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