4,g ની બુલેટ સમક્ષિતિજ દિશામાં 300,m/s ઝડપથી | vedclass

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Question

$\begin{array}{l}
Given,\,{m_1} = 4g,{u_1} = 300m/s\
{m_2} = 0.8\,kg = 800\,g,\,{u_2} = 0\,m/s\
From\,law\,of\,conservation\,of\,momentum,\
{m_

Vedclass · Accepted Answer

$0.379\,m$

Vedclass · Answer

$\begin{array}{l}
Given,\,{m_1} = 4g,{u_1} = 300m/s\
{m_2} = 0.8\,kg = 800\,g,\,{u_2} = 0\,m/s\
From\,law\,of\,conservation\,of\,momentum,\
{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\
Let\,the\,velocity\,of\,combined\,system\
 = v\,m/s\
then,\
4 	imes 300 + 800 	imes 0 = \left( {800 + 4} ight) 	imes v\
v = \frac{{1200}}{{804}} = 1.49\,m/s
\end{array}$

$\begin{array}{l}
Now,\,\mu  = 0.3\left( {given} ight)\
a = \mu g\
a = 0.3 	imes 10\left( {take\,g = 10\,m/{s^2}} ight)\
 = 3\,m/{s^2}\
then,\,from\,{v^2} = {u^2} + 2as\
{\left( {1.49} ight)^2} = 0 + 2 	imes 3 	imes s\
s = \frac{{\left( {{{1.49}^2}} ight)}}{6}\,;\,\,s = \frac{{2.22}}{6} = 0.379\,m
\end{array}$

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