A bag of sand of mass m is suspended by a rope. A bullet of mass `(m)/(20)` is fired at it with a velocity vand gets embedded into it. The velocity of the bag finally is

Options

A. `(v)/(20)xx21`
B. `(20v)/(21)`
C. `(v)/(20)`
D. `(v)/(21)`

Explanation

To solve the problem, we will use the principle of conservation of linear momentum.

### Step-by-Step Solution:

1. **Identify the masses and initial velocities**:
- Mass of the bag of sand, \( m \).
- Mass of the bullet, \( \frac{m}{20} \).
- Initial velocity of the bag, \( 0 \) (since it is at rest).
- Initial velocity of the bullet, \( v \).

2. **Calculate the initial momentum**:
The initial momentum \( p_{initial} \) is the sum of the momentum of the bullet and the bag:
\[ p_{initial} = \text{momentum of bullet} + \text{momentum of bag} = \left(\frac{m}{20} \cdot v\right) + (m \cdot 0) = \frac{m}{20} v \]

3. **Determine the final velocity after the collision**:
After the bullet embeds into the bag, the total mass becomes:
\[ m + \frac{m}{20} = \frac{20m}{20} + \frac{m}{20} = \frac{21m}{20} \]
Let the final velocity of the combined mass (bag + bullet) be \( v' \).

4. **Calculate the final momentum**:
The final momentum \( p_{final} \) is given by:
\[ p_{final} = \text{total mass} \cdot \text{final velocity} = \left(\frac{21m}{20}\right) v' \]

5. **Apply the conservation of momentum**:
According to the conservation of momentum:
\[ p_{initial} = p_{final} \]
Substituting the expressions we derived:
\[ \frac{m}{20} v = \left(\frac{21m}{20}\right) v' \]

6. **Solve for \( v' \)**:
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{20} v = \frac{21}{20} v' \]
To isolate \( v' \), multiply both sides by \( \frac{20}{21} \):
\[ v' = \frac{1}{21} v \]

### Final Answer:
The final velocity of the bag (with the bullet embedded in it) is:
\[ v' = \frac{v}{21} \]