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Question Number 71939 by ajfour last updated on 22/Oct/19

Commented by ajfour last updated on 22/Oct/19

$$FindvandVintermsofgiven$$$$parameters.coefficientof$$$$restitutioniseandtheplaneis$$$$frictionlesshorizontalground.$$

Answered by mr W last updated on 22/Oct/19

Commented by mr W last updated on 22/Oct/19

$$let\mu =\frac{M}{m}$$$$\sin \theta =\frac{b}{R+r}$$$$u_{1x}=u\sin \theta$$$$u_{1y}=u\cos \theta$$$$u_{1x}=v_{1x}+v_{2x}$$$${mu}_{1x}={mv}_{1x}-{Mv}_{2x}$$$$u_{1x}=v_{1x}-\mu v_{2x}$$$$\Rightarrow v_{2x}=0$$$$\Rightarrow v_{1x}=u_{1x}=u\sin \theta$$$${eu}_{1y}=v_{2y}-v_{1y}$$$${mu}_{1y}={mv}_{1y}+{Mv}_{2y}$$$$u_{1y}=v_{1y}+\mu v_{2y}$$$$(1+e)u_{1y}=(1+\mu )v_{2y}$$$$\Rightarrow v_{2y}=\frac{1+e}{1+\mu }{u}_{1y}=\frac{(1+e)u\cos \theta }{1+\mu }$$$$\Rightarrow v_{1y}=\frac{(1-e\mu )u\cos \theta }{1+\mu }$$$$v=\sqrt{v_{1x}^2+v_{1y}^2}=u\sqrt{{\sin }^2\theta +{\left(\frac{1-e\mu }{1+\mu }\right)}^2{\cos }^2\theta }$$$$V=v_{2y}=\frac{(1+e)u\cos \theta }{1+\mu }$$

Commented by ajfour last updated on 23/Oct/19

$$Thankssir,beautifulsolution.$$

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