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  • A bullet looses <img alt="\left ( \frac{1}{n} \right )^{th}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cdpi%7B100%7D%20%5Cleft%20%28%20%5Cfrac%7B1%7D%7Bn%7D%20%5Cright%20%29%5E%7Bth%7D

A bullet looses \left ( \frac{1}{n} \right )^{th}of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be :

Option: 1

\frac{n^{2}}{2n-1}


Option: 2

\frac{2n^{2}}{n-1}


Option: 3

Infinite


Option: 4

n


Answers (1)

best_answer

 Let the bullet be traveling at a speed v before entering the first plank.

If it loses (\frac{1}{n})^{th} velocity its new velocity will become v-\frac{v}{n}

Kinetic energy lost

\Delta \mathrm{E}=\frac{1}{2} \mathrm{m} v^2-\frac{1}{2} \mathrm{~m}\left(v-\frac{v}{\mathrm{n}}\right)^2=\frac{1}{2} m v^2\left(1-\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)^2\right)

Now, let N be the number of planks after which the velocity of the bullet becomes 0.
Thus we get
\mathrm{N} \times(\text{ Energy lost in one plank} )= \text{Total Energy}

or
 
\mathrm{N} \times \Delta \mathrm{E}=\frac{1}{2} \mathrm{mv}^2

\begin{aligned} &\frac{\mathrm{mv}}{2 \Delta \mathrm{E}}=\mathrm{N}=\frac{1}{1-\left(\frac{\mathrm{n}-1}{\mathrm{n}}\right)^2}\\ &\Rightarrow \mathrm{N}=\frac{\mathrm{n}^2}{2 \mathrm{n}-1} \end{aligned} 

Posted by

Devendra Khairwa

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