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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


Option: 2


Option: 3


Option: 4


Answers (1)


 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}

\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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