A bullet passes through a plank and loses `(1)/(20)` of its initial velocity. The minimum number of such planks which can just stop the bullet is

Options

A. 20
B. 19
C. 12
D. 11

Explanation

To solve the problem of how many planks a bullet can pass through before it stops, we can follow these steps:

### Step 1: Understand the velocity loss
When the bullet passes through one plank, it loses \( \frac{1}{20} \) of its initial velocity \( u \). Therefore, the remaining velocity after passing through one plank is:
\[ v = u - \frac{1}{20}u = \frac{19}{20}u \]

### Step 2: Determine the relationship of velocity loss over multiple planks
If the bullet passes through \( n \) planks, we can express the final velocity after passing through \( n \) planks as:
\[ v_n = \left(\frac{19}{20}\right)^n u \]
We want to find \( n \) such that the bullet comes to a stop, meaning \( v_n = 0 \). However, since it cannot reach exactly zero, we can consider the condition where the bullet's velocity becomes negligibly small.

### Step 3: Set up the equation for stopping condition
The bullet will stop when its velocity becomes effectively zero. For practical purposes, we can set:
\[ \left(\frac{19}{20}\right)^n u \approx 0 \]
This indicates that we need to find \( n \) such that the bullet loses all its kinetic energy.

### Step 4: Use the concept of energy loss
The kinetic energy of the bullet is given by:
\[ KE = \frac{1}{2} m u^2 \]
After passing through \( n \) planks, the kinetic energy becomes:
\[ KE_n = \frac{1}{2} m \left(\left(\frac{19}{20}\right)^n u\right)^2 = \frac{1}{2} m \left(\frac{19}{20}\right)^{2n} u^2 \]
We want \( KE_n \) to be negligible, which means:
\[ \left(\frac{19}{20}\right)^{2n} \approx 0 \]

### Step 5: Calculate the minimum number of planks
To find \( n \), we can use logarithms. Setting up the equation:
\[ \left(\frac{19}{20}\right)^n = \frac{1}{20} \]
Taking the logarithm of both sides:
\[ n \log\left(\frac{19}{20}\right) = \log\left(\frac{1}{20}\right) \]
Solving for \( n \):
\[ n = \frac{\log\left(\frac{1}{20}\right)}{\log\left(\frac{19}{20}\right)} \]

### Step 6: Calculate the values using logarithms
Using a calculator:
\[ \log\left(\frac{1}{20}\right) \approx -1.3010 \]
\[ \log\left(\frac{19}{20}\right) \approx -0.0223 \]
Thus:
\[ n \approx \frac{-1.3010}{-0.0223} \approx 58.3 \]
Since \( n \) must be a whole number, we round up to the nearest whole number, which gives \( n = 59 \).

### Conclusion
The minimum number of planks that can just stop the bullet is approximately \( 59 \).