A bullet of mass 10 g moving horizontally with a velocity of `400ms^(-1)` strikes a wooden block of mass 2kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

Options

A. `100ms^(-1)`
B. `80ms^(-1)`
C. `120ms^(-1)`
D. `160ms^(-1)`

Answer

To solve the problem, we will use the principles of conservation of momentum and energy.

### Step 1: Understand the scenario
A bullet of mass \( m_b = 10 \, \text{g} = 0.01 \, \text{kg} \) is moving with a velocity \( v_b = 400 \, \text{m/s} \) and strikes a wooden block of mass \( m_B = 2 \, \text{kg} \). After the bullet passes through the block, the block rises a vertical distance of \( h = 10 \, \text{cm} = 0.1 \, \text{m} \).

### Step 2: Calculate the potential energy gained by the block
When the block rises to a height \( h \), it gains potential energy given by:
\[ PE = m_B \cdot g \cdot h \]
where \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).

Substituting the values:
\[ PE = 2 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 \cdot 0.1 \, \text{m} = 1.962 \, \text{J} \]

### Step 3: Relate the potential energy to the kinetic energy of the block
The potential energy gained by the block is equal to the kinetic energy it had just after the bullet passed through it. The kinetic energy (KE) of the block can be expressed as:
\[ KE = \frac{1}{2} m_B v_B^2 \]
where \( v_B \) is the velocity of the block just after the bullet passes through it.

Setting the potential energy equal to the kinetic energy:
\[ 1.962 \, \text{J} = \frac{1}{2} (2 \, \text{kg}) v_B^2 \]

### Step 4: Solve for the velocity of the block
Rearranging the equation:
\[ v_B^2 = \frac{1.962 \, \text{J} \cdot 2}{2 \, \text{kg}} = 1.962 \, \text{m}^2/\text{s}^2 \]
\[ v_B = \sqrt{1.962} \approx 1.4 \, \text{m/s} \]

### Step 5: Apply conservation of momentum
Before the collision, the momentum of the system is given by the bullet alone:
\[ p_{\text{initial}} = m_b v_b = 0.01 \, \text{kg} \cdot 400 \, \text{m/s} = 4 \, \text{kg m/s} \]

After the bullet passes through the block, the momentum is the sum of the momenta of the bullet and the block:
\[ p_{\text{final}} = m_b v_{b'} + m_B v_B \]
where \( v_{b'} \) is the velocity of the bullet after it exits the block.

Setting the initial momentum equal to the final momentum:
\[ 4 = 0.01 v_{b'} + 2 \cdot 1.4 \]
\[ 4 = 0.01 v_{b'} + 2.8 \]

### Step 6: Solve for the final velocity of the bullet
Rearranging gives:
\[ 0.01 v_{b'} = 4 - 2.8 = 1.2 \]
\[ v_{b'} = \frac{1.2}{0.01} = 120 \, \text{m/s} \]

### Conclusion
The speed of the bullet after it emerges horizontally from the block is approximately \( 120 \, \text{m/s} \).

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