/*! 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 27 ssm A golf ball bounces down a f... [FREE SOLUTION] | 91Ó°ÊÓ

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ssm A golf ball bounces down a flight of steel stairs, striking several steps on the way down, but never hitting the edge of a step. The ball starts at the top step with a vertical velocity component of zero. If all the collisions with the stairs are elastic, and if the vertical height of the staircase is \(3.00 \mathrm{~m}\), determine the bounce height when the ball reaches the bottom of the stairs. Neglect air resistance.

Short Answer

Expert verified
The bounce height is 3.00 m.

Step by step solution

01

Understanding the Problem

The ball starts from the top step with an initial vertical velocity of zero and bounces down the stairs with elastic collisions. We are asked to find the bounce height of the ball once it reaches the bottom of the stair after a total vertical drop of 3.00 meters.
02

Consider Elastic Collisions

Since the collisions are elastic, the mechanical energy of the ball is conserved. This means that the potential energy lost by the ball as it falls is converted into kinetic energy, which is then converted back to potential energy when it bounces to its maximum height.
03

Calculate Potential Energy Loss

The potential energy at the top is given by \( PE = mgh \) where \( h = 3.00 \) m and \( g = 9.81 \text{ m/s}^2 \). This energy is converted to kinetic energy at the bottom of the stairs before the bounce.
04

Determine Kinetic Energy Before Bounce

At the bottom of the stairs, the kinetic energy \( KE \) equals the initial potential energy. Thus, \( KE = mgh = m \cdot 9.81 \cdot 3.00 \). Since this energy is carried over during the bounce, the height it will reach is calculated next.
05

Calculate Bounce Height

After the bounce at the bottom, the ball's potential energy at the peak of its bounce height is \( PE_{bounce} = mgh_{bounce} \). Equating the potential energy gained to the kinetic energy before bounce gives us: \( mgh_{bounce} = mgh \). This implies \( h_{bounce} = h \).
06

Conclusion

Therefore, the height reached after the bounce is equal to the initial drop height, \( h_{bounce} = 3.00 \text{ m} \), due to the conservation of energy in elastic collisions.

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Key Concepts

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

Conservation of Energy
In physics, the principle of conservation of energy is a fundamental concept that states that energy cannot be created or destroyed in an isolated system, only transformed from one form to another. Imagine it as a closed bank where your amount of money stays the same, only moving from your savings to your checking account or vice versa. During the golf ball's journey down the stairs, energy shifts between potential energy and kinetic energy.
When the ball is at the top step, it holds a certain amount of potential energy due to its position above the ground. As it moves down, this potential energy converts into kinetic energy as the ball gains speed. Upon each elastic collision, the kinetic energy endured is transferred back into potential energy as the ball bounces up again.
The elastic nature of the collisions ensures that no energy is lost as heat or sound, which is a key point in maintaining the total mechanical energy throughout the entire motion of the ball. In our example, this means that by the time the ball reaches the bottom, and after rolling back up, it will reach approximately the same height it had at the beginning.
Potential Energy
Potential energy is like energy on hold, similar to a stretched rubber band that holds energy as you pull it taut. For our golf ball, when it's positioned at the top of the staircase, it has maximum potential energy thanks to its height above the ground. Potential energy due to gravitational forces can be calculated using the formula: \[PE = mgh\]where \( m \) stands for mass, \( g \) for gravitational acceleration \( (9.81 \text{ m/s}^2) \), and \( h \) for height above the ground. As the ball rolls down, the potential energy decreases while converting into kinetic energy, demonstrating that potential energy is closely tied to an object's position.
In situations where all potential energy converts without losses into kinetic energy, due to factors like elastic collisions here, the ball, once subjected to gravity, will strike back up to the very height it originated from at the top, ensuring the cycle continues smoothly.
Kinetic Energy
Kinetic energy is the energy of motion. Picture yourself running down a hill: the faster you go, the more kinetic energy you have. Kinetic energy can be calculated with the equation: \[KE = \frac{1}{2}mv^2\]where \( m \) is the mass of the object, and \( v \) is its velocity. For the golf ball, its kinetic energy is zero at the top of the stairs because it's not moving vertically yet.
As it descends, this kinetic energy grows in proportion to its increasing speed. By the time the ball reaches the bottom of the stairs, all the initial potential energy has converted into kinetic energy. The amazing thing about elastic collisions is that energy is conserved during each bounce, permitting the ball to convert that kinetic energy back into potential energy effortlessly, reaching the peak of its bounce at the same height it initially started. Thus, the ball continuously alternates this energy between potential and kinetic, courtesy of gravity and the perfectly elastic interactions with the stairs.

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Most popular questions from this chapter

In a performance test two cars take the same time to accelerate from rest up to the same speed. Car A has a mass of \(1400 \mathrm{~kg}\), and car \(\mathrm{B}\) has a mass of \(1900 \mathrm{~kg}\). During the test, which car (a) has the greater change in momentum, (b) experiences the greater impulse, and (c) is acted upon by the greater net average force? In each case, give your reasoning.

Each of these problems consists of Concept Questions followed by a related quantitative Problem. The Concept Questions involve little or no mathematics. They focus on the concepts with which the problems deal. Recognizing the concepts is the essential initial step in any problem-solving technique. Concept Questions A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon throws a rock. Does the momentum of the wagon increase, decrease, or remain the same (a) when the rock is thrown directly forward and (b) when the rock is thrown directly backward? (c) In which case does the wagon have the greater speed after the rock is thrown? Problem The total mass of the wagon, rider, and rock is \(95.0 \mathrm{~kg} .\) The mass of the rock is \(0.300 \mathrm{~kg} .\) Initially the wagon is rolling forward at a speed of \(0.500 \mathrm{~m} / \mathrm{s}\). Then the rock is thrown with a speed of \(16.0 \mathrm{~m} / \mathrm{s}\). Both speeds are relative to the ground. Find the speed of the wagon after the rock is thrown directly forward in one case and directly backward in another. Check to see that your answers are consistent with your answers to the Concept Questions.

A \(0.500-\mathrm{kg}\) ball is dropped from rest at a point \(1.20 \mathrm{~m}\) above the floor. The ball rebounds straight upward to a height of \(0.700 \mathrm{~m}\). What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

ssm Starting with an initial speed of \(5.00 \mathrm{~m} / \mathrm{s}\) at a height of \(0.300 \mathrm{~m}\), a \(1.50-\mathrm{kg}\) ball swings downward and strikes a \(4.60\) -kg ball that is at rest, as the drawing shows. (a) Using the principle of conservation of mechanical energy, find the speed of the \(1.50-\mathrm{kg}\) ball just before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision. (c) How high does each ball swing after the collision, ignoring air resistance?

A golfer, driving a golf ball off the tee, gives the ball a velocity of \(+38 \mathrm{~m} / \mathrm{s}\). The mass of the ball is \(0.045 \mathrm{~kg},\) and the duration of the impact with the golf club is \(3.0 \times 10^{-3} \mathrm{~s}\). (a) What is the change in momentum of the ball? (b) Determine the average force applied to the ball by the club.

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