A ball is dropped form a height of 20 cm . Ball rebounds to a height of 10 cm . What is the loss of energy ?

Options

A. 0.25
B. 0.75
C. 0.5
D. 1

Explanation

To find the loss of energy when a ball is dropped from a height of 20 cm and rebounds to a height of 10 cm, we can follow these steps:

### Step 1: Convert Heights to Meters
First, we need to convert the heights from centimeters to meters for consistency in calculations.
- Height from which the ball is dropped (h1) = 20 cm = 0.2 m
- Height to which the ball rebounds (h2) = 10 cm = 0.1 m

### Step 2: Calculate Initial Potential Energy (PE_initial)
The potential energy (PE) of the ball at the height of 20 cm can be calculated using the formula:
\[ \text{PE} = mgh \]
Where:
- \( m \) = mass of the ball (we will keep it as \( m \) since it will cancel out later)
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( h \) = height (0.2 m)

So, the initial potential energy is:
\[ \text{PE_initial} = mg(0.2) \]

### Step 3: Calculate Final Potential Energy (PE_final)
After rebounding to a height of 10 cm, the potential energy can be calculated similarly:
\[ \text{PE_final} = mg(0.1) \]

### Step 4: Calculate Loss of Energy
The loss of energy (ΔE) can be calculated as the difference between the initial and final potential energies:
\[ \Delta E = \text{PE_initial} - \text{PE_final} \]
Substituting the values we calculated:
\[ \Delta E = mg(0.2) - mg(0.1) \]
\[ \Delta E = mg(0.2 - 0.1) \]
\[ \Delta E = mg(0.1) \]

### Step 5: Conclusion
The loss of energy is given by:
\[ \Delta E = 0.1mg \]
This indicates that the energy lost is proportional to the mass of the ball and the acceleration due to gravity.