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Question Number 43585 by ajfour last updated on 12/Sep/18

Commented by ajfour last updated on 12/Sep/18

$$Thesmallballhasradius\mathit{r},$$$$whilethelargerballhasradius$$$$\mathit{R},theycollideonsmooth$$$$horizontalground,impact$$$$parameterisequalto\mathit{r}.$$$$Thesmallballhitsthelarger$$$$ball\left(atrestinitially\right)with$$$$speed\mathit{u}.$$$$\mathit{F}\mathit{i}\mathit{n}\mathit{d}\mathit{\theta },\mathit{\varphi },\mathit{v},and\mathit{V}intermsof$$$$\mathit{m},\mathit{M},\mathit{r},\mathit{R}.$$$$\left(collisioniscompletlyelastic\right).$$

Commented by MrW3 last updated on 12/Sep/18

$$NowIknowyouareacurlingsport$$$$fan.$$

Answered by MrW3 last updated on 14/Sep/18

$$\theta ={\sin }^{-1}\frac{r}{r+R}$$$$\lambda =\frac{M}{m}$$$$mu=mv\cos \varphi +MV\cos \theta$$$$\Rightarrow u=v\cos \varphi +\lambda V\cos \theta ...\left(i\right)$$$$0=mv\sin \varphi -MV\sin \theta$$$$\Rightarrow v\sin \varphi =\lambda V\sin \theta ...\left(ii\right)$$$$V-v\cos (\varphi +\theta )=eu\cos \theta ...\left(iii\right)$$$$$$$$from\left(iii\right):$$$$V-v\cos \varphi \cos \theta +v\sin \varphi \sin \theta =u\cos \theta$$$$V-(u-\lambda V\cos \theta )\cos \theta +\lambda V{\sin }^2\theta =u\cos \theta$$$$V-u\cos \theta +\lambda V=u\cos \theta$$$$\Rightarrow V=\frac{2u\cos \theta }{1+\lambda }$$$$\tan \varphi =\frac{v\sin \varphi }{v\cos \varphi }=\frac{\lambda V\sin \theta }{u-\lambda V\cos \theta }=\frac{\frac{2\lambda u\cos \theta \sin \theta }{1+\lambda }}{u-\frac{2\lambda u{\cos }^2\theta }{1+\lambda }}$$$$\Rightarrow \tan \varphi =\frac{\lambda \sin 2\theta }{1-\lambda \cos 2\theta }$$$$\Rightarrow \varphi ={\tan }^{-1}\frac{\lambda \sin 2\theta }{1-\lambda \cos 2\theta }$$$$\Rightarrow v=\frac{\lambda V\sin \theta }{\sin \varphi }=\frac{u\sqrt{{\lambda }^2+1-2\lambda \cos 2\theta }}{1+\lambda }$$

Commented by ajfour last updated on 14/Sep/18

$$Webothare\left(hopefully\right)correct$$$$now,Sir!$$

Commented by MrW3 last updated on 14/Sep/18

$$thanksforhelpingmetogetthe$$$$rightsolution!$$

Answered by ajfour last updated on 14/Sep/18

$$\sin \theta =\frac{r}{R+r}$$$$MV\cos \theta +mv\cos \varphi =mu...\left(i\right)$$$$MV\sin \theta =mv\sin \varphi ....\left(ii\right)$$$$V-v\cos (\theta +\varphi )=u\cos \theta ....\left(iii\right)$$$$using\left(ii\right)in\left(i\right)$$$$MV\cos \theta +MV\sin \theta \cot \varphi =mu$$$$let\frac{m}{M}=q$$$$\Rightarrow V(\cos \theta +\sin \theta \cot \varphi )=qu...\left(a\right)$$$$using\left(ii\right)in\left(iii\right)$$$$V-\frac{MV\sin \theta \cos (\theta +\varphi )}{m\sin \varphi }=u\cos \theta$$$$V(1-\frac{\sin \theta \cos (\theta +\varphi )}{q\sin \varphi })=u\cos \theta ..\left(b\right)$$$$eq.\left(b\right)\mathbf{÷}eq.\left(a\right)gives$$$$\frac{[1-\frac{\sin \theta \cos (\theta +\varphi )}{q\sin \varphi }]}{\cos \theta +\sin \theta \cot \varphi }=\frac{\cos \theta }{q}$$$$$$$$\Rightarrow q\sin \varphi -\sin \theta \cos \theta \cos \varphi$$$$+\sin {}^2\theta \sin \varphi =\cos {}^2\theta \sin \varphi$$$$+\sin \theta \cos \theta \cos \varphi$$$$\Rightarrow (q-\cos {}^2\theta +\sin {}^2\theta )\sin \varphi$$$$=2\sin \theta \cos \theta \cos \varphi$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$or\tan \varphi =\frac{\sin 2\theta }{q-\cos 2\theta }$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$NowV=\frac{qu}{\cos \theta +\sin \theta \cot \varphi }\left[see\left(a\right)\right]$$$$=\frac{qu\sin 2\theta }{\cos \theta \sin 2\theta +q\sin \theta -\sin \theta \cos 2\theta }$$$$$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\Rightarrow \mathit{V}=\frac{2qu\cos \theta }{1+q}$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$v=\frac{MV\sin \theta }{m\sin \varphi }$$$$=\frac{2u\sin \theta \cos \theta }{1+q}\times \frac{\sqrt{\sin {}^{2}2\theta +{(q-\cos 2\theta )}^2}}{\sin 2\theta }$$$$$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}$$$$\mathit{v}=\mathit{u}\left(\frac{\sqrt{1+q^2-2q\cos 2\theta }}{1+q}\right)$$$$\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

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