Question Number 196697 by mr W last updated on 29/Aug/23
Commented by mr W last updated on 29/Aug/23
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Commented by mr W last updated on 29/Aug/23
$${a}\:{ball}\:{is}\:{released}\:{from}\:{a}\:{height}\:{of}\:\boldsymbol{{h}}\: \\ $$$${above}\:{a}\:{long}\:{inclined}\:{plane}\:{as}\:{shown}. \\ $$$${if}\:{the}\:{coefficient}\:{of}\:{restitution}\:{is}\:\boldsymbol{{e}}, \\ $$$${find}\:{the}\:{velocity}\:{of}\:{the}\:{ball}\:{just}\:{after} \\ $$$${the}\:{n}^{{th}} \:{collision}. \\ $$
Commented by mahdipoor last updated on 29/Aug/23
$${obtained}\:{fron}\:{the}\:{same}\:{question}\:{as}\:{before}: \\ $$$$\begin{cases}{{v}_{{i}} ={v}_{\mathrm{1}} {e}^{{i}−\mathrm{1}} }\\{{u}_{{i}} ={u}_{\mathrm{1}} −\mathrm{2}{v}_{\mathrm{1}} \left(\mathrm{1}+{e}+{e}^{\mathrm{2}} +…+{e}^{{i}−\mathrm{2}} \right){tan}\theta}\end{cases} \\ $$$${in}\:{this}\:{case}\:\Rightarrow\begin{cases}{{v}_{\mathrm{1}} ={e}\sqrt{{agh}}×{cos}\theta}\\{{u}_{\mathrm{1}} =−\sqrt{\mathrm{2}{gh}}×{sin}\theta}\end{cases} \\ $$$$\begin{cases}{{v}_{{i}} =\sqrt{\mathrm{2}{gh}}.{cos}\theta×{e}^{{i}} }\\{{u}_{{i}} =−\sqrt{\mathrm{2}{gh}}.{sin}\theta\left(\frac{\mathrm{1}+\mathrm{3}{e}−\mathrm{2}{e}^{{i}} }{\mathrm{1}−{e}}\right)}\end{cases} \\ $$
Answered by mr W last updated on 30/Aug/23
