A bullet losses 1/n of its velocity in passing through a plank. What is the least number of plancks required to stop the bullet ? (Assuming constant retardation)

Options

A. `(n ^(2))/(2n-1)`
B. `(2n^(2))/(n -1)`
C. Inifinite
D. n

Answer

To solve the problem of determining the least number of planks required to stop a bullet that loses \( \frac{1}{n} \) of its velocity when passing through each plank, we can follow these steps:

### Step-by-Step Solution:

1. **Define Variables**:
- Let the initial velocity of the bullet be \( u \).
- Let the thickness of each plank be \( s \).
- The bullet loses \( \frac{1}{n} \) of its velocity after passing through one plank.

2. **Calculate Velocity After First Plank**:
- After passing through the first plank, the velocity of the bullet becomes:
\[ v_1 = u - \frac{u}{n} = u \left(1 - \frac{1}{n}\right) = u \frac{n-1}{n} \]

3. **Use the Equation of Motion**:
- We can use the equation of motion:
\[ v^2 = u^2 + 2as \]
- Here, \( v \) is the final velocity after passing through the plank, \( u \) is the initial velocity, \( a \) is the retardation (negative acceleration), and \( s \) is the thickness of the plank.

4. **Substituting Values**:
- For the first plank, substituting \( v_1 \) into the equation gives:
\[ \left(u \frac{n-1}{n}\right)^2 = u^2 + 2(-a)s \]
- Simplifying this:
\[ \frac{u^2(n-1)^2}{n^2} = u^2 - 2as \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ 2as = u^2 - \frac{u^2(n-1)^2}{n^2} \]
- This simplifies to:
\[ 2as = u^2 \left(1 - \frac{(n-1)^2}{n^2}\right) \]
- Further simplifying:
\[ 2as = u^2 \left(\frac{n^2 - (n-1)^2}{n^2}\right) = u^2 \left(\frac{n^2 - (n^2 - 2n + 1)}{n^2}\right) = u^2 \left(\frac{2n - 1}{n^2}\right) \]

6. **Finding Total Distance for n Planks**:
- If the bullet passes through \( n \) planks, the total distance traveled is \( n \times s \).
- Setting the final velocity \( v = 0 \) after passing through \( n \) planks, we apply the equation:
\[ 0 = u^2 + 2(-a)(ns) \]
- Rearranging gives:
\[ 2ans = u^2 \]

7. **Equating the Two Expressions for 2as**:
- From the earlier step, we have:
\[ 2as = \frac{u^2(2n - 1)}{n^2} \]
- Substituting this into the equation \( 2ans = u^2 \):
\[ n \cdot \frac{u^2(2n - 1)}{n^2} = u^2 \]
- This simplifies to:
\[ \frac{n(2n - 1)}{n^2} = 1 \]

8. **Solving for n**:
- This leads to:
\[ n(2n - 1) = n^2 \]
- Simplifying gives:
\[ 2n^2 - n - n^2 = 0 \implies n^2 - n = 0 \implies n(n - 1) = 0 \]
- Thus, \( n = 1 \) or \( n = 0 \) (not valid in this context).

9. **Final Expression**:
- The least number of planks required to stop the bullet is given by:
\[ n = \frac{n^2}{2n - 1} \]

### Conclusion:
The least number of planks required to stop the bullet is \( \frac{n^2}{2n - 1} \).