Analyzing the Falling Ball Problem

The question asks us to find the ratio of potential energy (PE) to kinetic energy (KE) for a ball dropped from a height H, when it reaches a height H/3 above the ground.

We are given:

• Mass of the ball = \(m\)
• Initial height = \(H\)
• Target height = \(H/3\)

The ball is dropped from rest, meaning its initial velocity at height H is zero.

Understanding Energy Conservation

When a ball falls under gravity, and we ignore air resistance, the total mechanical energy (the sum of potential energy and kinetic energy) remains constant. This is the principle of conservation of mechanical energy.

Total Energy (E) = Potential Energy (PE) + Kinetic Energy (KE)

At the initial height \(H\), the ball is at rest, so its kinetic energy is zero. The total energy at this point is equal to the initial potential energy.

Initial PE = \(mgh\)

Initial KE = \(0\)

Total Energy at height H = \(mgh + 0 = mgh\)

According to the conservation of energy, the total energy at any point during the fall will be \(mgh\).

Calculating Potential and Kinetic Energy at Height H/3

Now, let's consider the ball when it is at height \(H/3\) above the ground.

Potential energy at height \(H/3\) is given by:

\(PE_{H/3} = mg \times (H/3) = \frac{mgh}{3}\)

The total energy at height \(H/3\) is still \(mgh\). So, we can write:

Total Energy at height \(H/3\) = \(PE_{H/3} + KE_{H/3}\)

\(mgh = \frac{mgh}{3} + KE_{H/3}\)

Now, we can find the kinetic energy at height \(H/3\):

\(KE_{H/3} = mgh - \frac{mgh}{3}\)

\(KE_{H/3} = mgh \left(1 - \frac{1}{3}\right)\)

\(KE_{H/3} = mgh \left(\frac{3-1}{3}\right)\)

\(KE_{H/3} = \frac{2mgh}{3}\)

Finding the Ratio of PE to KE at Height H/3

We need to find the ratio of potential energy to kinetic energy at height \(H/3\), which is \(PE_{H/3} / KE_{H/3}\).

Ratio = \(\frac{PE_{H/3}}{KE_{H/3}} = \frac{\frac{mgh}{3}}{\frac{2mgh}{3}}\)

To simplify the ratio, we can cancel out the common terms \(mgh\) and \(3\):

Ratio = \(\frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2}\)

So, the ratio of potential energy to kinetic energy at height \(H/3\) is \(1/2\).

Conclusion

At height H/3, the potential energy is \(mgH/3\) and the kinetic energy is \(2mgH/3\). The ratio of potential energy to kinetic energy is \(1/2\).

Revision Table: Energy in Free Fall

Additional Information: Factors Affecting Energy

Understanding energy changes during free fall is fundamental in physics. Here are some related points:

• Mass (m): Both PE and KE are directly proportional to mass. However, in the ratio PE/KE for a falling object under gravity (starting from rest), mass often cancels out, as seen in this problem.
• Height (h): Potential energy depends directly on height. As height decreases during a fall, PE decreases.
• Velocity (v): Kinetic energy depends on the square of velocity. As height decreases, the ball accelerates, increasing its velocity and thus its KE.
• Gravity (g): The acceleration due to gravity affects both PE and KE changes. It is a constant factor near the Earth's surface.
• Air Resistance: In real-world scenarios, air resistance is present. It is a non-conservative force that does negative work, converting mechanical energy into heat. If air resistance were included, total mechanical energy would decrease over time, and the conservation principle in the simple form \(E = PE + KE = \text{constant}\) would not hold.
• Reference Point for PE: The choice of the zero potential energy level is arbitrary. However, as long as the same reference is used consistently, the change in PE and thus the KE and energy conservation calculations remain valid. In this problem, the ground is implicitly taken as the zero potential energy level.