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Question-64688




Question Number 64688 by peter frank last updated on 20/Jul/19
Answered by mr W last updated on 20/Jul/19
M_0 =original mass  −u(dm/dt)=m(dv/dt)  −u∫_M_0  ^( m) (dm/m)=∫_v_0  ^( v) dv  −u(ln m−ln M_0 )=v−v_0   ⇒v=v_0 −u ln (m/M_0 )  with m=(M_0 /2)  ⇒v=20−5 ln (1/2)=23.47 m/s
$${M}_{\mathrm{0}} ={original}\:{mass} \\ $$$$−{u}\frac{{dm}}{{dt}}={m}\frac{{dv}}{{dt}} \\ $$$$−{u}\int_{{M}_{\mathrm{0}} } ^{\:{m}} \frac{{dm}}{{m}}=\int_{{v}_{\mathrm{0}} } ^{\:{v}} {dv} \\ $$$$−{u}\left(\mathrm{ln}\:{m}−\mathrm{ln}\:{M}_{\mathrm{0}} \right)={v}−{v}_{\mathrm{0}} \\ $$$$\Rightarrow{v}={v}_{\mathrm{0}} −{u}\:\mathrm{ln}\:\frac{{m}}{{M}_{\mathrm{0}} } \\ $$$${with}\:{m}=\frac{{M}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\Rightarrow{v}=\mathrm{20}−\mathrm{5}\:\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{23}.\mathrm{47}\:{m}/{s} \\ $$
Commented by peter frank last updated on 20/Jul/19
thank you
$${thank}\:{you} \\ $$

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