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Question Number 54901 by rahul 19 last updated on 14/Feb/19

Commented by rahul 19 last updated on 14/Feb/19

$$Ans:\left(1\right)!$$

Answered by mr W last updated on 14/Feb/19

$$v_0=velocityofmassmbeforestrike$$$$v_1=velocityofmassmafterstrike$$$$u=velocityofmassMafterstrike$$$$v_0=u+v_1$$$${mv}_0=-{mv}_1+Mu$$$$\Rightarrow {mv}_0=-{mv}_1+M(v_0-v_1)$$$$\Rightarrow (M-m)v_0=(M+m)v_1$$$$\Rightarrow \frac{v_1}{v_0}=\frac{M-m}{M+m}$$$$\frac{{KE}_1}{{KE}_0}={\left(\frac{v_1}{v_0}\right)}^2=\frac{{PE}_1}{{PE}_0}=\frac{l(1-\cos {\theta }_1)}{l(1-\cos {\theta }_0)}=\frac{2l{\sin }^2\frac{{\theta }_1}{2}}{2l{\sin }^2\frac{{\theta }_0}{2}}\approx {\left(\frac{{\theta }_1}{{\theta }_0}\right)}^2$$$$\Rightarrow {\left(\frac{v_1}{v_0}\right)}^2={\left(\frac{M-m}{M+m}\right)}^2\approx {\left(\frac{{\theta }_1}{{\theta }_0}\right)}^2$$$$\Rightarrow \frac{M-m}{M+m}\approx \frac{{\theta }_1}{{\theta }_0}$$$$\Rightarrow 1-\frac{2m}{M+m}\approx \frac{{\theta }_1}{{\theta }_0}$$$$\frac{M+m}{2m}\approx \frac{{\theta }_0}{{\theta }_0-{\theta }_1}$$$$M\approx (\frac{2{\theta }_0}{{\theta }_0-{\theta }_1}-1)m=\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)m$$$$\Rightarrow answer\left(1\right)$$

Commented by Otchere Abdullai last updated on 14/Feb/19

$$Brilliantprof$$

Commented by rahul 19 last updated on 14/Feb/19

$$IntheaboveQ.iftherewasno$$$$blockofmassM,will{\theta }_0={\theta }_1?$$$$Imeanwillstringcomebacktosameposition$$$$ornot?$$

Commented by mr W last updated on 14/Feb/19

$$certainly.ifmassm{doesn}^{\prime }tgiveenergy$$$$toanotherobject,itwillcomeback$$$$toitsorginalposition,i.e.{\theta }_1={\theta }_0.$$

Commented by rahul 19 last updated on 14/Feb/19

$$WegotM=m\left(\frac{{\theta }_0+{\theta }_1}{{\theta }_0-{\theta }_1}\right)soWhen{\theta }_0={\theta }_1$$$$M=\mathrm{\infty }butyousaidifM=0then{\theta }_0={\theta }_1?$$$$howitispossible?$$

Commented by mr W last updated on 14/Feb/19

$$IfM\to \mathrm{\infty },i.e.Misawall,themassm$$$$willgetitswholeenergybackafterthe$$$$hitagainstthewall.IfM=0,i.e.there$$$$isnocollision,themassmwillnot$$$$changeitsmotionatall.itwillgothe$$$$position-{\theta }_{0}andthenbackto{\theta }_0.$$

Commented by rahul 19 last updated on 14/Feb/19

$$ThankyouSir!$$

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