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A ball is dropped from rest from the top of a 6.10 -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 make 5 bounces and still reach the windowsill.

Step by step solution

01

Determine Initial Potential Energy

The initial potential energy is calculated using the formula \( PE = mgh \), where \( m \) is the mass of the ball, \( g \) is the acceleration due to gravity (9.81 \( \text{m/s}^2 \)), and \( h \) is the initial height (6.10 m). The exact mass is not necessary as it will be canceled out in further calculations, so the initial potential energy is \( PE_0 = m \cdot 9.81 \cdot 6.10 \).
02

Calculate Initial Kinetic Energy

Just before the ball hits the ground for the first time, all the potential energy is converted into kinetic energy. Thus, the initial kinetic energy \( KE_0 \) is equal to the initial potential energy: \( KE_0 = m \cdot 9.81 \cdot 6.10 \).
03

Determine Kinetic Energy After Each Bounce

With each bounce, the ball loses 10% of its kinetic energy. Therefore, the kinetic energy after the first bounce, \( KE_1 = 0.9 \cdot KE_0 \). For subsequent bounces, \( KE_n = 0.9 \cdot KE_{n-1} \) or \( KE_n = (0.9)^n \cdot KE_0 \).
04

Link Kinetic Energy to Maximum Height

To find the maximum height after each bounce, equate the kinetic energy at the maximum ascend to the potential energy at that height: \( KE_n = mgh_n \). This gives \( h_n = \frac{KE_n}{mg} \).
05

Solve for When Height Equals Windowsill

Since you're looking for the bounce where the ball still reaches a windowsill of 2.44 m, set \( h_n = 2.44 \). Thus, \( \frac{(0.9)^n \cdot 9.81 \cdot 6.10}{9.81} = 2.44 \), simplifying gives \( (0.9)^n \cdot 6.10 = 2.44 \).
06

Solve for Number of Bounces

Rearrange the equation from Step 5 to find \( n \): \[ n = \frac{\ln(\frac{2.44}{6.10})}{\ln(0.9)} \]Compute this to find the number of bounces, \( n \), that allows the ball to reach the windowsill.

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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 key concept in physics, representing the energy stored in an object due to its position, condition, or arrangement. In the context of our exercise, it is the energy associated with the object's height from the ground.

The formula used to calculate potential energy is:
  • \( PE = mgh \)
Here, \( m \) is the mass of the object (in kilograms), \( g \) is the acceleration due to gravity (approximately 9.81 m/s² on Earth), and \( h \) is the height in meters above a reference point.

When the ball is at the top of the 6.10 m building, it has a maximum potential energy. As it falls, this potential energy is converted into kinetic energy up until the moment it hits the ground. Understanding potential energy helps us analyze how energy transforms as the ball moves through its trajectory.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. As the ball falls from the building, its potential energy is transformed into kinetic energy. The formula for kinetic energy is:
  • \( KE = \frac{1}{2}mv^2 \)
Here, \( v \) is the velocity of the object as it moves. However, in our exercise, an easier approach is to consider that the initial kinetic energy just before impact is equal to the initial potential energy at the top of the building: \( KE_0 = mgh \).

After each bounce, the ball loses 10% of its kinetic energy. This step-by-step decay of kinetic energy is important in determining how high the ball can bounce back. With each bounce, a lesser amount of kinetic energy means the ball reaches a lower height, until a point it cannot reach the 2.44 m windowsill.
Inelastic Collision
An inelastic collision is one where some of the object's kinetic energy is not conserved and is transformed into other forms of energy, like heat or sound, during the collision. In our roadmap, after every collision with the ground, the ball loses a portion of its kinetic energy and hence bounces back to a lower height.

In this case, it's stated that 10% of the kinetic energy is lost with every bounce. This is mathematically represented by multiplying the kinetic energy after each bounce by 0.9: \( KE_n = 0.9 \cdot KE_{n-1} \).

Understanding inelastic collisions is crucial when evaluating how many times the ball can bounce and still reach a particular height, such as the windowsill in the exercise.
Motion Dynamics
Motion dynamics involves studying the forces and energies involved in the movement of objects. This exercise primarily focuses on the transformation of energy forms as the ball moves under gravity, illustrating core principles of motion dynamics.

Here, the main forces acting are gravity, promoting the ball's movement downward and affecting its motion upon bouncing back. The interaction of these forces is quantified through the transition of potential energy into kinetic energy and the influence of inelastic collisions reducing the kinetic energy after each bounce.

By understanding motion dynamics, we can predict how far and how fast the ball moves, and with how much energy, in each part of its trajectory. Analyzing these factors is essential for solving how many bounces it takes for the ball to still meet the windowsill's height.

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Most popular questions from this chapter

ssm A projectile (mass \(=0.20 \mathrm{kg}\) is fired at and embeds itself in a stationary target (mass \(=2.50 \mathrm{kg}\) . With what percentage of the projectile's incident kinetic energy does the target (with the projectile in it) fly off after being struck?

Multiple-Concept Example 7 deals with some of the concepts that are used to solve this problem. A cue ball (mass \(=0.165 \mathrm{kg}\) ) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of \(+1.50 \mathrm{N}\) \cdots to the ball. The ball then slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck.

ssm A lumberjack (mass \(=98 \mathrm{kg} )\) is standing at rest on one end of a floating log (mass \(=230 \mathrm{kg}\) ) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of \(+3.6 \mathrm{m} / \mathrm{s}\) relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log just before the lumberjack jumps off? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

In a performance test, each of two cars takes 9.0 s to accelerate from rest to 27 m/s. Car A has a mass of 1400 kg, and car B has a mass of 1900 kg. Find the net average force that acts on each car during the test.

Two stars in a binary system orbit around their center of mass. The centers of the two stars are \(7.17 \times 10^{11} \mathrm{m}\) apart. The larger of the two stars has a mass of \(3.70 \times 10^{30} \mathrm{kg},\) and its center is \(2.08 \times 10^{11} \mathrm{m}\) from the system's center of mass. What is the mass of the smaller star?

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