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Question Number 81011 by mr W last updated on 08/Feb/20

Commented by mr W last updated on 08/Feb/20

$$Aballwithmassmandradiusrfalls$$$$fromaheightofhandhitstwowedges$$$$withmassMforeachasshown.$$$$Ifthecoefficientofrestitutionise$$$$andthefrictioncoefficientbetween$$$$wedgesandgroundis\mu ,findthe$$$$distancethewedgesmove.$$

Commented by ajfour last updated on 09/Feb/20

Commented by ajfour last updated on 09/Feb/20

$$u=\sqrt{2g(h-r\sec \alpha )}....\left(i\right)$$$$V\sin \alpha +v\cos \alpha =eu\cos \alpha ..\left(ii\right)$$$$J\sin \alpha =\sin \alpha \int Ndt=MV;$$$$2J\cos \alpha =m(v+u)$$$$\Rightarrow 2MV\cos \alpha =m(v+u)\sin \alpha$$$$using\left(ii\right):$$$$2MV\cos {}^2\alpha =m\{eu\cos \alpha -V\sin \alpha$$$$+u\cos \alpha \}\sin \alpha$$$$V=\frac{mu(e+1)\cos \alpha \sin \alpha }{2M\cos {}^2\alpha +m\sin {}^2\alpha }$$$$2\mu MgL={MV}^2$$$$\Rightarrow L=\frac{V^2}{2\mu g}$$$$L=\frac{u^2}{2\mu g}{\left\{\frac{m(e+1)\cos \alpha \sin \alpha }{2M\cos {}^2\alpha +m\sin {}^2\alpha }\right\}}^2$$$$L=\frac{(h-r\sec \alpha )}{\mu }{\left\{\frac{(e+1)\tan \alpha }{\frac{2M}{m}+\tan {}^2\alpha }\right\}}^2.$$

Commented by mr W last updated on 09/Feb/20

$$thanksalotsir!$$$$whyisthemomentumconservation$$$$inverticaldirectionnotappliable?$$

Commented by ajfour last updated on 09/Feb/20

$$whilecollisiontakesplace,$$$$normalreactionfromground$$$$onwedges,increasessubstan-$$$$ciallyandbacktoMg.$$

Commented by mr W last updated on 09/Feb/20

$${that}^{\prime }sit!$$

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