Can bullets and planks madden you? Maybe yes!

A bullet starts with some initial velocity uu.

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

Let nn be the number of planks required to stop it.

So on further calculations, I knew that n=x22x1 n = \dfrac{x^2}{2x-1}

Is it correct? I am unsure as

If x=20x = 20

Then n=11n = 11 approx.

If x=50x=50

Then n=26n = 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,av , u , a and ss.

Let the no. of planks required be nn

Let a plank decrease it's velocity uu by ux\dfrac{u}{x}

Let thickness of one plank be zz

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

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

v2=u2+2asv^2 = u^2 + 2as

(uux)2=u2+2az\left(u - \dfrac{u}{x}\right)^2 = u^2 + 2az

(uux)2u2=2az\left(u - \dfrac{u}{x}\right)^2 - u^2 = 2az

(uxux)2u2=2az\left(\dfrac{ux - u}{x}\right)^2 -u^2 = 2az

(ux)2+u22u2x(ux)2x2=2az\dfrac{(ux)^2 + u^2 - 2u^2 x - (ux)^2}{x^2} = 2az

u22u2x2x2z=a\dfrac{u^2 - 2u^2 x}{2x^2 z} = a

Now after crossing nn planks, it's velocity will be 0.

Now by again applying third equation of motion we get,

v2=u2+2asv^2 = u^2 + 2as

Here, v=0v = 0

0=u2+2(u22u2x2x2z)(n)(z)0 = u^2 + 2 (\dfrac{u^2 - 2u^2 x}{2x^2 z}) (n)(z)

0=u2+u22u2xx2(n)0 = u^2 + \dfrac{u^2 - 2u^2 x}{x^2} (n)

0=(ux)2+(u22u2x)nx20 = \dfrac{(ux)^2 + (u^2 - 2u^2 x)n}{x^2}

0=u2x2+(u22u2x)n0 = u^2 x^2 + (u^2 - 2u^2 x)n

0=u2(x2+[12x]n)0 = u^2(x^2 + [1 - 2x]n)

0=x2+(12x)n0 = x^2 + (1 - 2x)n

x2=(12x)n-x^2 = (1 - 2x)n

x212x=n\large\dfrac{-x^2}{1 - 2x} = n


n=x22x1\large n=\dfrac{x^2}{2x - 1}

I hope that this is correct.

Note by Vinayak Bansal
3 years, 5 months ago

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