A `15 gm` bullet is fired horizontally into a `3 kg` block of wood suspended by a string. The bullet sticks in the block, and the impact causes the block to swing `10 cm` above the initial level. The velocity of the bullet nearly is (in `ms^(-1)`)

Options

A. 281
B. 326
C. 184
D. 58

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a bullet of mass \( m_b = 15 \, \text{g} = 0.015 \, \text{kg} \) that strikes a block of wood of mass \( m = 3 \, \text{kg} \). After the bullet embeds itself in the block, they swing together to a height of \( h = 10 \, \text{cm} = 0.1 \, \text{m} \).

### Step 2: Calculate the potential energy gained
When the block and bullet swing to a height \( h \), they gain potential energy (PE) given by the formula:
\[ PE = (m_b + m)gh \]
where \( g = 9.8 \, \text{m/s}^2 \).

Substituting the values:
\[ PE = (0.015 + 3) \cdot 9.8 \cdot 0.1 = 3.015 \cdot 9.8 \cdot 0.1 \]
\[ PE = 3.015 \cdot 0.98 = 2.9537 \, \text{J} \]

### Step 3: Relate potential energy to kinetic energy
The potential energy gained by the block and bullet system is equal to the kinetic energy (KE) lost by the bullet just before the collision. The kinetic energy is given by:
\[ KE = \frac{1}{2}(m_b + m)v^2 \]
Setting \( KE = PE \):
\[ \frac{1}{2}(m_b + m)v^2 = PE \]

### Step 4: Solve for the velocity \( v \)
Substituting the values:
\[ \frac{1}{2}(0.015 + 3)v^2 = 2.9537 \]
\[ \frac{1}{2}(3.015)v^2 = 2.9537 \]
\[ 3.015v^2 = 5.9074 \]
\[ v^2 = \frac{5.9074}{3.015} \approx 1.96 \]
\[ v \approx \sqrt{1.96} \approx 1.4 \, \text{m/s} \]

### Step 5: Use conservation of momentum to find the bullet's initial velocity
Using the conservation of momentum before and after the collision:
\[ m_b v_b = (m_b + m)v \]
where \( v_b \) is the initial velocity of the bullet.

Substituting the values:
\[ 0.015 v_b = (0.015 + 3)(1.4) \]
\[ 0.015 v_b = 3.015 \cdot 1.4 \]
\[ 0.015 v_b = 4.221 \]
\[ v_b = \frac{4.221}{0.015} \approx 281.4 \, \text{m/s} \]

### Final Answer
The velocity of the bullet before impact is approximately \( 281.4 \, \text{m/s} \).

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