Two metal spheres (10 kg and 30 kg) moving towards each other at speeds 10 m/s and 20 m/s respectivley.They collide inelasticity. What is the rise in temperature of the combined body after collision if al the energy lost appears in the form of heat. Given that specific heat of the metal of spheres is `0.03 "cal"//g^(@)C`

Answer

To solve the problem step by step, we will follow the principles of conservation of momentum and energy.

### Step 1: Calculate the final velocity after the inelastic collision.

**Given:**
- Mass of sphere 1, \( m_1 = 10 \, \text{kg} \)
- Velocity of sphere 1, \( v_1 = 10 \, \text{m/s} \)
- Mass of sphere 2, \( m_2 = 30 \, \text{kg} \)
- Velocity of sphere 2, \( v_2 = -20 \, \text{m/s} \) (negative because it is moving towards sphere 1)

**Using conservation of momentum:**
\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) V \]
Substituting the values:
\[ 10 \times 10 + 30 \times (-20) = (10 + 30) V \]
\[ 100 - 600 = 40 V \]
\[ -500 = 40 V \]
\[ V = -\frac{500}{40} = -12.5 \, \text{m/s} \]

### Step 2: Calculate the initial kinetic energy before the collision.

**Using the formula for kinetic energy:**
\[ KE = \frac{1}{2} m v^2 \]
Calculating for both spheres:
\[ KE_1 = \frac{1}{2} \times 10 \times (10)^2 = 500 \, \text{J} \]
\[ KE_2 = \frac{1}{2} \times 30 \times (20)^2 = 6000 \, \text{J} \]
**Total initial kinetic energy:**
\[ KE_{initial} = KE_1 + KE_2 = 500 + 6000 = 6500 \, \text{J} \]

### Step 3: Calculate the final kinetic energy after the collision.

Using the final velocity \( V = -12.5 \, \text{m/s} \):
\[ KE_{final} = \frac{1}{2} (m_1 + m_2) V^2 = \frac{1}{2} \times 40 \times (-12.5)^2 \]
Calculating:
\[ KE_{final} = \frac{1}{2} \times 40 \times 156.25 = 1250 \, \text{J} \]

### Step 4: Calculate the energy lost in the collision.

**Energy lost:**
\[ \Delta KE = KE_{initial} - KE_{final} = 6500 - 1250 = 5250 \, \text{J} \]

### Step 5: Convert energy lost to calories.

**Using the conversion factor \( 1 \, \text{cal} = 4.184 \, \text{J} \):**
\[ \Delta KE_{cal} = \frac{5250}{4.184} \approx 1257.4 \, \text{cal} \]

### Step 6: Calculate the rise in temperature.

**Using the formula:**
\[ Q = mc\Delta T \]
Where:
- \( Q = 1257.4 \, \text{cal} \)
- \( m = 40 \, \text{kg} = 40000 \, \text{g} \)
- \( c = 0.03 \, \text{cal/g}^\circ C \)

Rearranging for \( \Delta T \):
\[ \Delta T = \frac{Q}{mc} \]
Substituting the values:
\[ \Delta T = \frac{1257.4}{40000 \times 0.03} = \frac{1257.4}{1200} \approx 1.05^\circ C \]

### Final Answer:
The rise in temperature of the combined body after the collision is approximately **1.05°C**.

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