/*! 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 88 A Superball is dropped from a he... [FREE SOLUTION] | 91Ó°ÊÓ

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A Superball is dropped from a height of \(2.5 \mathrm{~m}\) and rebounds off the floor to a height of \(2.1 \mathrm{~m}\). If the ball is in contact with the floor for \(0.70 \mathrm{~ms}\), determine (a) the direction and (b) magnitude of the ball's average acceleration due to the floor.

Short Answer

Expert verified
(a) Upward, (b) 19171.43 m/s².

Step by step solution

01

Calculate Initial Velocity Before Impact

To determine the velocity of the ball just before it impacts the floor, use the kinematic equation: \[ v^2 = u^2 + 2as \]where \(u = 0\) (initial velocity), \(a = 9.8\, \text{m/s}^2\) (acceleration due to gravity), and \(s = 2.5\, \text{m}\) (height).By solving, we get:\[ v = \sqrt{2 \times 9.8 \times 2.5} = 7\, \text{m/s}\] Thus, the velocity just before impact is \(7\, \text{m/s}\) downward.
02

Calculate Final Velocity After Rebound

To find the final velocity as the ball rebounds upward, again use the kinematic equation:\[ v^2 = u^2 + 2as \]where \(u = 0\), \(a = -9.8\, \text{m/s}^2\) (since gravity opposes the upward motion), and \(s = 2.1\, \text{m}\).Solving gives:\[ v = \sqrt{2 \times 9.8 \times 2.1} = 6.42\, \text{m/s}\]The velocity immediately after rebound is \(6.42\, \text{m/s}\) upward.
03

Determine Change in Velocity

Calculate the change in velocity by noting the direction of velocities before and after impact. The initial velocity is \(-7\, \text{m/s}\) (downward) and the final velocity is \(6.42\, \text{m/s}\) (upward).Change in velocity: \[ \Delta v = 6.42 - (-7) = 13.42\, \text{m/s} \]
04

Calculate Average Acceleration

Use the formula for average acceleration:\[ a_{\text{avg}} = \frac{\Delta v}{\Delta t} \]where \(\Delta v = 13.42\, \text{m/s}\) and \(\Delta t = 0.70 \times 10^{-3}\, \text{s}\).By solving, we get:\[ a_{\text{avg}} = \frac{13.42}{0.70 \times 10^{-3}} = 19171.43\, \text{m/s}^2\]
05

Determine Direction of Average Acceleration

The direction of the acceleration is the same as the direction of the change in velocity. Since the velocity changes from downward to upward, the average acceleration is directed upward.

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Key Concepts

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

Kinematic Equations
Kinematic equations are vital tools in physics that help us describe motion. By applying these equations, we can solve problems involving velocity, acceleration, displacement, and time. In our example, we use the kinematic equation \[ v^2 = u^2 + 2as \]where:
  • \( v \) is the final velocity,
  • \( u \) is the initial velocity,
  • \( a \) is the acceleration,
  • \( s \) is the displacement.
To calculate the velocity of the ball as it hits and leaves the ground, we set the displacement and initial velocities and compute the resulting final velocities. These calculations help us understand how fast the ball is moving at different stages. Knowing these speeds helps set the stage for solving further elements of the physics problem, like determining acceleration.
Average Acceleration
Average acceleration is a key concept that shows how quickly the velocity of an object changes over a period. It can be calculated from the equation:\[ a_{\text{avg}} = \frac{\Delta v}{\Delta t} \]where:
  • \( \Delta v \) is the change in velocity,
  • \( \Delta t \) is the time over which the change happens.
In this problem, the ball's average acceleration during contact with the floor reflects how its velocity shifts from a downward direction to an upward one. The incredibly high value of average acceleration, over 19,171 \( \text{m/s}^2 \), demonstrates the significant force exerted by the floor to reverse the ball's motion in such a brief time period of 0.70 ms. Understanding average acceleration offers insights into how fast an object's speed can change under influence of an external force, like a bounce.
Change in Velocity
Change in velocity is the measure of how an object's speed and direction vary from one point in time to another. We calculate it by subtracting the initial velocity from the final velocity, represented as:\[ \Delta v = v_{\text{final}} - v_{\text{initial}} \]In the problem at hand, the ball's velocity before hitting the floor is downward at \(-7\, \text{m/s}\), and after rebounding, it is upwards at \(6.42\, \text{m/s}\). The change in velocity \(\Delta v\) is thus negative to positive, resulting in \(13.42\, \text{m/s}\). This significant shift is necessary for assessing how much the velocity has altered due to impact with the floor. Understanding the change helps us link how initial motion conditions transform under different scenarios like impacts, enables us to further evaluate the forces involved.
Impact and Rebound Dynamics
Impact and rebound dynamics explore how objects interact with surfaces during collisions and the consequent motion changes. When the ball falls and hits the floor, an impact occurs that instantly changes its speed and direction. During this brief contact, the floor exerts a force that results in an acceleration large enough to deflect the ball upward. These dynamics are influenced by factors such as the material of the ball, the rigidness of the impacted surface, and existing mechanical energy states. In our exercise, one can see how nearly all kinetic energy is converted into potential energy as the ball begins its upward trajectory. Rebound height being slightly less than the initial drop height suggests energy loss, mainly due to air resistance and internal friction in the ball itself. By mastering rebound dynamics, predictions can be made about post-collision velocity and force events. Insights into these dynamics underpin numerous real-world applications, from sports to material science.

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Most popular questions from this chapter

An object initially at rest experiences an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) on a level surface. Under these conditions, it travels \(6.00 \mathrm{~m}\). Let's designate the first \(3.00 \mathrm{~m}\) as phase 1 with a subscript of 1 for those quantities, and the second \(3.00 \mathrm{~m}\) as phase 2 with a subscript of \(2 .\) (a) The times for traveling each phase should be related by which condition: (3) \(t_{1}>t_{2} ?\) (b) Now calculate the (1) \(t_{1}

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