \( h \) represents the height from which the ball is released.

Potential energy essentially tells us that the ball has the capability to do work as it falls. This form of energy will convert primarily into kinetic energy as the ball descends, demonstrating how energy changes forms yet remains conserved within a closed system.
Inelastic Collision
An inelastic collision is particularly important in this scenario because it helps us understand how much energy the ball retains after hitting the ground. In a totally inelastic collision, two objects collide and move together as one mass post-collision, but in this case where the ball promptly rebounds, it is partially inelastic.

During such a collision, the ball loses some kinetic energy—10% in this scenario—transferred to the ground or converted to other forms like sound or heat. Hence, the inelastic nature of the collision reduces the ball's energy retention, impacting how high it can bounce back after each impact.
  • Inelastic collisions in everyday terms mean that not all energy is 'saved' for the next movement.
  • The percentage of energy lost in this problem is a crucial variable.
Understanding these collisions helps in calculating how many bounces the ball can make before it no longer reaches a desired height.
Energy Conservation
The principle of energy conservation underpins much of physics, specifically indicating that within a closed system, energy cannot be created or destroyed, only transformed. In this ball problem, energy transitions from potential to kinetic as it falls, and then following the inelastic collision, part of it becomes potential energy again as the ball bounces upward.

Despite these energy transformations, the total energy lessens due to inefficiencies. These are seen here as energy 'losses' during the collision, however, the total energy immediately before and after the collision is still derivations of the initial energy minus the energy dissipated in each bounce.
  • The conservation of energy simplifies understanding how energy behaves as an object falls, hits and rebounds.
  • This principle implies that we should track energy flow step by step through the problem.
Recognizing these transformations helps track and solve problems related to the height the ball can achieve after each bounce.
Gravitational Energy
Gravitational energy is a subset of potential energy that specifically pertains to an object's height above ground. It is the energy possessed by an object due to its position within a gravitational field. The higher the object, the greater its gravitational energy.

In this exercise, as the ball falls from 6.10 meters, its gravitational energy decreases while its kinetic energy increases, reflecting the energy exchange between potential and kinetic forms.br>
  • Gravitational energy is directly proportional to both the height of the ball and the gravitational force.
  • This makes gravitational energy crucial in estimating subsequent bounce heights after energy loss in collisions.
By understanding gravitational energy, we can predict not only how high the ball can ascend but also calculate the number of bounces it will make before falling short of reaching a 2.44-meter windowsill.

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

Starting with an initial speed of \(5.00 \mathrm{m} / \mathrm{s}\) at a height of \(0.300 \mathrm{m}, \mathrm{a} 1.50-\mathrm{kg}\) ball swings downward and strikes a \(4.60-\mathrm{kg}\) ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

The drawing shows a human figure in a sitting position. For purposes of this problem, there are three parts to the figure,and the center of mass of each one is shown in the drawing. These parts are: (1) the torso, neck, and head (total mass \(=41 \mathrm{kg}\) ) with a center of mass located on the \(y\) axis at a point \(0.39 \mathrm{m}\) above the origin, (2) the upper legs (mass \(=17 \mathrm{kg}\) ) with a center of mass located on the \(x\) axis at a point \(0.17 \mathrm{m}\) to the right of the origin, and (3) the lower legs and feet (total mass \(=9.9 \mathrm{kg}\) ) with a center of mass located \(0.43 \mathrm{m}\) to the right of and \(0.26 \mathrm{m}\) below the origin. Find the \(x\) and \(y\) coordinates of the center of mass of the human figure. Note that the mass of the arms and hands (approximately 12\% of the whole-body mass) has been ignored to simplify the drawing.

Two ice skaters have masses \(m_{1}\) and \(m_{2}\) and are initially stationary. Their skates are identical. They push against one another, as in Figure \(7.9,\) and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater 2 . What is the ratio \(m_{1} / m_{2}\) of their masses?

One average force \(\overrightarrow{\mathbf{F}}_{1}\) has a magnitude that is three times as large as that of another average force \(\overrightarrow{\mathbf{F}}_{2} .\) Both forces produce the same impulse. The average force \(\overrightarrow{\mathbf{F}}_{1}\) acts for a time interval of \(3.2 \mathrm{ms}\). For what time interval does the average force \(\overrightarrow{\mathbf{F}}_{2}\) act?

A mine car (mass \(=440 \mathrm{kg}\) ) rolls at a speed of \(0.50 \mathrm{m} / \mathrm{s}\) on a horizontal track, as the drawing shows. A 150-kg chunk of coal has a speed of \(0.80 \mathrm{m} / \mathrm{s}\) when it leaves the chute. Determine the speed of the car-coal system after the coal has come to rest in the car.

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.