A glass ball is dropped from height 10 m. If there is 15% loss of energy due to impact, then after one impact, the ball will go upto.

Options

A. 2m
B. 4m
C. 6m
D. None of these

Answer

To solve the problem step by step, we will follow the principles of energy conservation and the effect of energy loss due to impact.

### Step 1: Calculate the initial potential energy (PE_initial)
The potential energy of the glass ball when it is dropped from a height \( h \) is given by the formula:
\[ PE_{\text{initial}} = mgh \]
Where:
- \( m \) is the mass of the ball (we will see that it cancels out later),
- \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)),
- \( h \) is the height from which the ball is dropped (10 m).

Substituting the values:
\[ PE_{\text{initial}} = mg \cdot 10 \]
\[ PE_{\text{initial}} = 10m \, \text{J} \quad \text{(since \( g \approx 10 \, \text{m/s}^2 \))} \]

### Step 2: Calculate the energy lost due to impact
The problem states that there is a 15% loss of energy due to impact. Therefore, the energy lost can be calculated as:
\[ \text{Energy lost} = 0.15 \times PE_{\text{initial}} = 0.15 \times 10m = 1.5m \, \text{J} \]

### Step 3: Calculate the remaining energy after impact
The remaining energy after the impact can be calculated by subtracting the energy lost from the initial potential energy:
\[ PE_{\text{remaining}} = PE_{\text{initial}} - \text{Energy lost} \]
\[ PE_{\text{remaining}} = 10m - 1.5m = 8.5m \, \text{J} \]

### Step 4: Calculate the height the ball will reach after the impact
The remaining potential energy will convert back into potential energy at the new height \( h' \):
\[ PE_{\text{remaining}} = mgh' \]
Setting the two expressions for potential energy equal gives:
\[ 8.5m = mgh' \]
Since \( m \) cancels out, we have:
\[ 8.5 = gh' \]
Substituting \( g = 10 \, \text{m/s}^2 \):
\[ 8.5 = 10h' \]
Solving for \( h' \):
\[ h' = \frac{8.5}{10} = 0.85 \, \text{m} \]

### Final Answer
The ball will go up to a height of **0.85 m** after one impact.

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