400; ms^{-1} ના સમક્ષિતિજ વેગથી ગતિ કરતી 10;g | vedclass

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Question

$\begin{array}{l}
\,\,\,\,\,Mass\,of\,bullet,\,m = 10\,g = 0.01kg\
Initial\,speed\,of\,bullet,\,u = 400\,m{s^{ - 1}}\
Mass\,of\,block,\,M = 2\,kg

Vedclass · Accepted Answer

$120$

Vedclass · Answer

$\begin{array}{l}
\,\,\,\,\,Mass\,of\,bullet,\,m = 10\,g = 0.01kg\
Initial\,speed\,of\,bullet,\,u = 400\,m{s^{ - 1}}\
Mass\,of\,block,\,M = 2\,kg\
Lenght\,of\,string\,l = 5\,m\
Speed\,of\,the\,block\,after\,collision = {v_1}\
Speed\,of\,the\,bullet\,on\,emerging\,from\
block,\,v = ?\
{m{Using}}\,energy\,conseravtion\,principel\
for\,the\,block,
\end{array}$

$\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,{(KE + PE)_{{\mathop{m Re}
olimits} ference}} = {\left( {KE + PE} ight)_h}\
 \Rightarrow \,\frac{1}{2}Mv_1^2 = Mgh\,\,\,or,\,\,\,{v_1} = \sqrt {2gh} \
\,\,\,\,\,{v_1} = \sqrt {2 	imes 10 	imes 0.1}  = \sqrt 2 \,m{s^{ - 1}}\
{m{Using}}\,momentum\,conservation\,principle\
for\,block\,and\,bullet\,system,
\end{array}$

$\begin{array}{l}
{\left( {M 	imes 0 + mu} ight)_{Before\,collision}} = {\left( {M 	imes {v_1} + mv} ight)_{After\,collision}}\
 \Rightarrow \,0.01 	imes 400 = 2\sqrt 2  + 0.01 	imes v\
 \Rightarrow v = \frac{{4 - 2\sqrt 2 }}{{0.01}} = 117.15\,m{s^{ - 1}} \approx 120\,m{s^{ - 1}}
\end{array}$

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