Problem 51 - College Physics

Question

A 0.20 -kg rubber ball is dropped from a height of \(1.0 \mathrm{~m}\) above the floor and it bounces back to a height of \(0.70 \mathrm{~m} .\) (a) What is the ball's speed just before hitting the floor? (b) What is the speed of the ball just as it leaves the ground? (c) How much energy was lost and where did it go?

Step-by-Step Solution

Answer

(a) 4.43 m/s, (b) 3.70 m/s, (c) 0.588 J lost to sound, heat, deformation.

Step 1: Set Up Given Problem Parameters

The mass of the rubber ball is given as 0.20 kg, the initial drop height is 1.0 m, and the bounce-back height is 0.70 m. We're going to use these to calculate speeds and energy changes.

Step 2: Calculate Speed Before Impact

Use the conservation of energy principle for the ball falling. The potential energy at the top converts to kinetic energy just before hitting the floor. The formula is:\[ mgh_1 = \frac{1}{2}mv_1^2\]where \(g = 9.8 \text{ m/s}^2\), \(h_1 = 1.0 \text{ m}\), and we solve for \(v_1\):\[v_1 = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 1.0} = \sqrt{19.6} \approx 4.43 \text{ m/s}\]

Step 3: Calculate Speed After Bounce

Now calculate the speed of the ball as it rebounds to a height of 0.70 m. Again, potential energy converts to kinetic energy, but this time for upward motion. The formula is:\[mgh_2 = \frac{1}{2}mv_2^2\]where \(h_2 = 0.70 \text{ m}\), and we solve for \(v_2\):\[v_2 = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 0.70} = \sqrt{13.72} \approx 3.70 \text{ m/s}\]

Step 4: Calculate Energy Lost

Calculating energy loss involves the difference between the initial and final mechanical energies. The initial potential energy is:\[E_{initial} = mgh_1 = 0.20 \times 9.8 \times 1.0 = 1.96 \text{ J}\]The final potential energy is:\[E_{final} = mgh_2 = 0.20 \times 9.8 \times 0.70 = 1.372 \text{ J}\]The energy lost is:\[E_{lost} = E_{initial} - E_{final} = 1.96 - 1.372 = 0.588 \text{ J}\]This energy was converted to other forms like sound, heat, and internal energy due to deformation.

Key Concepts

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. For our rubber ball example, kinetic energy becomes most important when the ball is just about to hit the ground or when it just leaves the ground.
To find the speed of the ball just before it hits the floor, we use the conservation of energy principle. The potential energy at the highest point is converted to kinetic energy at the lowest point.

• The formula for kinetic energy is: \( KE = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity.
• In calculations, the velocity just before impact is found using \( v_1 = \sqrt{2gh_1} \), which gives approximately \( 4.43 ext{ m/s} \).

This simple example shows how the speed and thus the kinetic energy of an object increases as it falls. Understanding kinetic energy is crucial for analyzing motion and energy transformations.

Potential Energy

Potential energy is related to the position of an object in a gravitational field. In our problem, the potential energy is at its maximum when the ball is held at a height of 1.0 m.
Potential energy can be thought of as stored energy because it has the potential to be transformed into kinetic energy as the object moves.

• The formula for potential energy is: \( PE = mgh \), where \( g \) is the acceleration due to gravity, typically \( 9.8 ext{ m/s}^2 \).
• Initially, the ball has a potential energy of \( 1.96 ext{ J} \). After the bounce to a height of 0.70 m, the potential energy reduces to \( 1.372 ext{ J} \).

The change in potential energy reflects the energy transformed into kinetic energy and other forms during the ball's motion, like sound and heat. It's important to remember that potential energy depends on both the height and mass of an object, helping us predict outcomes of different scenarios.

Energy Loss

Energy loss occurs when mechanical energy is transformed into other forms like sound, heat, or internal energy.
In our ball example, not all the initial energy is recovered after the bounce. Part of it has been lost.

• Initially, the rubber ball had an energy of \( 1.96 ext{ J} \).
• After it bounced back to 0.70 meters, the energy is \( 1.372 ext{ J} \).
• The energy loss is calculated as the difference: \( 0.588 ext{ J} \).

This energy was not destroyed; it was converted to other forms. Some of it converts to sound energy (why we hear the bounce), while others transform into heat through air resistance and internal deformation of the ball.
Understanding energy loss is vital, especially in engineering and environmental studies, as it helps in designing more efficient systems.