Question Number 109760 by ajfour last updated on 25/Aug/20
Commented by ajfour last updated on 25/Aug/20
$${Find}\:{u}\:{in}\:{terms}\:{of}\:{l},\:{b},\:{g}\:{such}\:{that} \\ $$$${bob}\:{hits}\:{hanging}\:{block}. \\ $$$$ \\ $$
Answered by mr W last updated on 25/Aug/20
Commented by mr W last updated on 27/Aug/20
![(1/2)mv^2 +mgl(1+sin θ)=(1/2)mu^2 ⇒v^2 =u^2 −2gl(1+sin θ) mg sin θ=((mv^2 )/l) ⇒u^2 =gl(2+3 sin θ) ⇒v^2 =gl sin θ R=l cos θ h=l(1−sin θ)−b t=(R/(v cos ϕ))=(l/(v tan θ)) h=v sin ϕ (l/(v tan θ))−(1/2)g ((l/(v tan θ)))^2 l(1−sin θ)−b=((l cos θ)/(tan θ))−((gl^2 )/(2v^2 tan^2 θ)) ((gl^2 )/(2v^2 tan^2 θ))=b+l((1/(sin θ))−1) v^2 =((gl^2 )/(2 tan^2 θ [b+l(((1−sin θ)/(sin θ)))])) gl sin θ=((gl^2 )/(2 tan^2 θ [b+l(((1−sin θ)/(sin θ)))])) sin θ=(1/(2 tan^2 θ ((b/l)−1+(1/(sin θ))))) 2 sin^3 θ ((b/l)−1+(1/(sin θ)))=1−sin^2 θ ⇒2(1−(b/l))sin^3 θ−3 sin^2 θ+1=0 ⇒(1/(sin θ))=2 sin [(1/3) sin^(−1) (1−(b/l))+((2π)/3)] ⇒u=(√({2+(3/(2 sin [(1/3) sin^(−1) (1−(b/l))+((2π)/3)]))}gl))](https://www.tinkutara.com/question/Q109832.png)
$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} +{mgl}\left(\mathrm{1}+\mathrm{sin}\:\theta\right)=\frac{\mathrm{1}}{\mathrm{2}}{mu}^{\mathrm{2}} \\ $$$$\Rightarrow{v}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}{gl}\left(\mathrm{1}+\mathrm{sin}\:\theta\right) \\ $$$${mg}\:\mathrm{sin}\:\theta=\frac{{mv}^{\mathrm{2}} }{{l}} \\ $$$$\Rightarrow{u}^{\mathrm{2}} ={gl}\left(\mathrm{2}+\mathrm{3}\:\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{v}^{\mathrm{2}} ={gl}\:\mathrm{sin}\:\theta \\ $$$${R}={l}\:\mathrm{cos}\:\theta \\ $$$${h}={l}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)−{b} \\ $$$${t}=\frac{{R}}{{v}\:\mathrm{cos}\:\varphi}=\frac{{l}}{{v}\:\mathrm{tan}\:\theta} \\ $$$${h}={v}\:\mathrm{sin}\:\varphi\:\frac{{l}}{{v}\:\mathrm{tan}\:\theta}−\frac{\mathrm{1}}{\mathrm{2}}{g}\:\left(\frac{{l}}{{v}\:\mathrm{tan}\:\theta}\right)^{\mathrm{2}} \\ $$$${l}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)−{b}=\frac{{l}\:\mathrm{cos}\:\theta}{\mathrm{tan}\:\theta}−\frac{{gl}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:\theta} \\ $$$$\frac{{gl}^{\mathrm{2}} }{\mathrm{2}{v}^{\mathrm{2}} \mathrm{tan}^{\mathrm{2}} \:\theta}={b}+{l}\left(\frac{\mathrm{1}}{\mathrm{sin}\:\:\theta}−\mathrm{1}\right) \\ $$$${v}^{\mathrm{2}} =\frac{{gl}^{\mathrm{2}} }{\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} \:\theta\:\left[{b}+{l}\left(\frac{\mathrm{1}−\mathrm{sin}\:\theta}{\mathrm{sin}\:\theta}\right)\right]} \\ $$$${gl}\:\mathrm{sin}\:\theta=\frac{{gl}^{\mathrm{2}} }{\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} \:\theta\:\left[{b}+{l}\left(\frac{\mathrm{1}−\mathrm{sin}\:\theta}{\mathrm{sin}\:\theta}\right)\right]} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{tan}^{\mathrm{2}} \:\theta\:\left(\frac{{b}}{{l}}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\right)} \\ $$$$\mathrm{2}\:\mathrm{sin}^{\mathrm{3}} \:\theta\:\left(\frac{{b}}{{l}}−\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\theta}\right)=\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\Rightarrow\mathrm{2}\left(\mathrm{1}−\frac{{b}}{{l}}\right)\mathrm{sin}^{\mathrm{3}} \:\theta−\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\theta+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{sin}\:\theta}=\mathrm{2}\:\mathrm{sin}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}−\frac{{b}}{{l}}\right)+\frac{\mathrm{2}\pi}{\mathrm{3}}\right] \\ $$$$\Rightarrow{u}=\sqrt{\left\{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}\:\mathrm{sin}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}−\frac{{b}}{{l}}\right)+\frac{\mathrm{2}\pi}{\mathrm{3}}\right]}\right\}{gl}} \\ $$
Commented by ajfour last updated on 27/Aug/20
$${l}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)−{b}=\frac{{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \varphi}{\mathrm{2}{g}} \\ $$$$\:\:{h}=\:\mathrm{1}−\mathrm{sin}\:\theta−\frac{{b}}{{l}}\:\:\:=\frac{\mathrm{sin}\:\theta\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{sin}\:\theta\left(\mathrm{2}+\mathrm{cos}\:^{\mathrm{2}} \theta\right)=\mathrm{2}\left(\mathrm{1}−\frac{{b}}{{l}}\right) \\ $$$$\:\:\:\:\mathrm{sin}\:\theta\left(\mathrm{3}−\mathrm{sin}\:^{\mathrm{2}} \theta\right)=\mathrm{2}\left(\mathrm{1}−\frac{{b}}{{l}}\right) \\ $$$$.. \\ $$$$\:\:{Sir}\:{i}\:{think}\:{eq}.\:{for}\:\mathrm{sin}\:\theta\:{should}\:{be} \\ $$$$\:\:\:{this}… \\ $$$$…. \\ $$
Commented by ajfour last updated on 27/Aug/20
$${mrW}\:{sir},\:{dont}\:{you}\:{think}\:{even}\:{this} \\ $$$${eq}.\:{is}\:{correct}..? \\ $$
Commented by mr W last updated on 27/Aug/20
$${in}\:{line}\:\mathrm{1}\:{you}\:{assumed}\:{that}\:{the}\:{ball} \\ $$$${reaches}\:{the}\:{highest}\:{point}\:{when}\:{it}\:{hits} \\ $$$${the}\:{block},\:{i}\:{think}\:{this}\:{is}\:{not}\:{correct}. \\ $$
Commented by mr W last updated on 27/Aug/20
Commented by ajfour last updated on 27/Aug/20
$${thanks}\:{sir},\:{confusion}\:{clear}. \\ $$
Commented by mr W last updated on 27/Aug/20
![u=(√({2+(3/(2 sin [(1/3) sin^(−1) (1−(b/l))+((2π)/3)]))}gl))](https://www.tinkutara.com/question/Q110196.png)
$${u}=\sqrt{\left\{\mathrm{2}+\frac{\mathrm{3}}{\mathrm{2}\:\mathrm{sin}\:\left[\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{1}−\frac{{b}}{{l}}\right)+\frac{\mathrm{2}\pi}{\mathrm{3}}\right]}\right\}{gl}} \\ $$
Commented by mr W last updated on 29/Aug/20
Commented by ajfour last updated on 29/Aug/20
$${nice}\:{observation},\:{Sir}. \\ $$