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

\(\begin{array}{l}
\left( a \right)\,\,\,\,\,\,Let\,u\,be\,the\,initial\,velocity\,of\,the\,bullet\,of\,\\
\,\,\,\,\,\,\,\,\,\,\,\,mass\,m.\\
\,\,\,\,\,\,\,\,\,\,\,\,After{\rm{ passing}}\,through\,a\,plank\,of\,width\,x,\\
\,\,\,\,\,\,\,\,\,\,\,\,its\,velocity\,decreases\,to\,v.\\
\,\,\,\,\,\,\,\,\,\,\,\,\therefore \,u - v = \frac{4}{n}\,{\kern 1pt} or,\,v = u - \frac{4}{n} = \frac{{u\left( {n - 1} \right)}}{n}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,IfF\,be\,the\,retarding\,force\,applied\,d\,by\,each\,\\
\,\,\,\,\,\,\,\,\,\,\,\,\,plank,\,then\,{\rm{ using}}\,work\, - \,energy\,theorem,\\
Fx = \frac{1}{2}m{u^2} - \frac{1}{2}m{v^2} = \frac{1}{2}m{u^2} - \frac{1}{2}m{u^2}\frac{{\left( {n - {1^2}} \right)}}{{{n^2}}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}m{u^2}\left[ {\frac{{1 - \left( {n - {1^2}} \right)}}{{{n^2}}}} \right]
\end{array}\)

\(\begin{array}{l}
Fx = \frac{1}{2}m{u^2}\left( {\frac{{2n - 1}}{{{n^2}}}} \right)\\
Let\,P\,be\,the\,number\,of\,planks\,required\,to\,\\
stope\,the\,bullet.\\
Total\,{\rm{distance}}\,travelled\,by\,the\,bullet\,be\,fore\\
{\rm{ coming}}\,torest = Px\\
{\rm{Using}}\,{\rm{work - energy}}\,{\rm{theorem}}\,{\rm{aging,}}\\
F\left( {Px} \right) = \frac{1}{2}m{u^2} - 0\\
or,\,P\left( {Fx} \right) = P\left[ {\frac{1}{2}m{u^2}\frac{{\left( {2n - 1} \right)}}{{{n^2}}}} \right] = \frac{1}{2}m{u^2}\\
\therefore \,P = \,\frac{{{n^2}}}{{2n - 1}}\,
\end{array}\)