Question-90923

Question Number 90923 by ajfour last updated on 26/Apr/20

Commented by ajfour last updated on 26/Apr/20

$${A}\:{ball}\:{of}\:{mass}\:{m}\:{hits}\:{horizontally} \\ $$$${and}\:{elastically}\:{the}\:{centre}\:{of}\:{a} \\ $$$${hinged}\:{square}\:{plate}\:{of}\:{side}\:{s}\:{and} \\ $$$${mass}\:{M}.\:{Find}\:{velocity}\:{v}\:{of}\:{ball} \\ $$$${juzt}\:{after}\:{impact}. \\ $$

Answered by mr W last updated on 26/Apr/20

I=((Ms^2 )/3) before the hit: E=(1/2)mu^2 after the hit: E=(1/2)mv^2 +(1/2)Iω^2 =(1/2)mu^2 ⇒v^2 +((Ms^2 )/(3m))ω^2 =u^2 Iω−(s/2)mv=(s/2)mu ⇒ω=(u+v)((3m)/(2Ms)) v^2 +(u+v)^2 ((3m)/(4M))=u^2 3m(u+v)=4M(u−v) ⇒v=((4M−3m)/(4M+3m))u (backwards +)

$${I}=\frac{{Ms}^{\mathrm{2}} }{\mathrm{3}} \\ $$$${before}\:{the}\:{hit}: \\ $$$${E}=\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} \\ $$$${after}\:{the}\:{hit}: \\ $$$${E}=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}{I}\omega^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} \\ $$$$\Rightarrow{v}^{\mathrm{2}} +\frac{{Ms}^{\mathrm{2}} }{\mathrm{3}{m}}\omega^{\mathrm{2}} ={u}^{\mathrm{2}} \\ $$$${I}\omega−\frac{{s}}{\mathrm{2}}{mv}=\frac{{s}}{\mathrm{2}}{mu} \\ $$$$\Rightarrow\omega=\left({u}+{v}\right)\frac{\mathrm{3}{m}}{\mathrm{2}{Ms}} \\ $$$${v}^{\mathrm{2}} +\left({u}+{v}\right)^{\mathrm{2}} \frac{\mathrm{3}{m}}{\mathrm{4}{M}}={u}^{\mathrm{2}} \\ $$$$\mathrm{3}{m}\left({u}+{v}\right)=\mathrm{4}{M}\left({u}−{v}\right) \\ $$$$\Rightarrow{v}=\frac{\mathrm{4}{M}−\mathrm{3}{m}}{\mathrm{4}{M}+\mathrm{3}{m}}{u}\:\:\:\:\left({backwards}\:+\right) \\ $$

Commented by ajfour last updated on 27/Apr/20

$${Very}\:{Nice}\:{and}\:{correct},\:{Sir}! \\ $$

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Question-90923

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