A ball is dropped from a height of 10 m. It strikes the ground and rebounds up to a height of 2.5 m. During the collision, the per cent loss in the kinetic energy is:
75%
Calculating Energy Loss in Ball Collision
The problem asks for the percentage loss in kinetic energy when a ball is dropped from a height of 10 m and rebounds to a height of 2.5 m. This involves understanding the transformation between potential energy and kinetic energy and how energy is lost during the collision with the ground.
Understanding Energy Before and After Collision
Let's consider the energy of the ball at different points in its motion. We will assume that air resistance is negligible.
- Before Collision: Just before the ball hits the ground, all its initial potential energy (from the drop height) is converted into kinetic energy. The initial height is $h_1 = 10$ m.
- After Collision: Just after the ball rebounds from the ground, it has kinetic energy that allows it to reach the rebound height. The rebound height is $h_2 = 2.5$ m. As the ball rises to its maximum rebound height, this kinetic energy is converted into potential energy.
Energy Conversion at Maximum Height and Just Before/After Collision
The potential energy (PE) of an object of mass $m$ at a height $h$ is given by $PE = mgh$, where $g$ is the acceleration due to gravity. The kinetic energy (KE) of an object of mass $m$ with velocity $v$ is given by $KE = \frac{1}{2}mv^2$.
- At the initial height ($h_1$), the ball is momentarily at rest, so its initial kinetic energy is zero. The total mechanical energy is approximately equal to its potential energy. As it falls, potential energy converts to kinetic energy. Just before hitting the ground (height $\approx 0$), its potential energy is nearly zero, and its kinetic energy is maximum, approximately equal to the initial potential energy.
- Similarly, after rebounding, the ball starts with maximum kinetic energy and zero potential energy (at height $\approx 0$). As it rises to the rebound height ($h_2$), its kinetic energy converts to potential energy. At the maximum rebound height, its kinetic energy is zero, and its potential energy is maximum, approximately equal to the kinetic energy it had just after rebounding.
Therefore, we can relate the kinetic energy just before the collision ($KE_{before}$) and just after the collision ($KE_{after}$) to the potential energies at the initial and rebound heights, respectively:
- Kinetic energy just before collision ($KE_{before}$): This energy comes from the initial potential energy at height $h_1$. Assuming energy conservation during the fall, $KE_{before} = mgh_1$.
- Kinetic energy just after collision ($KE_{after}$): This energy determines how high the ball rebounds. As it reaches height $h_2$, this kinetic energy is converted into potential energy. So, $KE_{after} = mgh_2$.
Calculating the Percentage Loss in Kinetic Energy
The loss in kinetic energy during the collision is the difference between the kinetic energy just before and just after the collision:
\(\Delta KE = KE_{before} - KE_{after}\)
\(\Delta KE = mgh_1 - mgh_2\)
The percentage loss in kinetic energy is calculated with respect to the kinetic energy just before the collision:
\(\text{Percentage Loss} = \left( \frac{\Delta KE}{KE_{before}} \right) \times 100\%\)
Substitute the expressions for $\Delta KE$ and $KE_{before}$:
\(\text{Percentage Loss} = \left( \frac{mgh_1 - mgh_2}{mgh_1} \right) \times 100\%\)
We can simplify this expression:
\(\text{Percentage Loss} = \left( \frac{mgh_1}{mgh_1} - \frac{mgh_2}{mgh_1} \right) \times 100\%\)
\(\text{Percentage Loss} = \left( 1 - \frac{h_2}{h_1} \right) \times 100\%\)
Now, plug in the given values: $h_1 = 10$ m and $h_2 = 2.5$ m.
\(\text{Percentage Loss} = \left( 1 - \frac{2.5 \, \text{m}}{10 \, \text{m}} \right) \times 100\%\)
\(\text{Percentage Loss} = \left( 1 - 0.25 \right) \times 100\%\)
\(\text{Percentage Loss} = \left( 0.75 \right) \times 100\%\)
\(\text{Percentage Loss} = 75\%\)
Summary of Calculation
| Parameter | Value |
|---|---|
| Initial Height ($h_1$) | 10 m |
| Rebound Height ($h_2$) | 2.5 m |
| Formula for Percentage Loss | \(\left( 1 - \frac{h_2}{h_1} \right) \times 100\%\) |
| Calculation | \(\left( 1 - \frac{2.5}{10} \right) \times 100\% = (1 - 0.25) \times 100\% = 0.75 \times 100\% = 75\%\) |
| Percentage Loss | 75% |
The percentage loss in kinetic energy during the collision is 75%.
Revision Table: Energy Loss Concepts
| Concept | Description |
|---|---|
| Potential Energy (PE) | Energy stored due to position or height. Formula: \(PE = mgh\). |
| Kinetic Energy (KE) | Energy possessed due to motion. Formula: \(KE = \frac{1}{2}mv^2\). |
| Energy Conservation (Ideal Case) | In the absence of non-conservative forces (like air resistance, friction), the total mechanical energy (PE + KE) remains constant. |
| Energy Loss in Collision | Real-world collisions often involve conversion of mechanical energy into other forms like heat, sound, or deformation, resulting in a loss of mechanical energy. This loss is often significant in inelastic collisions. |
| Coefficient of Restitution (e) | A measure of how 'bouncy' a collision is. For a ball bouncing vertically, \(e = \sqrt{\frac{h_{rebound}}{h_{drop}}}\). The percentage kinetic energy remaining after collision is related to $e^2$. Percentage loss is \( (1-e^2) \times 100\% \). In this problem, \(e = \sqrt{\frac{2.5}{10}} = \sqrt{0.25} = 0.5\). Percentage loss \( = (1 - (0.5)^2) \times 100\% = (1 - 0.25) \times 100\% = 75\% \). |
Additional Information: Types of Collisions and Energy
Collisions can be broadly classified based on whether kinetic energy is conserved:
- Elastic Collision: Kinetic energy is conserved during the collision. This is an ideal case often used in simplified models, especially for atomic or subatomic particle interactions. Momentum is also conserved.
- Inelastic Collision: Kinetic energy is NOT conserved during the collision. Some kinetic energy is converted into other forms (heat, sound, deformation). Momentum is still conserved in the absence of external forces. The ball bouncing off the ground is a common example of an inelastic collision.
- Perfectly Inelastic Collision: This is an extreme case of an inelastic collision where the colliding objects stick together after impact. The loss of kinetic energy is maximum in such a collision, though not necessarily 100% unless one of the objects was initially at rest or they move together at zero velocity afterwards.
In this ball bouncing problem, the collision with the ground is inelastic because kinetic energy is lost, as evidenced by the ball not returning to its original height.
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