/*! 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 87 A ball having a mass of \(150 \m... [FREE SOLUTION] | 91Ó°ÊÓ

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A ball having a mass of \(150 \mathrm{~g}\) strikes a wall with a speed of \(5.2 \mathrm{~m} / \mathrm{s}\) and rebounds with only \(50 \%\) of its initial kinetic energy. (a) What is the speed of the ball immediately after rebounding? (b) What is the magnitude of the impulse on the wall from the ball? (c) If the ball is in contact with the wall for \(7.6 \mathrm{~ms}\), what is the magnitude of the average force on the ball from the wall during this time interval?

Short Answer

Expert verified
The speed after rebounding is approximately 3.676 m/s, the impulse is 1.309 kg*m/s, and the average force is about 172.23 N.

Step by step solution

01

Convert Units

First, convert the mass of the ball from grams to kilograms, as standard SI units are required for calculations: \[ m = 150\,\text{g} \times \frac{1\,\text{kg}}{1000\,\text{g}} = 0.150\,\text{kg} \]
02

Calculate Initial Kinetic Energy

Calculate the initial kinetic energy of the ball using the formula:\[ KE_i = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.150\,\text{kg} \times (5.2\,\text{m/s})^2 = 2.028\,\text{J} \]
03

Calculate Final Kinetic Energy after Rebound

The ball rebounds with only 50% of its initial kinetic energy, so calculate the final kinetic energy:\[ KE_f = 0.5 \times KE_i = 0.5 \times 2.028\,\text{J} = 1.014\,\text{J} \]
04

Calculate Speed After Rebounding

Use the kinetic energy formula to find the speed of the ball after rebounding:\[ KE_f = \frac{1}{2} m v_f^2 \]\[ 1.014\,\text{J} = \frac{1}{2} \times 0.150\,\text{kg} \times v_f^2 \]Solving for \( v_f \):\[ v_f^2 = \frac{2 \times 1.014}{0.150} \text{m}^2/\text{s}^2 \]\[ v_f = \sqrt{13.52}\,\text{m/s} \approx 3.676\,\text{m/s} \]
05

Calculate Change in Momentum

The change in momentum (impulse) can be calculated given the change in velocity:\[ \Delta p = m(v_f - (-v_i)) = m(v_f + v_i) \]\[ \Delta p = 0.150\,\text{kg} \times (3.676 + 5.2)\,\text{m/s} = 1.3089\,\text{kg m/s} \]
06

Calculate Average Force on the Ball

To find the average force, use the formula:\[ F_{\text{avg}} = \frac{\Delta p}{\Delta t} \]where \( \Delta t = 7.6\,\text{ms} = 7.6 \times 10^{-3}\,\text{s} \):\[ F_{\text{avg}} = \frac{1.3089\,\text{kg m/s}}{7.6 \times 10^{-3}\,\text{s}} \approx 172.23\,\text{N} \]

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fundamental concept that expresses the energy an object has due to its motion. In this problem, we start by calculating the initial kinetic energy of the ball. The formula to calculate kinetic energy is given by:\[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass and \( v \) is the velocity. For our ball:- Mass (\( m \)): 0.150 kg (after converting from grams)- Velocity (\( v \)): 5.2 m/sPlacing these values in the formula, we find that the initial kinetic energy is 2.028 Joules.
This energy tells us how much work the ball can do due to its motion.

After rebounding, the ball retains only \( 50\% \) of its initial kinetic energy. This new kinetic energy (\( KE_f \)) is:\[ KE_f = 0.5 \times KE_i = 1.014 \text{ J} \]The reduced kinetic energy accounts for energy lost in the form of sound and heat.
To find the speed after rebounding, we rearrange the kinetic energy formula, solving for velocity, resulting in \( 3.676 \text{ m/s} \).
This new speed reflects the energy lost during impact.
Average Force
Average force is crucial when determining how strong an interaction is during a short time interval.
It's the force applied consistently over the interaction’s duration. In this example, the focus is on the force during the ball's contact with the wall.We start with the impulse, which can be calculated using:\[ \text{Impulse} = \Delta p = m(v_f - (-v_i)) \]The impulse is the product of mass and change in velocity.- Initial velocity (\( v_i \)): 5.2 m/s (towards the wall, so negative in reversal logic)- Final velocity (\( v_f \)): 3.676 m/sPut into calculations, the impulse is 1.3089 kg m/s.
To discover the average force exerted by the wall on the ball, we use:\[ F_{\text{avg}} = \frac{\Delta p}{\Delta t} \]Where \( \Delta t \) is the contact time, converted to seconds as \( 7.6 \times 10^{-3} \text{ s} \).
The result, \( 172.23 \text{ N} \), shows how intensely the wall pushed back over this period.
Rebounding Ball
Understanding how balls rebound involves both physics and real-world observation.
"Rebounding" in physics often involves a change in direction and speed due to interaction with a surface, like in our case, with a wall.
  • Initially, regarding our ball, it impacts the wall, then rebounds off, with the velocity decreasing due to kinetic energy loss.
  • By describing the initial and final kinetic energies, we can depict the scenario of rebound; the ball left the wall less energetic than when it hit.
  • The rebound speed determined is due to the measured energy after half of it was lost. This conservation of energy principle is central to defining how the world functions on an atomic level.
The ball's journey before and after striking a surface elegantly displays how momentum and energy interact. It encompasses not just reductions in speed but also highlights the role of external forces, like friction or internal deformations, that play critical roles in real-world applications.

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Most popular questions from this chapter

A rocket is moving away from the solar system at a speed of \(6.0 \times 10^{3} \mathrm{~m} / \mathrm{s} .\) It fires its engine, which ejects exhaust with a speed of \(3.0 \times 10^{3} \mathrm{~m} / \mathrm{s}\) relative to the rocket. The mass of the rocket at this time is \(4.0 \times 10^{4} \mathrm{~kg},\) and its acceleration is \(2.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) What is the thrust of the engine? (b) At what rate, in kilograms per second, is exhaust ejected during the firing?

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