A bullet of mass `20 g` moving horizontally strikes a block of mass `480 g` suspended by a string of length `2 m`. If the collision is completely inelastic , the minimum velocity of bullet , so that the combined mass complete vertical circle should be

Options

A. `100 m//s`
B. `150 m//s`
C. `200 m//s`
D. `250 m//s`

Explanation

To solve the problem step by step, we will follow the principles of conservation of momentum and energy.

### Step 1: Understand the System
We have a bullet of mass \( m_1 = 20 \, \text{g} = 0.02 \, \text{kg} \) moving horizontally, striking a block of mass \( m_2 = 480 \, \text{g} = 0.48 \, \text{kg} \) that is suspended by a string. The collision is completely inelastic, meaning the bullet and block stick together after the collision.

### Step 2: Apply Conservation of Momentum
Before the collision, the block is at rest, so its initial momentum is zero. The initial momentum of the bullet is:

\[ p_{\text{initial}} = m_1 \cdot U + m_2 \cdot 0 = 0.02U \]

After the collision, the combined mass is \( m_1 + m_2 = 0.02 + 0.48 = 0.50 \, \text{kg} \) and moves with a velocity \( V \). Thus, the final momentum is:

\[ p_{\text{final}} = (m_1 + m_2) \cdot V = 0.50V \]

Setting the initial momentum equal to the final momentum gives:

\[ 0.02U = 0.50V \]

From this equation, we can solve for \( V \):

\[ V = \frac{0.02U}{0.50} = \frac{U}{25} \]

### Step 3: Determine the Minimum Velocity for Vertical Circle
To complete a vertical circle, the combined mass must have enough kinetic energy at the bottom to reach the top of the circle. At the top of the circle, the minimum velocity \( V_t \) required is given by:

\[ V_t = \sqrt{gR} \]

where \( R \) is the radius of the circle (which is equal to the length of the string, \( L = 2 \, \text{m} \)), and \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)).

Thus,

\[ V_t = \sqrt{gL} = \sqrt{10 \cdot 2} = \sqrt{20} = 2\sqrt{5} \, \text{m/s} \]

### Step 4: Set Kinetic Energy Equal to Potential Energy
The kinetic energy (KE) at the bottom must equal the potential energy (PE) at the top:

\[ \text{KE} = \frac{1}{2}(m_1 + m_2)V^2 \]
\[ \text{PE} = (m_1 + m_2)g(2L) = (0.50)(10)(4) = 20 \, \text{J} \]

Setting KE equal to PE:

\[ \frac{1}{2}(0.50)V^2 = 20 \]

Solving for \( V^2 \):

\[ 0.25V^2 = 20 \implies V^2 = \frac{20}{0.25} = 80 \implies V = \sqrt{80} = 4\sqrt{5} \, \text{m/s} \]

### Step 5: Relate \( V \) to \( U \)
From our earlier equation \( V = \frac{U}{25} \):

\[ \frac{U}{25} = 4\sqrt{5} \]

Solving for \( U \):

\[ U = 25 \cdot 4\sqrt{5} = 100\sqrt{5} \, \text{m/s} \]

### Final Step: Calculate \( U \)
Calculating \( U \):

\[ U \approx 100 \cdot 2.236 = 223.6 \, \text{m/s} \]

### Conclusion
The minimum velocity of the bullet \( U \) required for the combined mass to complete a vertical circle is approximately:

\[ U \approx 250 \, \text{m/s} \]