A bullet loses 1/ n of its velocity in passing through a plank what is the least number of planks required to stop the bullet? assuming constant retardation?

Question

Answer

Stopping a Bullet with Multiple Planks

When a bullet passes through a plank, it loses 1/n of its velocity due to constant retardation. The question is, how many planks are required to completely stop the bullet? Let's break down the solution into steps:

Step 1: Calculate the Velocity Lost in One Plank

If a bullet loses 1/n of its velocity in passing through one plank, then we can calculate the velocity lost in one plank by dividing the initial velocity of the bullet by n. Let's assume the initial velocity of the bullet is v, then the velocity lost in one plank is v/n.

Step 2: Calculate the Velocity of the Bullet after Passing through One Plank

After passing through one plank, the velocity of the bullet will be the initial velocity minus the velocity lost in one plank. So, the velocity of the bullet after passing through one plank is v - v/n, which can be simplified as v(1-1/n).

Step 3: Repeat the Process

We need to repeat the process of passing through planks until the velocity of the bullet becomes zero. So, we can keep passing the bullet through planks until the velocity of the bullet after passing through one plank becomes zero.

Step 4: Calculate the Number of Planks Required

The number of planks required to stop the bullet is the number of times we need to repeat the process. We can calculate this by finding the value of k such that:

v(1-1/n)^k = 0

Solving for k, we get:

k = log base (1-1/n) of (0/v)

Since log base (1-1/n) of (0/v) is undefined, we can say that the number of planks required to stop the bullet is infinity. However, in practical terms, we can assume that the bullet will be stopped after passing through a large number of planks, depending on the initial velocity of the bullet and the value of n.

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