Question

A ball loses \(15.0 \%\) of its kinetic energy when it bounces back from a concrete walk. With what speed must you throw it vertically down from a height of \(12.4 \mathrm{~m}\) to have it bounce back to that same height? Ignore air resistance.

Short answer

Following these steps should result in the solution to the problem. Plug the known values into the equation derived from the law of energy conversation and solve for the initial speed required for the ball to bounce back to the original height.

Step-by-step solution

Analyze the Energy Loss

The ball loses \(15.0\%\) of its kinetic energy when it bounces back. Therefore, the loss fraction will be \(15.0\% / 100 = 0.15\).

Apply the Law of Conservation of Energy

Let's denote the initial speed as \( v \). The potential energy \( mgh \) right before hitting the ground is converted into kinetic energy \(0.5 * m * v^2 \) when it hits the ground. After hitting, the ball retains \( (1 - 0.15) = 0.85 \) or \(85\%\) of its kinetic energy. This fraction of its energy is enough to take it back to the original height. Therefore, we can write the energy conservation equation as \( mgh = 0.85 * 0.5 * m * v^2 \). The mass \( m \) of the ball cancels out.

Evaluation

The height \( h \) is given as \( 12.4m \) and gravitational acceleration \( g \) as \( 9.8m/s^2 \). Substituting the known values and solving for \( v \), we get \( v = \sqrt{2gh/0.85} \).

Calculation

Substituting \( h = 12.4m \) and \( g = 9.8m/s^2 \) into the formula and performing the calculation, we obtain the initial speed \( v \) required for the ball to bounce back to the original height.

Key concepts

Energy Conservation

Energy conservation is a fundamental concept in physics that describes how energy can be transferred or transformed but never created or destroyed. In our problem, the ball's energy goes through a change when it is thrown, hits the ground, and bounces back. Initially, the ball has potential energy due to its height above the ground. As it moves downward, this potential energy is converted into kinetic energy as the ball gains speed. After hitting the ground, a portion of that kinetic energy is lost due to the impact, and the remaining is converted back to potential energy as it ascends. This energy transformation follows the principle of energy conservation.

• Initial potential energy gets converted to kinetic energy.
• Kinetic energy gets converted back to potential energy minus energy losses.

This cyclical conversion underscores the conservation of energy in isolated systems.

Potential Energy

Potential energy, particularly gravitational potential energy in this case, is the energy possessed by an object because of its position relative to a lower point. It is given by the formula \( PE = mgh \), where:\

• \(m\) is the mass of the object,
• \(g\) is the acceleration due to gravity,
• \(h\) is the height above the ground.

In the exercise, the ball's potential energy is highest at the starting height of 12.4 meters. As the ball falls, this potential energy is transformed into kinetic energy as it speeds up. When analyzing problems like these, it is essential to understand how potential energy is dependent on both the height and the mass of the object.

Law of Conservation of Energy

The law of conservation of energy is a principle stating that the total energy of an isolated system remains constant, although it may change forms. In this exercise, this law is crucial for determining how the ball's energy changes during its motion. The key steps involving the law are:

• The potential energy at the top changes to kinetic energy as it hits the ground.
• Only a portion of the kinetic energy is used for returning the ball to the initial height due to energy loss through bouncing.

Formally, in mathematical terms for the exercise, it shows that the potential energy lost (\(mgh\)) is balanced by the kinetic energy at impact, and conservation allows us to equate these forms to solve for the initial speed needed.

Gravitational Acceleration

Gravitational acceleration is a constant force acting on all objects with mass near the Earth's surface and is usually denoted as \( g \). It measures approximately \( 9.8 \, \text{m/s}^2 \) and plays a vital role in computations involving motion under gravity. In the given problem, gravitational acceleration influences both the potential and kinetic energy of the ball.
When calculating potential energy \( (mgh) \) and kinetic energy \( (0.5mv^2) \), \( g \) ensures that the motion is predictable and consistent. It's the reason why the ball accelerates downward, converting its stored energy into kinetic form. Understanding \( g \) helps us comprehend how objects move in freefall and how energy conversions occur in gravitational fields.