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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
$${Ans}:\:\left(\mathrm{1}\right)! \\ $$
Answered by mr W last updated on 14/Feb/19
$${v}_{\mathrm{0}} ={velocity}\:{of}\:{mass}\:{m}\:{before}\:{strike} \\ $$$${v}_{\mathrm{1}} ={velocity}\:{of}\:{mass}\:{m}\:{after}\:{strike} \\ $$$${u}={velocity}\:{of}\:{mass}\:{M}\:{after}\:{strike} \\ $$$${v}_{\mathrm{0}} ={u}+{v}_{\mathrm{1}} \\ $$$${mv}_{\mathrm{0}} =−{mv}_{\mathrm{1}} +{Mu} \\ $$$$\Rightarrow{mv}_{\mathrm{0}} =−{mv}_{\mathrm{1}} +{M}\left({v}_{\mathrm{0}} −{v}_{\mathrm{1}} \right) \\ $$$$\Rightarrow\left({M}−{m}\right){v}_{\mathrm{0}} =\left({M}+{m}\right){v}_{\mathrm{1}} \\ $$$$\Rightarrow\frac{{v}_{\mathrm{1}} }{{v}_{\mathrm{0}} }=\frac{{M}−{m}}{{M}+{m}} \\ $$$$\frac{{KE}_{\mathrm{1}} }{{KE}_{\mathrm{0}} }=\left(\frac{{v}_{\mathrm{1}} }{{v}_{\mathrm{0}} }\right)^{\mathrm{2}} =\frac{{PE}_{\mathrm{1}} }{{PE}_{\mathrm{0}} }=\frac{{l}\left(\mathrm{1}−\mathrm{cos}\:\theta_{\mathrm{1}} \right)}{{l}\left(\mathrm{1}−\mathrm{cos}\:\theta_{\mathrm{0}} \right)}=\frac{\mathrm{2}{l}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta_{\mathrm{1}} }{\mathrm{2}}}{\mathrm{2}{l}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta_{\mathrm{0}} }{\mathrm{2}}}\approx\left(\frac{\theta_{\mathrm{1}} }{\theta_{\mathrm{0}} }\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{v}_{\mathrm{1}} }{{v}_{\mathrm{0}} }\right)^{\mathrm{2}} =\left(\frac{{M}−{m}}{{M}+{m}}\right)^{\mathrm{2}} \approx\left(\frac{\theta_{\mathrm{1}} }{\theta_{\mathrm{0}} }\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{M}−{m}}{{M}+{m}}\approx\frac{\theta_{\mathrm{1}} }{\theta_{\mathrm{0}} } \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{2}{m}}{{M}+{m}}\approx\frac{\theta_{\mathrm{1}} }{\theta_{\mathrm{0}} } \\ $$$$\frac{{M}+{m}}{\mathrm{2}{m}}\approx\frac{\theta_{\mathrm{0}} }{\theta_{\mathrm{0}} −\theta_{\mathrm{1}} } \\ $$$${M}\approx\left(\frac{\mathrm{2}\theta_{\mathrm{0}} }{\theta_{\mathrm{0}} −\theta_{\mathrm{1}} }−\mathrm{1}\right){m}=\left(\frac{\theta_{\mathrm{0}} +\theta_{\mathrm{1}} }{\theta_{\mathrm{0}} −\theta_{\mathrm{1}} }\right){m} \\ $$$$\Rightarrow{answer}\:\left(\mathrm{1}\right) \\ $$
Commented by Otchere Abdullai last updated on 14/Feb/19
$${Brilliant}\:{prof} \\ $$
Commented by rahul 19 last updated on 14/Feb/19
$${In}\:{the}\:{above}\:{Q}.\:{if}\:{there}\:{was}\:{no} \\ $$$${block}\:{of}\:{mass}\:{M}\:,\:{will}\:\theta_{\mathrm{0}} =\theta_{\mathrm{1}} \:? \\ $$$${I}\:{mean}\:{will}\:{string}\:{come}\:{back}\:{to}\:{same}\:{position} \\ $$$${or}\:{not}? \\ $$
Commented by mr W last updated on 14/Feb/19
$${certainly}.\:{if}\:{mass}\:{m}\:{doesn}'{t}\:{give}\:{energy} \\ $$$${to}\:{an}\:{other}\:{object},\:{it}\:{will}\:{come}\:{back} \\ $$$${to}\:{its}\:{orginal}\:{position},\:{i}.{e}.\:\theta_{\mathrm{1}} =\theta_{\mathrm{0}} . \\ $$
Commented by rahul 19 last updated on 14/Feb/19
$${We}\:{got}\:{M}=\:{m}\left(\frac{\theta_{\mathrm{0}} +\theta_{\mathrm{1}} }{\theta_{\mathrm{0}} −\theta_{\mathrm{1}} }\right)\:{so}\:{When}\:\theta_{\mathrm{0}} =\theta_{\mathrm{1}} \\ $$$${M}=\infty\:{but}\:{you}\:{said}\:{if}\:{M}=\mathrm{0}\:{then}\:\theta_{\mathrm{0}} =\theta_{\mathrm{1}} \:? \\ $$$$\:{how}\:{it}\:{is}\:{possible}? \\ $$
Commented by mr W last updated on 14/Feb/19
$${If}\:{M}\rightarrow\infty,\:{i}.{e}.\:{M}\:{is}\:{a}\:{wall},\:{the}\:{mass}\:{m} \\ $$$${will}\:{get}\:{its}\:{whole}\:{energy}\:{back}\:{after}\:{the} \\ $$$${hit}\:{against}\:{the}\:{wall}.\:{If}\:{M}=\mathrm{0},\:{i}.{e}.\:{there} \\ $$$${is}\:{no}\:{collision},\:{the}\:{mass}\:{m}\:{will}\:{not} \\ $$$${change}\:{its}\:{motion}\:{at}\:{all}.\:{it}\:{will}\:{go}\:{the} \\ $$$${position}\:−\theta_{\mathrm{0}} \:{and}\:{then}\:{back}\:{to}\:\theta_{\mathrm{0}} . \\ $$
Commented by rahul 19 last updated on 14/Feb/19
$${Thank}\:{you}\:{Sir}! \\ $$