A bullet loses ` 1//20` of its velocity in passing through a plank. What is the least number of plank required to stop the bullet .

Options

A. ` 8`
B. `7`
C. ` 11`
D. ` 14`

Explanation

To solve the problem of how many planks are required to stop a bullet that loses \( \frac{1}{20} \) of its velocity when passing through each plank, we can follow these steps:

### Step 1: Define the initial velocity
Let the initial velocity of the bullet be \( u \).

### Step 2: Determine the velocity after passing through one plank
When the bullet passes through one plank, it loses \( \frac{1}{20} \) of its velocity. Therefore, the velocity after passing through one plank, \( v \), can be calculated as:
\[ v = u - \frac{u}{20} = u \left(1 - \frac{1}{20}\right) = u \left(\frac{19}{20}\right) \]

### Step 3: Calculate the change in velocity
The change in velocity after passing through one plank is:
\[ \Delta v = u - v = u - u \left(\frac{19}{20}\right) = \frac{u}{20} \]

### Step 4: Use the kinematic equation
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where \( v \) is the final velocity (0, since we want to stop the bullet), \( u \) is the initial velocity, \( a \) is the acceleration (which will be negative since the bullet is decelerating), and \( s \) is the distance traveled.

### Step 5: Set up the equation for multiple planks
If \( n \) is the number of planks, the total distance \( s \) the bullet travels through \( n \) planks can be expressed as:
\[ s = n \cdot s_1 \]
where \( s_1 \) is the distance the bullet travels through one plank.

### Step 6: Substitute values into the kinematic equation
Substituting the values into the kinematic equation:
\[ 0 = \left(\frac{19}{20}u\right)^2 + 2(-a)(ns) \]
We know that \( a \) can be expressed in terms of the change in velocity and distance. The acceleration \( a \) can be calculated from the change in velocity:
\[ a = \frac{\Delta v}{s_1} = \frac{\frac{u}{20}}{s_1} \]
Thus, we can rewrite the kinematic equation as:
\[ 0 = \left(\frac{19}{20}u\right)^2 - 2\left(\frac{u}{20s_1}\right)(ns_1) \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ \left(\frac{19}{20}u\right)^2 = \frac{2u}{20}ns_1 \]
This simplifies to:
\[ \frac{361}{400}u^2 = \frac{2u}{20}ns_1 \]

### Step 8: Solve for \( n \)
To find \( n \), we can isolate it:
\[ n = \frac{361}{400} \cdot \frac{20}{2s_1} = \frac{361 \cdot 10}{400 \cdot s_1} \]

### Step 9: Calculate the minimum number of planks
Since \( s_1 \) is the distance through one plank, we can assume \( s_1 = 1 \) for simplicity. Thus:
\[ n = \frac{3610}{400} = 9.025 \]
Since \( n \) must be an integer, we round up to the next whole number:
\[ n = 10 \]

### Step 10: Conclusion
To ensure the bullet stops completely, we need at least \( 11 \) planks.