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 need to follow these steps:

### Step 1: Calculate the initial kinetic energy of the bullet.
The formula for kinetic energy (KE) is given by:

\[ KE = \frac{1}{2} mv^2 \]

Where:
- \( m \) = mass of the bullet = 10 g = 0.01 kg (convert grams to kilograms)
- \( v \) = speed of the bullet = 20 m/s

Substituting the values:

\[ KE = \frac{1}{2} \times 0.01 \, \text{kg} \times (20 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 0.01 \times 400 = 2 \, \text{J} \]

### Step 2: Determine the lost kinetic energy.
Since the bullet gets stuck in the ice, all of its kinetic energy is lost. Therefore, the lost kinetic energy is:

\[ \text{Lost KE} = 2 \, \text{J} \]

### Step 3: Calculate the energy transferred to the ice.
According to the problem, 50% of the lost kinetic energy goes to the ice:

\[ \text{Energy to ice} = 0.5 \times \text{Lost KE} = 0.5 \times 2 \, \text{J} = 1 \, \text{J} \]

### Step 4: Calculate the amount of ice melted.
The latent heat of fusion of ice is approximately \( L = 334 \, \text{kJ/kg} = 334,000 \, \text{J/kg} \).

To find the mass of ice melted (\( m \)), we use the formula:

\[ Q = mL \]

Where:
- \( Q \) = energy absorbed by the ice = 1 J
- \( L \) = latent heat of fusion of ice = 334,000 J/kg

Rearranging the formula to find \( m \):

\[ m = \frac{Q}{L} = \frac{1 \, \text{J}}{334,000 \, \text{J/kg}} \approx 0.00000299 \, \text{kg} = 2.99 \, \text{g} \]

### Final Answer:
The amount of ice that will melt is approximately **2.99 g**.

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