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Question Number 119645 by aurpeyz last updated on 26/Oct/20
A particle of mass m_1  lies on a smooth  horizontal table and its connected to  a freely hanging particle of mass m_2   by a light inextensible string passing  over a smooth fixed pulley situated  at the edge of the table. obtain the  expression for the time taken for  mass m_(1 ) to reach the edge of the table.
$${A}\:{particle}\:{of}\:{mass}\:{m}_{\mathrm{1}} \:{lies}\:{on}\:{a}\:{smooth} \\ $$$${horizontal}\:{table}\:{and}\:{its}\:{connected}\:{to} \\ $$$${a}\:{freely}\:{hanging}\:{particle}\:{of}\:{mass}\:{m}_{\mathrm{2}} \\ $$$${by}\:{a}\:{light}\:{inextensible}\:{string}\:{passing} \\ $$$${over}\:{a}\:{smooth}\:{fixed}\:{pulley}\:{situated} \\ $$$${at}\:{the}\:{edge}\:{of}\:{the}\:{table}.\:{obtain}\:{the} \\ $$$${expression}\:{for}\:{the}\:{time}\:{taken}\:{for} \\ $$$${mass}\:{m}_{\mathrm{1}\:} {to}\:{reach}\:{the}\:{edge}\:{of}\:{the}\:{table}. \\ $$
Commented by aurpeyz last updated on 26/Oct/20
pls help
$${pls}\:{help} \\ $$
Answered by TANMAY PANACEA last updated on 26/Oct/20
m_2 gāˆ’T=m_2 a  T=tensin in string  T=m_1 a  a=((m_2 g)/(m_1 +m_2 ))  L=(1/2)at^2 ā†’t=(āˆš((2L(m_1 +m_2 ))/(m_2 g)))   L=distance between edge and initial position  of m_1
$${m}_{\mathrm{2}} {g}āˆ’{T}={m}_{\mathrm{2}} {a}\:\:{T}={tensin}\:{in}\:{string} \\ $$$${T}={m}_{\mathrm{1}} {a} \\ $$$${a}=\frac{{m}_{\mathrm{2}} {g}}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} } \\ $$$${L}=\frac{\mathrm{1}}{\mathrm{2}}{at}^{\mathrm{2}} \rightarrow{t}=\sqrt{\frac{\mathrm{2}{L}\left({m}_{\mathrm{1}} +{m}_{\mathrm{2}} \right)}{{m}_{\mathrm{2}} {g}}}\: \\ $$$${L}={distance}\:{between}\:{edge}\:{and}\:{initial}\:{position} \\ $$$${of}\:{m}_{\mathrm{1}} \\ $$
Commented by aurpeyz last updated on 28/Oct/20
thanks
$${thanks} \\ $$
Commented by TANMAY PANACEA last updated on 28/Oct/20
most welcome
$${most}\:{welcome} \\ $$

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