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 specific heat capacity of the ball is `800J kg^(-1) K^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial potential energy of the ball
The potential energy (PE) when the ball is at height \( h_1 = 2.0 \, \text{m} \) is given by the formula:
\[ PE = mgh_1 \]
where \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

### Step 2: Calculate the velocity of the ball just before it hits the ground
Using the conservation of energy, the potential energy at height \( h_1 \) is converted into kinetic energy (KE) just before hitting the ground:
\[ KE = \frac{1}{2} mv^2 \]
Setting the potential energy equal to kinetic energy:
\[ mgh_1 = \frac{1}{2} mv^2 \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ gh_1 = \frac{1}{2} v^2 \implies v^2 = 2gh_1 \implies v = \sqrt{2gh_1} \]

### Step 3: Calculate the velocity just after the collision
After the collision, the ball rises to a height \( h_2 = 1.5 \, \text{m} \). The velocity just after the collision can be calculated similarly:
\[ v' = \sqrt{2gh_2} \]

### Step 4: Calculate the change in kinetic energy
The change in kinetic energy (ΔKE) during the collision is given by:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} = \frac{1}{2} m v^2 - \frac{1}{2} m v'^2 \]
Factoring out \( \frac{1}{2} m \):
\[ \Delta KE = \frac{1}{2} m (v^2 - v'^2) \]

### Step 5: Calculate the energy converted to thermal energy
According to the problem, \( 40\% \) of the mechanical energy lost goes into thermal energy of the ball:
\[ \text{Thermal energy} = 0.4 \times \Delta KE \]

### Step 6: Relate thermal energy to temperature change
The thermal energy gained by the ball can also be expressed in terms of the specific heat capacity \( c \) and the change in temperature \( \Delta T \):
\[ \text{Thermal energy} = mc\Delta T \]
Setting the two expressions for thermal energy equal gives:
\[ 0.4 \times \Delta KE = mc\Delta T \]

### Step 7: Solve for the change in temperature
Rearranging the equation to find \( \Delta T \):
\[ \Delta T = \frac{0.4 \times \Delta KE}{mc} \]

### Step 8: Substitute known values and calculate
Now, we can substitute the values we calculated and the specific heat capacity \( c = 800 \, \text{J/kg/K} \) to find \( \Delta T \).

### Final Calculation
1. Calculate \( v \) and \( v' \):
- \( v = \sqrt{2 \times 10 \times 2} = \sqrt{40} \, \text{m/s} \)
- \( v' = \sqrt{2 \times 10 \times 1.5} = \sqrt{30} \, \text{m/s} \)

2. Calculate \( \Delta KE \):
- \( \Delta KE = \frac{1}{2} m (40 - 30) = \frac{1}{2} m \times 10 = 5m \)

3. Substitute into the equation for \( \Delta T \):
\[ \Delta T = \frac{0.4 \times 5m}{mc} = \frac{0.4 \times 5}{800} = \frac{2}{800} = 0.0025 \, \text{K} \]

Thus, the rise in temperature of the ball is:
\[ \Delta T = 0.0025 \, \text{K} \text{ or } 2.5 \times 10^{-3} \, \text{K} \]