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%
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.
Let's consider the energy of the ball at different points in its motion. We will assume that air resistance is negligible.
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 $KE = \frac{1}{2}mv^2$.
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:
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.
\(\text{Percentage Loss} = \left( 1 - \frac{2.5 \, \text{m}}{10 \, \text{m}} \right) \times 100\%\)
\(\text{Percentage Loss} = \left( 1 - 0.25 \right) \times 100\%\)
\(\text{Percentage Loss} = \left( 0.75 \right) \times 100\%\)
\(\text{Percentage Loss} = 75\%\)
| 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%.
| 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 \( = (1 - (0.5)^2) \times 100\% = (1 - 0.25) \times 100\% = 75\% \). |
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