Question

A basketball of mass \(0.624 \mathrm{~kg}\) is shot from a vertical height of \(1.20 \mathrm{~m}\) and at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). After reaching its maximum height, the ball moves into the hoop on its downward path, at \(3.05 \mathrm{~m}\) above the ground. Using the principle of energy conservation, determine how fast the ball is moving just before it enters the hoop.

Short answer

Answer: The velocity of the basketball just before it enters the hoop is 17.06 m/s.

Step-by-step solution

Identify what is given and what needs to be found

We are given the mass of the basketball, its initial height, speed, and the height at which the hoop is located. We need to find the velocity of the basketball just before it enters the hoop.

Consider energy conservation

The total mechanical energy of the basketball remains constant during its motion (ignoring any air resistance). The total mechanical energy of an object consists of its kinetic energy (KE) and its potential energy (PE). The equation for energy conservation is: Total Mechanical Energy at initial point = Total Mechanical Energy at the hoop Initially, the basketball has both kinetic and potential energy; while at the hoop, it will also have both kinetic and potential energy.

Write down the equations for kinetic energy and potential energy

Kinetic energy (KE) is given by the formula: \(KE = \frac{1}{2}mv^2\) Potential energy (PE) due to gravity is given by the formula: \(PE = mgh\)

Write the energy conservation equation

We have the total mechanical energy (TME) at the initial point and at the hoop: \(TME_{initial} = KE_{initial} + PE_{initial}\) \(TME_{hoop} = KE_{hoop} + PE_{hoop}\) By applying the energy conservation law, we get \(TME_{initial} = TME_{hoop}\).

Substitute the values and solve for the final velocity

Using the given values, let's calculate the initial kinetic energy, initial potential energy, and potential energy at the hoop. \(KE_{initial} = \frac{1}{2}(0.624 \mathrm{~kg})(20.0 \mathrm{~m/s})^2 = 124.8 \mathrm{~J}\) \(PE_{initial} = (0.624 \mathrm{~kg})(9.81 \mathrm{~m/s^2})(1.20 \mathrm{~m}) = 7.32464 \mathrm{~J}\) \(PE_{hoop} = (0.624 \mathrm{~kg})(9.81 \mathrm{~m/s^2})(3.05 \mathrm{~m}) = 18.36972 \mathrm{~J}\) Now substitute these values into the energy conservation equation: \(124.8 \mathrm{~J} + 7.32464 \mathrm{~J} = KE_{hoop} + 18.36972 \mathrm{~J}\) Now, we need to solve for \(KE_{hoop}\). Rearrange the equation: \(KE_{hoop} = 124.8 \mathrm{~J} + 7.32464 \mathrm{~J} - 18.36972 \mathrm{~J} = 113.755 \mathrm{~J}\) Now, we need to find the velocity of the basketball as it enters the hoop. We'll use the kinetic energy equation: \(v_{hoop} = \sqrt{\frac{2 * KE_{hoop}}{m}}\) \(v_{hoop} = \sqrt{\frac{2 * 113.755 \mathrm{~J}}{0.624 \mathrm{~kg}}} = 17.06 \mathrm{~m/s}\)

Final answer

The velocity of the basketball just before it enters the hoop is 17.06 m/s.

Key concepts

Mechanical Energy

Mechanical energy is the sum of kinetic and potential energies in a system. It's a snapshot of an object's capability to do work due to its motion or position. For instance, when our basketball in the exercise is hurtling through the air, it holds kinetic energy because of its speed and potential energy because of its height above the ground.

During its flight, mechanical energy will manifest as potential energy when the ball reaches the peak of its arc and kinetic energy when it moves at maximum speed. If we disregard air resistance and other non-conservative forces, the total mechanical energy of the basketball remains unchanged throughout its journey. This conservation is a powerful tool as it allows us to analyze movement and predict speeds at different points without knowing all the intricate details of the path taken by the basketball.

Kinetic Energy

Kinetic energy (KE) is related to the motion of an object. The amount of kinetic energy an object has is determined by its mass and the speed at which it's moving. The formula for kinetic energy is expressed as
\(KE = \frac{1}{2}mv^2\)
, where \(m\) is the mass and \(v\) is the velocity.

In the case of the exercise example, the kinetic energy can be calculated at any point during the basketball's flight. Initially, the basketball possesses a significant amount of kinetic energy because of its initial speed. As it rises, its speed decreases, so its kinetic energy decreases as well. Understanding kinetic energy helps predict how fast the basketball will be going at different points in its trajectory, such as right before it enters the hoop.

Potential Energy

Potential energy (PE) is the stored energy in an object due to its position or state. For objects in a gravitational field, like our basketball example, gravitational potential energy is most relevant. It's calculated with the formula:
\(PE = mgh\)
, where \(m\) is the mass, \(g\) is the acceleration due to gravity, and \(h\) is the height from a reference level (often the ground).

As the basketball rises after being shot, its potential energy increases since it gains height. The highest point in its arc has the maximum potential energy. When the ball falls towards the hoop, it loses potential energy, converted back into kinetic energy. This exchange allows us to use the ball's height at different points to calculate its potential energy and understand how this energy transference affects the ball's speed.

Principle of Energy Conservation

The principle of energy conservation states that in a closed system with no external work being done, the total amount of energy remains constant, though it may change forms, such as from potential to kinetic energy and vice versa.

In our basketball scenario, by considering the ball's total mechanical energy at the beginning and at the hoop, we demonstrate conservation of energy. This principle simplifies the problem because, regardless of the exact path the ball takes, we know that the initial mechanical energy equals the mechanical energy at the hoop. If we know the mass, the height of the hoop, and the initial conditions, we can figure out the speed of the basketball as it enters the hoop by rearranging the conservation of energy equation.

Applying the principle of energy conservation leads to the solution provided in the exercise, showing that the ball moves at a speed of 17.06 m/s just before entering the hoop. This fundamental physics principle allows us to solve complex problems by focusing on initial and final states rather than the details in between.