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Question Number 43187 by ajfour last updated on 08/Sep/18

Commented by ajfour last updated on 08/Sep/18

$$massof(toy+fuel)=M+m$$$$exhaustspeed\left(relative\right)=u$$$$frictioncoefficient=\mu$$$$Find\left(\mathit{i}\right)\mathit{m}\mathit{a}\mathit{x}\mathit{i}\mathit{m}\mathit{u}\mathit{m}\mathit{v}\mathit{e}\mathit{l}\mathit{o}\mathit{c}\mathit{i}\mathit{t}\mathit{y}$$$$\left(\mathit{i}\mathit{i}\right)\mathit{D}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{n}\mathit{e}\mathit{h}\mathit{o}\mathit{w}\mathit{f}\mathit{a}\mathit{r}\mathit{c}\mathit{a}\mathit{n}\mathit{i}\mathit{t}\mathit{g}\mathit{o}.$$

Answered by ajfour last updated on 08/Sep/18

$$Case1:(fueloveratt=t_1,$$$$toystopsatt=t_2)$$$$letinitialmassoffuelbe{\mathit{m}}_0$$$$M+m_0=M_0$$$$(M+m)\frac{dv}{dt}=-\mu N+q\left(u\cos \theta \right)$$$$N=(M+m)g+qu\sin \theta$$$$\Rightarrow \frac{dv}{dt}=-\mu g+\frac{qu(\cos \theta -\mu \sin \theta )}{M+m_0-qt}$$$$let\lambda =\cos \theta -\mu \sin \theta$$$$\Rightarrow {\int }_0^{v}dv=-\mu gt+\lambda u\ln \left(\frac{M_0}{M_0-qt}\right)$$$$displacement\mathit{s}forthisaccelerating$$$$periodis$$$$\mathit{s}=-\frac{\mathit{\mu }{\mathit{g}\mathit{t}}^2}{2}+\mathit{\lambda }\mathit{u}{\int }_0^{\mathit{t}}\ln \left(\frac{{\mathit{M}}_0}{{\mathit{M}}_0-\mathit{q}\mathit{t}}\right)\mathit{d}\mathit{t}$$$${\int }_0^t\ln \left(\frac{M_0}{M_0-qt}\right)dt=t\ln \left(\frac{M_0}{M_0-qt}\right){\mid }_0^t$$$$-{\int }_0^{t}t\left(\frac{q}{M_0-qt}\right)dt$$$$=t\ln \left(\frac{M_0}{M_0-qt}\right)+t+\frac{M_0}{q}\ln \left(\frac{M_0-qt}{M_0}\right)$$$$\Rightarrow {\mathit{s}}_1=\lambda u[\frac{m_0}{q}-\frac{(M_0-m_0)}{q}\ln \left(\frac{M_0}{M}\right)]$$$$-\frac{\mu {gm}_0^2}{2q^2}$$$$Now{t}_1=\frac{m_0}{q}$$$${\mathit{v}}_{\mathit{m}\mathit{a}\mathit{x}}=-\frac{\mathit{\mu }{\mathit{g}\mathit{m}}_0}{\mathit{q}}+\mathit{\lambda }\mathit{u}\ln \left(\frac{\mathit{M}+{\mathit{m}}_0}{\mathit{M}}\right)$$$$$$$$hereafter$$$${\mathit{s}}_2=\frac{{\mathit{v}}_{\mathit{m}\mathit{a}\mathit{x}}^2}{2\mathit{\mu }\mathit{g}}$$$$Totaldistancetraveledis$$$$={\mathit{s}}_1+{\mathit{s}}_2.$$

Commented by MrW3 last updated on 08/Sep/18

$$allcorrectsir!$$

Answered by MrW3 last updated on 08/Sep/18

$$let{m}_0=massoffuelatt=0$$$$$$$${let}^{\prime }ssayatt=t_{1}tankisempty.$$$$anditstopsatt=t_2$$$$0⩽t⩽t_1:motionwithvariablemassM+m$$$$t_1⩽t⩽t_2:motionwithconstantmassM$$$$$$$$infollowingitisassumedthatthe$$$$frictionissosmallaswellasuand$$$$qaresolargethatthetoystartsto$$$$moveatt=0,seecondition\left(I\right).$$$$$$$$(M+m)g-N-u\sin \theta \frac{d(M+m)}{dt}=0$$$$\Rightarrow N=(M+m)g+uq\sin \theta$$$$-N\mu -u\cos \theta \frac{d(M+m)}{dt}=(M+m)\frac{dv}{dt}$$$$\Rightarrow -(M+m)\mu g+(\cos \theta -\mu \sin \theta )uq=(M+m)\frac{dv}{dt}$$$$\Rightarrow (M+m)(\mu g+\frac{dv}{dt})=uq(\cos \theta -\mu \sin \theta )$$$$m=m_0-qt$$$$\Rightarrow (M+m_0-qt)(\mu g+\frac{dv}{dt})=uq(\cos \theta -\mu \sin \theta )$$$$\Rightarrow \frac{dv}{dt}=\frac{uq(\cos \theta -\mu \sin \theta )}{M+m_0-qt}-\mu g$$$$let{M}_0=M+m_0,\lambda =\cos \theta -\mu \sin \theta$$$$\Rightarrow a=\frac{dv}{dt}=\frac{uq\lambda }{M_0-qt}-\mu g$$$$sothatthetoystartstomoveatt=0,$$$$\frac{uq\lambda }{M_0}-\mu g⩾0$$$$\Rightarrow uq⩾\frac{\mu {gM}_0}{\lambda }=\frac{\mu g(M+m_0)}{\cos \theta -\mu \sin \theta }...\left(I\right)$$$$$$$${\int }_0^{v}dv={\int }_0^t(\frac{uq\lambda }{M_0-qt}-\mu g)dt$$$$\Rightarrow v={[-u\lambda \ln (M_0-qt)-\mu gt]}_0^t$$$$\Rightarrow v\left(t\right)=u\lambda \ln \frac{M_0}{M_0-qt}-\mu gt$$$$thisisvalidfort⩽t_1=\frac{m_0}{q}=momentwhentankisempty$$$$v_1=v\left(t_1\right)=u\lambda \ln \frac{M_0}{M}-\frac{\mu {gm}_0}{q}$$$$$$$$fort>t_1:$$$$-\mu Mg=M\frac{dv}{dt}$$$$\Rightarrow \frac{dv}{dt}=-\mu g$$$${\int }_{v_1}^{v}dv=-{\int }_{t_1}^t\mu gdt$$$$v\left(t\right)-v_1=-\mu g(t-t_1)$$$$\Rightarrow v\left(t\right)=v_1-\mu g(t-t_1)$$$$att=t_2:v=0$$$$u\lambda \ln \frac{M_0}{M}-\frac{\mu {gm}_0}{q}-\mu g(t_2-\frac{m_0}{q})=0$$$$t_2=\frac{u\lambda }{\mu g}\ln (1+\frac{m_0}{M})$$$$$$$$sincea⩾0fort⩽t_1,themax.visreachedat{t}_1:$$$$v_{max}=v_1=u\lambda \ln (1+\frac{m_0}{M})-\frac{\mu {gm}_0}{q}$$$$$$$$\frac{ds}{dt}=u\lambda \ln \frac{M_0}{M_0-qt}-\mu gt$$$${\int }_0^{s}ds={\int }_0^t(u\lambda \ln \frac{M_0}{M_0-qt}-\mu gt)dt$$$$s\left(t\right)=\frac{u\lambda }{q}{[M_0\ln (M_0-qt)+qt\ln \left(\frac{M_0}{M_0-qt}\right)+qt]}_0^t-\frac{\mu {gt}^2}{2}$$$$s\left(t\right)=\frac{u\lambda }{q}[qt-(M_0-qt)\ln \left(\frac{M_0}{M_0-qt}\right)]-\frac{\mu {gt}^2}{2}$$$$s\left(t\right)=u\lambda [t-(\frac{M_0}{q}-t)\ln \left(\frac{M_0}{M_0-qt}\right)]-\frac{\mu {gt}^2}{2}$$$$att=t_1:$$$$s_1=\frac{u\lambda }{q}[m_0-M\ln \left(\frac{M_0}{M}\right)]-\frac{\mu {gm}_0^2}{2q^2}$$$$distancecoveredin{t}_1⩽t⩽t_2:$$$$\mathrm{\Delta }{s}_2=\frac{v_1^2}{2\mu g}=\frac{1}{2\mu g}{[u\lambda \ln \frac{M_0}{M}-\frac{\mu {gm}_0}{q}]}^2$$$$$$$$totaldistance:$$$$s_{max}=s_2=s_1+\mathrm{\Delta }{s}_2$$$$\Rightarrow s_{max}=\frac{u\lambda }{q}[m_0-M\ln (1+\frac{m_0}{M})]-\frac{\mu {gm}_0^2}{2q^2}+\frac{1}{2\mu g}{[u\lambda \ln (1+\frac{m_0}{M})-\frac{\mu {gm}_0}{q}]}^2$$

Commented by ajfour last updated on 08/Sep/18

$$ThanksSir,v_{max}matches,rest$$$$shouldagreetoo.$$

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