Question

A ball falls from a height of 20 m, and then bounces back up to 17 m height. The loss of energy is

Hint:

The ratio of the final height to the initial height corresponds to the square of the coefficient of restitution, $e^2 = \frac{h_2}{h_1}$.
The fractional loss of mechanical energy is directly given by $1 - e^2$.

Answer 1

Step 1: Understanding the Concept:
When a ball falls from an initial height and bounces back up to a lower final height, it has lost a portion of its initial mechanical energy.
This energy loss is typically due to non-conservative forces like air resistance and the inelastic nature of the collision with the ground.
Step 2: Key Formula or Approach:
The potential energy of the ball at any given height is $E = mgh$.
The percentage loss of energy is calculated using the formula:
\[ \text{Percentage Loss} = \left( \frac{E_i - E_f}{E_i} \right) \times 100% \] Step 3: Detailed Explanation:
Let the initial height be $h_1 = 20\text{ m}$.
Let the final height reached after the bounce be $h_2 = 17\text{ m}$.
The initial potential energy before falling is $E_i = mgh_1$.
The final potential energy after the bounce is $E_f = mgh_2$.
The loss in energy is $\Delta E = mgh_1 - mgh_2 = mg(h_1 - h_2)$.
Substituting the values, the percentage energy loss is:
\[ \text{Percentage Loss} = \frac{mg(20 - 17)}{mg(20)} \times 100% \] \[ \text{Percentage Loss} = \frac{3}{20} \times 100% \] \[ \text{Percentage Loss} = 3 \times 5% = 15% \] Step 4: Final Answer:
The percentage loss of energy is $15%$.