A bullet looses {left( {frac{1}{n}} right)^{t | vedclass

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Question

$\begin{array}{l}
\left( a ight)\,\,\,\,\,\,Let\,u\,be\,the\,initial\,velocity\,of\,the\,bullet\,of\,\
\,\,\,\,\,\,\,\,\,\,\,\,mass\,m.\
\,\,\,

Vedclass · Accepted Answer

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

Vedclass · Answer

$\begin{array}{l}
\left( a ight)\,\,\,\,\,\,Let\,u\,be\,the\,initial\,velocity\,of\,the\,bullet\,of\,\
\,\,\,\,\,\,\,\,\,\,\,\,mass\,m.\
\,\,\,\,\,\,\,\,\,\,\,\,After{m{ passing}}\,through\,a\,plank\,of\,width\,x,\
\,\,\,\,\,\,\,\,\,\,\,\,its\,velocity\,decreases\,to\,v.\
\,\,\,\,\,\,\,\,\,\,\,\,	herefore \,u - v = \frac{4}{n}\,{\kern 1pt} or,\,v = u - \frac{4}{n} = \frac{{u\left( {n - 1} ight)}}{n}\
\,\,\,\,\,\,\,\,\,\,\,\,\,IfF\,be\,the\,retarding\,force\,applied\,d\,by\,each\,\
\,\,\,\,\,\,\,\,\,\,\,\,\,plank,\,then\,{m{ using}}\,work\, - \,energy\,theorem,\
Fx = \frac{1}{2}m{u^2} - \frac{1}{2}m{v^2} = \frac{1}{2}m{u^2} - \frac{1}{2}m{u^2}\frac{{\left( {n - {1^2}} ight)}}{{{n^2}}}\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2}m{u^2}\left[ {\frac{{1 - \left( {n - {1^2}} ight)}}{{{n^2}}}} ight]
\end{array}$

$\begin{array}{l}
Fx = \frac{1}{2}m{u^2}\left( {\frac{{2n - 1}}{{{n^2}}}} ight)\
Let\,P\,be\,the\,number\,of\,planks\,required\,to\,\
stope\,the\,bullet.\
Total\,{m{distance}}\,travelled\,by\,the\,bullet\,be\,fore\
{m{ coming}}\,torest = Px\
{m{Using}}\,{m{work - energy}}\,{m{theorem}}\,{m{aging,}}\
F\left( {Px} ight) = \frac{1}{2}m{u^2} - 0\
or,\,P\left( {Fx} ight) = P\left[ {\frac{1}{2}m{u^2}\frac{{\left( {2n - 1} ight)}}{{{n^2}}}} ight] = \frac{1}{2}m{u^2}\
	herefore \,P = \,\frac{{{n^2}}}{{2n - 1}}\,
\end{array}$

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

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