Question

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)

• A. n
• B. n-1
• C. n^2/(2n-1)
• D. ceil(n^2/(2n-1))

Answer 1

Answer: (D)

ceil(n^2/(2n-1))

Answer 2

Let the initial velocity of the bullet be $v_0$. After passing through one plank, the velocity becomes $v_1 = v_0(1 - \frac{1}{n}) = v_0 \frac{n-1}{n}$. Let the retardation be $a$ and the thickness of one plank be $d$. Using the equation $v^2 - u^2 = 2as$: For one plank: $v_1^2 - v_0^2 = 2(-a)d$ $(v_0 \frac{n-1}{n})^2 - v_0^2 = -2ad$ $v_0^2 \left( \frac{(n-1)^2}{n^2} - 1 \right) = -2ad$ $v_0^2 \left( \frac{n^2 - 2n + 1 - n^2}{n^2} \right) = -2ad$ $v_0^2 \left( \frac{1 - 2n}{n^2} \right) = -2ad$ $v_0^2 \left( \frac{2n - 1}{n^2} \right) = 2ad \quad (*)$

Let $N$ be the least number of planks required to stop the bullet. For $N$ planks: $0^2 - v_0^2 = 2(-a)(Nd)$ $v_0^2 = 2aNd \quad (**)$

Divide $(**)$ by $(*)$: $\frac{v_0^2}{v_0^2 \left( \frac{2n - 1}{n^2} \right)} = \frac{2aNd}{2ad}$ $\frac{n^2}{2n - 1} = N$

Since the number of planks must be an integer, we take the ceiling of the result. $N = \left\lceil \frac{n^2}{2n - 1} \right\rceil$