A wooden block of mass 2m is hung with the help of a light string of length / in the vertical plane. A bullet of mass m/4 moving horizontally with velocity `V_0{V_0 = sqrt(5gl)}` penetrates the block and comes out with velocity `v_0/2` The maximum height reached by the block is (Assume string remains vertical till bullet passes through the block)

Options

A. `v_0^2 /(256g)`
B. `v_0^2 /(128g)`
C. `v_0^2 /(512g)`
D. `v_0^2 /(32g)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the initial momentum of the bullet
The bullet of mass \( m/4 \) is moving horizontally with a velocity \( V_0 = \sqrt{5gl} \). The initial momentum \( p_{initial} \) of the bullet can be calculated using the formula:
\[ p_{initial} = \text{mass} \times \text{velocity} = \frac{m}{4} \cdot V_0 \]

### Step 2: Determine the final momentum of the bullet after passing through the block
After penetrating the block, the bullet comes out with a velocity of \( \frac{V_0}{2} \). The final momentum \( p_{final} \) of the bullet is:
\[ p_{final} = \frac{m}{4} \cdot \frac{V_0}{2} = \frac{m \cdot V_0}{8} \]

### Step 3: Use conservation of momentum to find the velocity of the block after the bullet exits
According to the conservation of momentum, the initial momentum of the bullet must equal the final momentum of the bullet plus the momentum of the block. Let \( V_b \) be the velocity of the block immediately after the bullet exits. The equation can be set up as:
\[ p_{initial} = p_{final} + p_{block} \]
\[ \frac{m}{4} \cdot V_0 = \frac{m \cdot V_0}{8} + 2m \cdot V_b \]

### Step 4: Solve for the velocity of the block \( V_b \)
Rearranging the equation gives:
\[ \frac{m}{4} \cdot V_0 - \frac{m \cdot V_0}{8} = 2m \cdot V_b \]
\[ \left(\frac{2m \cdot V_0}{8} - \frac{m \cdot V_0}{8}\right) = 2m \cdot V_b \]
\[ \frac{m \cdot V_0}{8} = 2m \cdot V_b \]
Dividing both sides by \( 2m \):
\[ V_b = \frac{V_0}{16} \]

### Step 5: Calculate the maximum height reached by the block
The block will convert its kinetic energy into potential energy at the maximum height. The kinetic energy \( KE \) of the block is:
\[ KE = \frac{1}{2} \cdot 2m \cdot V_b^2 = m \cdot V_b^2 \]
Substituting \( V_b = \frac{V_0}{16} \):
\[ KE = m \cdot \left(\frac{V_0}{16}\right)^2 = m \cdot \frac{V_0^2}{256} \]

The potential energy \( PE \) at the maximum height \( h \) is given by:
\[ PE = mgh \]
Setting \( KE = PE \):
\[ m \cdot \frac{V_0^2}{256} = 2mgh \]
Cancelling \( m \) from both sides:
\[ \frac{V_0^2}{256} = 2gh \]
Solving for \( h \):
\[ h = \frac{V_0^2}{512g} \]

### Step 6: Substitute \( V_0 = \sqrt{5gl} \) into the height equation
\[ h = \frac{(\sqrt{5gl})^2}{512g} = \frac{5gl}{512g} = \frac{5l}{512} \]

Thus, the maximum height reached by the block is:
\[ \boxed{\frac{5l}{512}} \]