A body of mass M moves in outer space with vleocity v. It is desired to break the body into two parts so that the mass of one part is one-tenth of the total mass. After the explosion, the heavier part comes to rest whole the lighter part continues to move in the original direction of motion. The velocity of the small part will be

Options

A. `v`
B. `((v)/(2))`
C. `((v)/(10))`
D. `10 v`

Explanation

To solve the problem step by step, we will apply the principle of conservation of momentum.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Total mass of the body, \( M \).
- Initial velocity of the body, \( v \).
- The mass of the lighter part after the explosion is \( \frac{M}{10} \).
- Therefore, the mass of the heavier part is \( M - \frac{M}{10} = \frac{9M}{10} \).

2. **Apply the Conservation of Momentum:**
- The initial momentum of the system (before the explosion) is given by:
\[ P_{\text{initial}} = M \cdot v \]
- After the explosion, the heavier part comes to rest, so its velocity is \( 0 \).
- The momentum of the lighter part (mass \( \frac{M}{10} \)) moving with velocity \( V \) is:
\[ P_{\text{final}} = \left(\frac{M}{10}\right) \cdot V \]

3. **Set Up the Equation:**
- According to the conservation of momentum:
\[ P_{\text{initial}} = P_{\text{final}} \]
- Therefore, we can write:
\[ M \cdot v = 0 + \left(\frac{M}{10}\right) \cdot V \]

4. **Simplify the Equation:**
- Since the mass \( M \) is present on both sides, we can cancel it out (assuming \( M \neq 0 \)):
\[ v = \frac{1}{10} V \]

5. **Solve for \( V \):**
- Rearranging the equation gives:
\[ V = 10v \]

6. **Conclusion:**
- The velocity of the lighter part after the explosion is \( 10v \).

### Final Answer:
The velocity of the lighter part after the explosion is \( 10v \).