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Can bullets and planks madden you? Maybe yes!

A bullet starts with some initial velocity \(u\).

It hits a plank which reduces it's velocity by \(\dfrac{u}{x}\).

Let \(n\) be the number of planks required to stop it.

So on further calculations, I knew that \( n = \dfrac{x^2}{2x-1}\)

Is it correct? I am unsure as

If \(x = 20\)

Then \(n = 11\) approx.

If \(x=50\)

Then \(n = 26\) approx .

This means that the larger the velocity the plank will decrease, the larger will be the no. of planks required. This is quite weird as it should be the opposite.

Now, this is how I got my formula.

Note: All symbols have their usual meaning like \(v , u , a\) and \(s\).

Let the no. of planks required be \(n\)

Let a plank decrease it's velocity \(u\) by \(\dfrac{u}{x}\)

Let thickness of one plank be \(z\)

After crossing the first plank , \(v = u - \dfrac{u}{x}\)

Now, by third equation of motion for the first plank,

\(v^2 = u^2 + 2as\)

\(\left(u - \dfrac{u}{x}\right)^2 = u^2 + 2az\)

\(\left(u - \dfrac{u}{x}\right)^2 - u^2 = 2az\)

\(\left(\dfrac{ux - u}{x}\right)^2 -u^2 = 2az\)

\(\dfrac{(ux)^2 + u^2 - 2u^2 x - (ux)^2}{x^2} = 2az\)

\(\dfrac{u^2 - 2u^2 x}{2x^2 z} = a\)

Now after crossing \(n\) planks, it's velocity will be 0.

Now by again applying third equation of motion we get,

\(v^2 = u^2 + 2as\)

Here, \(v = 0\)

\(0 = u^2 + 2 (\dfrac{u^2 - 2u^2 x}{2x^2 z}) (n)(z)\)

\(0 = u^2 + \dfrac{u^2 - 2u^2 x}{x^2} (n)\)

\(0 = \dfrac{(ux)^2 + (u^2 - 2u^2 x)n}{x^2}\)

\(0 = u^2 x^2 + (u^2 - 2u^2 x)n\)

\(0 = u^2(x^2 + [1 - 2x]n)\)

\(0 = x^2 + (1 - 2x)n\)

\(-x^2 = (1 - 2x)n\)

\(\large\dfrac{-x^2}{1 - 2x} = n\)

      OR

\(\large n=\dfrac{x^2}{2x - 1}\)

I hope that this is correct.

Note by Vinayak Bansal
2 weeks, 2 days ago

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