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When a tennis ball bounces from a racket, the ball loses approximately \(30 \%\) of its kinetic energy to thermal energy. A ball that hits a racket at a speed of \(10 \mathrm{m} / \mathrm{s}\) will rebound with approximately what speed? A. \(8.5 \mathrm{m} / \mathrm{s}\) B. \(7.0 \mathrm{m} / \mathrm{s}\) C. \(4.5 \mathrm{m} / \mathrm{s}\) D. \(3.0 \mathrm{m} / \mathrm{s}\)

Short Answer

Expert verified
The speed at which the ball will rebound is approximately \(8.37 \mathrm{m/s}\), so the closest answer is A. \(8.5 \mathrm{m/ s}\).

Step by step solution

01

Understand the initial conditions

The tennis ball is initially moving at a speed of \(10 \mathrm{m/s}\) and hits the racket. After hitting, it loses \(30\%\) of its kinetic energy.
02

Formula for kinetic energy and loss

The formula for kinetic energy is \(\frac{1}{2}mv^2\), where \(m\) is the mass of the object and \(v\) is its speed. With \(30\%\) loss, the new kinetic energy becomes \(70\%\) of the initial. Hence, the initial and final kinetic energy relation can be written as \(\frac{1}{2}mv_i^2=0.70\times \frac{1}{2}mv_f^2\), where \(v_i\) is the initial speed and \(v_f\) is the final speed.
03

Solve for the final speed

From the equation above, we can solve for the final speed. The \(m\) and \(\frac{1}{2}\) terms can be cancelled on both sides, leaving us with \(v_i^2=0.70\times v_f^2\). Taking the square root, we get the final speed \(v_f=\sqrt{0.70}\times v_i = 0.837 \times 10 \mathrm{m/s} = 8.37 \mathrm{m/s}\).

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Key Concepts

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

Kinetic Energy Formula
Kinetic energy is the energy an object has due to its motion. It can be calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
where:
  • KE is the kinetic energy,

  • m is the mass of the object, and
  • v is the velocity (speed with a direction) of the object.
To understand kinetic energy loss, such as when a tennis ball bounces from a racket, we recognize that a portion of this kinetic energy is converted into other forms of energy, typically thermal energy. If a ball has an initial speed (v_i) and after impact its speed is reduced, we can apply the kinetic energy formula to both the initial and final states to quantify the change in kinetic energy.
Elastic and Inelastic Collisions
Collisions between objects can be categorized as either elastic or inelastic based on how the kinetic energy changes during the event.

Elastic Collisions

In an elastic collision, no kinetic energy is lost; it is conserved. This means the total kinetic energy before and after the collision remains the same, although individually the objects involved may alter their energy distribution.

Inelastic Collisions

In contrast, inelastic collisions do result in a loss of kinetic energy. This is the case with the tennis ball and racket example, where the ball does not bounce back with the same energy it had before striking the racket. Instead, a part of the ball's kinetic energy is converted into other forms, particularly thermal energy, leading to a decrease in the kinetic energy after the interaction.
Energy Conversion in Physics
Energy conversion is a fundamental concept in physics that refers to the transformation of energy from one form to another. The law of conservation of energy states that the total energy in a closed system remains constant; it can neither be created nor destroyed, only transformed from one form to another. These transformations are evident in many physical processes, including collisions. When a tennis ball collides with a racket and slows down, the kinetic energy of the ball is not destroyed but converted. Most of this energy transforms into thermal energy due to friction and deformation, and sometimes sound, explaining why the ball cannot rebound with the same speed it had before the impact.
Thermal Energy in Collisions
During a collision, some of the kinetic energy of the objects involved is often converted to thermal energy. This occurs due to friction and the deformation of the objects during impact. For instance, when a tennis ball hits a racket, the surface of the ball deforms and heats up due to the internal friction of its material. The amount of kinetic energy converted into thermal energy can be significant, as shown in the tennis ball example where approximately 30% of the ball's kinetic energy is lost in this form. This conversion is crucial in understanding why objects do not bounce back with the same speed after a collision, accounting for the observable loss of motion in inelastic collisions.

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Most popular questions from this chapter

A 25 kg child slides down a playground slide at a constant speed. The slide has a height of \(3.0 \mathrm{m}\) and is \(7.0 \mathrm{m}\) long. Using the law of conservation of energy, find the magnitude of the kinetic friction force acting on the child.

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a. How much work does an elevator motor do to lift a \(1000 \mathrm{kg}\) elevator a height of \(100 \mathrm{m}\) at a constant speed? b. How much power must the motor supply to do this in \(50 \mathrm{s}\) at constant speed?

When you ride a bicycle at constant speed, almost all of the energy you expend goes into the work you do against the drag force of the air. In this problem, assume that all of the energy expended goes into working against drag. As we saw in Section \(5.6,\) the drag force on an object is approximately proportional to the square of its speed with respect to the air. For this problem, assume that \(F \propto v^{2}\) exactly and that the air is motionless with respect to the ground unless noted Suppose a cyclist and her bicycle have a combined mass of \(60 \mathrm{kg}\) and she is cycling along at a speed of \(5 \mathrm{m} / \mathrm{s}\). If the drag force on the cyclist is \(10 \mathrm{N},\) how much energy does she use in cycling \(1 \mathrm{km} ?\) A. \(6 \mathrm{kJ}\) B. \(10 \mathrm{kJ}\) C. \(50 \mathrm{kJ}\) D. \(100 \mathrm{kJ}\)

In an amusement park water slide, people slide down an essentially frictionless tube. The top of the slide is \(3.0 \mathrm{m}\) above the bottom where they exit the slide, moving horizontally, \(1.2 \mathrm{m}\) above a swimming pool. What horizontal distance do they travel from the exit point before hitting the water? Does the mass of the person make any difference?

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