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Question-71939




Question Number 71939 by ajfour last updated on 22/Oct/19
Commented by ajfour last updated on 22/Oct/19
Find v and V in terms of given  parameters. coefficient of  restitution is e and the plane is  frictionless horizontal ground.
$${Find}\:{v}\:{and}\:{V}\:{in}\:{terms}\:{of}\:{given} \\ $$$${parameters}.\:{coefficient}\:{of} \\ $$$${restitution}\:{is}\:{e}\:{and}\:{the}\:{plane}\:{is} \\ $$$${frictionless}\:{horizontal}\:{ground}. \\ $$
Answered by mr W last updated on 22/Oct/19
Commented by mr W last updated on 22/Oct/19
let μ=(M/m)  sin θ=(b/(R+r))  u_(1x) =u sin θ  u_(1y) =u cos θ  u_(1x) =v_(1x) +v_(2x)   mu_(1x) =mv_(1x) −Mv_(2x)   u_(1x) =v_(1x) −μv_(2x)   ⇒v_(2x) =0  ⇒v_(1x) =u_(1x) =u sin θ  eu_(1y) =v_(2y) −v_(1y)   mu_(1y) =mv_(1y) +Mv_(2y)   u_(1y) =v_(1y) +μv_(2y)   (1+e)u_(1y) =(1+μ)v_(2y)   ⇒v_(2y) =((1+e)/(1+μ))u_(1y) =(((1+e)u cos θ)/(1+μ))  ⇒v_(1y) =(((1−eμ)u cos θ)/(1+μ))  v=(√(v_(1x) ^2 +v_(1y) ^2 ))=u(√(sin^2  θ+(((1−eμ)/(1+μ)))^2 cos^2  θ))  V=v_(2y) =(((1+e)u cos θ)/(1+μ))
$${let}\:\mu=\frac{{M}}{{m}} \\ $$$$\mathrm{sin}\:\theta=\frac{{b}}{{R}+{r}} \\ $$$${u}_{\mathrm{1}{x}} ={u}\:\mathrm{sin}\:\theta \\ $$$${u}_{\mathrm{1}{y}} ={u}\:\mathrm{cos}\:\theta \\ $$$${u}_{\mathrm{1}{x}} ={v}_{\mathrm{1}{x}} +{v}_{\mathrm{2}{x}} \\ $$$${mu}_{\mathrm{1}{x}} ={mv}_{\mathrm{1}{x}} −{Mv}_{\mathrm{2}{x}} \\ $$$${u}_{\mathrm{1}{x}} ={v}_{\mathrm{1}{x}} −\mu{v}_{\mathrm{2}{x}} \\ $$$$\Rightarrow{v}_{\mathrm{2}{x}} =\mathrm{0} \\ $$$$\Rightarrow{v}_{\mathrm{1}{x}} ={u}_{\mathrm{1}{x}} ={u}\:\mathrm{sin}\:\theta \\ $$$${eu}_{\mathrm{1}{y}} ={v}_{\mathrm{2}{y}} −{v}_{\mathrm{1}{y}} \\ $$$${mu}_{\mathrm{1}{y}} ={mv}_{\mathrm{1}{y}} +{Mv}_{\mathrm{2}{y}} \\ $$$${u}_{\mathrm{1}{y}} ={v}_{\mathrm{1}{y}} +\mu{v}_{\mathrm{2}{y}} \\ $$$$\left(\mathrm{1}+{e}\right){u}_{\mathrm{1}{y}} =\left(\mathrm{1}+\mu\right){v}_{\mathrm{2}{y}} \\ $$$$\Rightarrow{v}_{\mathrm{2}{y}} =\frac{\mathrm{1}+{e}}{\mathrm{1}+\mu}{u}_{\mathrm{1}{y}} =\frac{\left(\mathrm{1}+{e}\right){u}\:\mathrm{cos}\:\theta}{\mathrm{1}+\mu} \\ $$$$\Rightarrow{v}_{\mathrm{1}{y}} =\frac{\left(\mathrm{1}−{e}\mu\right){u}\:\mathrm{cos}\:\theta}{\mathrm{1}+\mu} \\ $$$${v}=\sqrt{{v}_{\mathrm{1}{x}} ^{\mathrm{2}} +{v}_{\mathrm{1}{y}} ^{\mathrm{2}} }={u}\sqrt{\mathrm{sin}^{\mathrm{2}} \:\theta+\left(\frac{\mathrm{1}−{e}\mu}{\mathrm{1}+\mu}\right)^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta} \\ $$$${V}={v}_{\mathrm{2}{y}} =\frac{\left(\mathrm{1}+{e}\right){u}\:\mathrm{cos}\:\theta}{\mathrm{1}+\mu} \\ $$
Commented by ajfour last updated on 23/Oct/19
Thanks sir, beautiful solution.
$${Thanks}\:{sir},\:{beautiful}\:{solution}. \\ $$

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