/*! 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 59 A dropped rubber ball hits the f... [FREE SOLUTION] | 91Ó°ÊÓ

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A dropped rubber ball hits the floor with a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) and rebounds to a height of \(0.25 \mathrm{~m} .\) What fraction of the initial kinetic energy was lost in the collision?

Short Answer

Expert verified
92.3% of the initial kinetic energy was lost.

Step by step solution

01

Understand Initial Kinetic Energy

The initial kinetic energy of the ball is calculated using the formula: \( KE_i = \frac{1}{2}mv^2 \), where \( m \) is the mass of the ball and \( v = 8.0 \mathrm{~m/s} \) is the speed. Since mass isn't provided, we'll compare ratios.
02

Calculate Initial Kinetic Energy

The initial kinetic energy per unit mass is given by \( KE_i = \frac{1}{2} \times 8^2 = 32 \mathrm{~m^2/s^2} \).
03

Calculate Potential Energy at Rebound Height

The potential energy at the maximum rebound height \( h = 0.25 \mathrm{~m} \) is \( PE_r = mgh \), which per unit mass is \( g \times 0.25 \), where \( g = 9.8 \mathrm{~m/s^2} \).
04

Calculate Kinetic Energy at Rebound Height

Since potential energy at its peak is converted from kinetic energy, \( KE_r = PE_r = g \times 0.25 = 2.45 \mathrm{~m^2/s^2} \).
05

Determine Energy Lost in Collision

Energy lost is \( KE_i - KE_r = 32 - 2.45 = 29.55 \mathrm{~m^2/s^2} \).
06

Calculate Fraction of Energy Lost

The fraction of the initial kinetic energy lost is \( \frac{29.55}{32} \approx 0.923 \).
07

Final Step: Conclusion

Approximately 92.3% of the initial kinetic energy was lost in the collision.

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Key Concepts

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

Potential Energy
Potential energy is a fundamental concept in physics that relates to the position or state of an object. It primarily refers to energy stored due to an object's position in a gravitational field. Consider a ball at a certain height above the ground. This ball has gravitational potential energy because of Earth's gravity pulling it downwards.

The potential energy (\( PE \)) of an object at height \( h \) is calculated using the formula:
  • \( PE = mgh \)
Here, \( m \) is the mass of the object, \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \mathrm{m/s^2} \)), and \( h \) is the height.

For example, in the exercise, the ball rebounds to a height of \( 0.25 \) meters. At this point, its potential energy is determined by the height and gravitational force. This energy represents the capability of the object to do work when it falls back down.
Energy Conservation
Energy conservation is a key principle in physics stating that energy cannot be created or destroyed; it can only change forms. In the context of a bouncing ball, energy shifts between kinetic and potential forms.When the ball initially drops, it possesses kinetic energy (\( KE \)), which is the energy of motion. As it reaches the ground and starts to bounce back up, its energy converts to potential energy at its peak height.
  • Initially: All energy is kinetic (\( KE_i \))
  • At peak height: All energy becomes potential (\( PE_r \))
Energy lost in an ideal situation should only transform between these two forms without loss. However, due to real-world factors like air resistance and heat generated in the collision, there is some energy you cannot recover after a bounce. Thus, some kinetic energy is lost in the ball's real-world bounces.
Collision Physics
Collision physics examines the interaction forces between bodies when they collide. This concept is vital in understanding how energy dissipates in interactions.

In a collision, bodies trade kinetic energy, but not all of it might be conserved as kinetic. The exercise highlights an inelastic collision where a rubber ball loses energy when it bounces off the ground.
  • Elastic collision: Total kinetic energy is conserved.
  • Inelastic collision: Some kinetic energy is lost.
In the context of the exercise, the ball starts with a certain kinetic energy. However, due to the collision's nature, not all kinetic energy gets converted back when it rebounds. As analyzed, around 92.3% of the initial kinetic energy is lost. This energy either transforms into other forms like heat or sound during the impact. Understanding these principles helps illustrate why the ball doesn’t bounce back to its original height post-collision.

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Most popular questions from this chapter

Two skaters with masses of \(65 \mathrm{~kg}\) and \(45 \mathrm{~kg}\), respectively, stand \(8.0 \mathrm{~m}\) apart, each holding one end of a piece of rope. (a) If they pull themselves along the rope until they meet, how far does each skater travel? (Neglect friction.) (b) If only the 45 -kg skater pulls along the rope until she meets her friend (who just holds onto the rope), how far does each skater travel?

An incoming 0.14 -kg baseball has a speed of \(45 \mathrm{~m} / \mathrm{s}\). The batter hits the ball, giving it a speed of \(60 \mathrm{~m} / \mathrm{s}\). If the contact time is \(0.040 \mathrm{~s},\) what is the average force of the bat on the ball?

A \(10-\mathrm{g}\) bullet moving horizontally at \(400 \mathrm{~m} / \mathrm{s}\) penetrates a \(3.0-\mathrm{kg}\) wood block resting on a horizontal surface. If the bullet slows down to \(300 \mathrm{~m} / \mathrm{s}\) after emerging from the block, what is the speed of the block immediately after the bullet emerges ( \(\mathbf{v}\) Fig. 6.33 )?

A fellow student states that the total momentum of a three-particle system \(\left(m_{1}=0.25 \mathrm{~kg}, m_{2}=0.20 \mathrm{~kg},\right.\) and \(m_{3}=0.33 \mathrm{~kg}\) ) is initially zero. He calculates that after an inelastic triple collision the particles have velocities of \(4.0 \mathrm{~m} / \mathrm{s}\) at \(0^{\circ}, 6.0 \mathrm{~m}\) at \(120^{\circ},\) and \(2.5 \mathrm{~m} / \mathrm{s}\) at \(230^{\circ},\) respec- tively, with angles measured from the \(+x\) -axis. Do you agree with his calculations? If not, assuming the first two answers to be correct, what should be the momentum of the third particle so the total momentum is zero?

A ballistic pendulum is a device used to measure the velocity of a projectile- for example, the muzzle velocity of a rifle bullet. The projectile is shot horizontally into, and becomes embedded in, the bob of a pendulum, as illustrated in \(>\) Fig. \(6.35 .\) The pendulum swings upward to some height \(h,\) which is measured. The masses of the block and the bullet are known. Using the laws of momentum and energy, show that the initial velocity of the projectile is given by \(v_{\mathrm{o}}=[(m+M) / m] 22 g h\).

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