A bullet of mass 10 g moving with a speed of `20 m//s` hits an ice block of mass 990 g kept on a frictionless floor and gets stuck in it. How much ice will melt if `50%` of the lost kinetic energy goes to ice?
(Temperature of ice block `= 0^@C`.)

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
We have a bullet of mass \( m_1 = 10 \, \text{g} \) moving with a speed \( v_1 = 20 \, \text{m/s} \) that hits an ice block of mass \( m_2 = 990 \, \text{g} \) at rest. The bullet gets stuck in the ice, and we need to find out how much ice will melt if \( 50\% \) of the lost kinetic energy goes to the ice.

### Step 2: Convert Mass to Kilograms
Convert the masses from grams to kilograms for consistency in SI units:
- \( m_1 = 10 \, \text{g} = 0.01 \, \text{kg} \)
- \( m_2 = 990 \, \text{g} = 0.99 \, \text{kg} \)

### Step 3: Calculate the Final Velocity After Collision
Using the conservation of momentum:
\[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f \]
Since the ice block is at rest, \( v_2 = 0 \):
\[ 0.01 \times 20 + 0.99 \times 0 = (0.01 + 0.99) v_f \]
\[ 0.2 = 1 \times v_f \implies v_f = 0.2 \, \text{m/s} \]

### Step 4: Calculate the Initial and Final Kinetic Energy
Calculate the initial kinetic energy \( KE_i \) of the bullet:
\[ KE_i = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 0.01 \times (20)^2 = 2 \, \text{J} \]

Calculate the final kinetic energy \( KE_f \) of the combined system:
\[ KE_f = \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} \times 1 \times (0.2)^2 = 0.02 \, \text{J} \]

### Step 5: Calculate the Lost Kinetic Energy
The lost kinetic energy \( KE_{lost} \) is:
\[ KE_{lost} = KE_i - KE_f = 2 \, \text{J} - 0.02 \, \text{J} = 1.98 \, \text{J} \]

### Step 6: Calculate the Energy Transferred to Ice
Since \( 50\% \) of the lost kinetic energy goes to the ice:
\[ E_{ice} = 0.5 \times KE_{lost} = 0.5 \times 1.98 \, \text{J} = 0.99 \, \text{J} \]

### Step 7: Convert Energy to Calories
Convert joules to calories (1 calorie = 4.18 joules):
\[ E_{ice} \text{ (in calories)} = \frac{0.99}{4.18} \approx 0.236 \, \text{cal} \]

### Step 8: Calculate the Mass of Ice Melted
The latent heat of fusion of ice is \( 80 \, \text{cal/g} \). The mass of ice melted \( m_{melted} \) is:
\[ m_{melted} = \frac{E_{ice}}{80} = \frac{0.236}{80} \approx 0.00295 \, \text{g} \]

### Final Answer
Approximately \( 0.00295 \, \text{g} \) of ice will melt.

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