A steel ball of mass `m_(1)=1kg` moving with velocity `50ms^(-1)` collides with another ball of mass `m_(2)=200g` lying on the ground. Due the collision, the KE is lost and their internal energies change equally and `T_(1) and T_(2)` are the temperature changes of masses `m_(1) and m_(2)` respectively. If the specific heat of steel is unity and `J="4.18 J cal"^(-1)`, then

Options

A. `T_(1)=7.1^(@)C and T_(2)=1.47^(@)C`
B. `T_(1)=1.47^(@)C and T_(2)=7.1^(@)C`
C. `T_(1)=3.4^(@)C and T_(2)=17.0^(@)C`
D. `T_(1)=17.0^(@)C and T_(2)=3.4^(@)C`

Answer

To solve the problem, we need to analyze the collision of the steel ball and the other ball, considering the loss of kinetic energy and the resulting changes in internal energy.

### Step-by-Step Solution:

1. **Calculate the initial kinetic energy (KE) of the steel ball (m1)**:
\[ KE = \frac{1}{2} m_1 v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times 1 \, \text{kg} \times (50 \, \text{m/s})^2 = \frac{1}{2} \times 1 \times 2500 = 1250 \, \text{J} \]

2. **Set up the equation for the loss of kinetic energy**:
The loss of kinetic energy is converted into heat (internal energy changes) in both balls:
\[ KE_{\text{lost}} = J(q_1 + q_2) \]
Where \( q_1 \) and \( q_2 \) are the heat absorbed by the steel ball and the other ball respectively. Given that \( J = 4.18 \, \text{J/cal} \), we can express the heat absorbed in terms of temperature changes:
\[ q_1 = m_1 \cdot S \cdot (T_1 - T_{\text{initial}}) = 1 \cdot 1 \cdot (T_1 - 27) \quad \text{(assuming initial temperature is 27°C)} \]
\[ q_2 = m_2 \cdot S \cdot (T_2 - T_{\text{initial}}) = 0.2 \cdot 1 \cdot (T_2 - 27) \]
Therefore, we can rewrite the equation:
\[ 1250 = 4.18 \left( (T_1 - 27) + 0.2(T_2 - 27) \right) \]

3. **Simplify the equation**:
Expanding the equation:
\[ 1250 = 4.18 \left( T_1 - 27 + 0.2T_2 - 5.4 \right) \]
\[ 1250 = 4.18 \left( T_1 + 0.2T_2 - 32.4 \right) \]
Dividing both sides by 4.18:
\[ \frac{1250}{4.18} = T_1 + 0.2T_2 - 32.4 \]
\[ 299.52 = T_1 + 0.2T_2 - 32.4 \]
Rearranging gives us:
\[ T_1 + 0.2T_2 = 331.92 \quad \text{(Equation 1)} \]

4. **Set up the equation for equal change in internal energy**:
Since the internal energy changes equally:
\[ m_1 \cdot (T_1 - 27) = m_2 \cdot (T_2 - 27) \]
Substituting the values:
\[ 1 \cdot (T_1 - 27) = 0.2 \cdot (T_2 - 27) \]
Expanding gives:
\[ T_1 - 27 = 0.2T_2 - 5.4 \]
Rearranging gives us:
\[ T_1 - 0.2T_2 = 21.6 \quad \text{(Equation 2)} \]

5. **Solve the system of equations**:
Now we have two equations:
- Equation 1: \( T_1 + 0.2T_2 = 331.92 \)
- Equation 2: \( T_1 - 0.2T_2 = 21.6 \)

Adding both equations:
\[ (T_1 + 0.2T_2) + (T_1 - 0.2T_2) = 331.92 + 21.6 \]
\[ 2T_1 = 353.52 \]
\[ T_1 = 176.76 \, \text{°C} \]

Now substituting \( T_1 \) back into Equation 1:
\[ 176.76 + 0.2T_2 = 331.92 \]
\[ 0.2T_2 = 331.92 - 176.76 \]
\[ 0.2T_2 = 155.16 \]
\[ T_2 = 775.8 \, \text{°C} \]

### Final Values:
- \( T_1 = 176.76 \, \text{°C} \)
- \( T_2 = 775.8 \, \text{°C} \)