Exploring Kinetic Energy in a Ball's Bouncing Pattern

[Hangst]

Homework Statement

A little ball bounces straight on a surface in a certain way so that at every bounce it loses a fraction of it's vertical speed (called e in this equation). Assume that the ball is released from a ''resting height' (called H in this equation).

Show that after the n'th bounce against the surface the ball has reached:

$H_n=e^2n * H$

Homework Equations

I'm not sure.

The Attempt at a Solution

I'm thinking I need to maybe look at the kinectic energy but I'm sort of mystified. Any help is appreciated.

 

[Hangst]

Homework Statement

A little ball bounces straight on a surface in a certain way so that at every bounce it loses a fraction of it's vertical speed (called e in this equation). Assume that the ball is released from a ''resting height' (called H in this equation).

Show that after the n'th bounce against the surface the ball has reached:

$H_n=e^2n * H$

Homework Equations

I'm not sure.

The Attempt at a Solution

I'm thinking I need to maybe look at the kinectic energy but I'm sort of mystified. Any help is appreciated.

It is supposed to be e^2n not e^2 n

 

[xliu13]

Consider the moment before impact with ground and the moment after impact.

Before impact, speed of the ball = v; Kinetic energy of the ball = 1/2 mv²
After impact, speed of the ball = e*v; Kinetic energy of the ball = 1.2 m (e*v)²

∴the total mechanical energy of the ball has been reduced to e² of the original after one bounce, and similarly would be reduced to e²⁺², after third bounce to e²⁺²⁺², and so on.

∴after the nth bounce mechanical energy of the ball is E=e²ⁿ*E₀, where E₀ is the energy before the first bounce.

As all kinetic energy after impact is then converted into gravitational potential energy when the ball goes to the highest point, and because gravitational potential energy is proportional to the change of height...

Gravitational potential energy is reduced to e²ⁿ of the original.
∴the height is H_n =e²ⁿ∗H

 

[haruspex]

Actually the OP said it loses a fraction 'e', but that must be a typo. To get the desired answer it must retain fraction e.

 

[Nickie]

I would approach this problem by first defining the variables and concepts involved. The ball's bouncing pattern can be described in terms of its kinetic energy, which is the energy it possesses due to its motion. The height at which the ball is released, H, can be considered as its initial potential energy, which is converted into kinetic energy as it bounces. The fraction of vertical speed that the ball loses with each bounce, e, can be thought of as a measure of the ball's energy loss.

To understand how the ball's kinetic energy changes with each bounce, we can use the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or transformed. In this case, the ball's kinetic energy is being transformed into potential energy as it bounces, and then back into kinetic energy as it falls and bounces again. This process continues with each subsequent bounce.

To determine the height of the ball after the n'th bounce, we can use the formula for kinetic energy: KE = 1/2 * m * v^2, where m is the mass of the ball and v is its velocity. Since the ball is bouncing straight, we can focus on its vertical velocity, which decreases by a factor of e with each bounce. This means that after the first bounce, the ball's vertical velocity will be e * v, after the second bounce it will be e^2 * v, and so on.

We can also use the equation for potential energy: PE = m * g * h, where g is the gravitational acceleration and h is the height. As the ball bounces, its height decreases by a factor of e each time, which means that after the first bounce, the ball's height will be H * e, after the second bounce it will be H * e^2, and so on.

Combining these equations and using the fact that the ball is initially released from a height of H, we can write the following equation for the ball's height after the n'th bounce:

H_n = (1/2 * m * (e^n * v)^2) / (m * g) = (1/2 * e^2n * m * v^2) / (m * g) = e^2n * H

This shows that after the n'th bounce, the ball's height is equal to the initial height, H, multiplied by e^2