A ball is dropped on a floor from a height of `2.0m`. After the collision it rises up to a height of `1.5m`. Assume that `40%` of the mechanical energy lost goes as thermal energy into the ball.Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is `800J K^(-1)`

Explanation

To solve the problem step-by-step, we will follow these calculations:

### Step 1: Calculate the initial and final heights
- The ball is dropped from a height \( h_1 = 2.0 \, m \).
- After the collision, it rises to a height \( h_2 = 1.5 \, m \).

### Step 2: Calculate the potential energy at the initial and final heights
- The potential energy (PE) at height \( h_1 \) is given by:
\[ PE_1 = mgh_1 = mg \times 2.0 \]
- The potential energy at height \( h_2 \) is:
\[ PE_2 = mgh_2 = mg \times 1.5 \]

### Step 3: Calculate the change in potential energy
- The change in potential energy (which corresponds to the mechanical energy lost) is:
\[ \Delta PE = PE_1 - PE_2 = mg \times 2.0 - mg \times 1.5 = mg(2.0 - 1.5) = mg \times 0.5 \]

### Step 4: Calculate the energy converted to thermal energy
- According to the problem, 40% of the mechanical energy lost goes into thermal energy:
\[ \text{Thermal Energy} = 0.40 \times \Delta PE = 0.40 \times mg \times 0.5 = 0.20 mg \]

### Step 5: Relate thermal energy to temperature rise
- The thermal energy absorbed by the ball can also be expressed in terms of its heat capacity and temperature change:
\[ \text{Thermal Energy} = C \Delta T \]
where \( C = 800 \, J/K \) is the heat capacity of the ball and \( \Delta T \) is the rise in temperature.

### Step 6: Set the equations equal to each other
- Equating the two expressions for thermal energy gives:
\[ 0.20 mg = C \Delta T \]

### Step 7: Solve for the temperature rise \( \Delta T \)
- Rearranging the equation:
\[ \Delta T = \frac{0.20 mg}{C} \]

### Step 8: Substitute values
- We can substitute \( C = 800 \, J/K \):
\[ \Delta T = \frac{0.20 mg}{800} \]

### Step 9: Simplify the expression
- The mass \( m \) cancels out:
\[ \Delta T = \frac{0.20 g}{800} \]
- Using \( g = 10 \, m/s^2 \):
\[ \Delta T = \frac{0.20 \times 10}{800} = \frac{2}{800} = 0.0025 \, K \]

### Final Answer
The rise in the temperature of the ball in the collision is \( \Delta T = 0.0025 \, K \).

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