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A-particle-of-mass-m-strikes-on-ground-with-angle-of-incidence-45-If-coefficient-of-restitution-e-1-2-find-the-velocity-after-impact-and-angle-of-reflection-

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Question Number 23049 by Tinkutara last updated on 25/Oct/17
A particle of mass m strikes on ground with angle of incidence 45°. If coefficient of restitution, e = (1/( (√2))) , find the velocity after impact and angle of reflection.
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{strikes}\:\mathrm{on}\:\mathrm{ground} \\ $$$$\mathrm{with}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{incidence}\:\mathrm{45}°.\:\mathrm{If}\:\mathrm{coefficient} \\ $$$$\mathrm{of}\:\mathrm{restitution},\:{e}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{velocity} \\ $$$$\mathrm{after}\:\mathrm{impact}\:\mathrm{and}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{reflection}. \\ $$
Answered by ajfour last updated on 25/Oct/17
parallel to ground (if friction be zero then) we conserve linear momentum. let initial speed be v_0 , and final speed be v. mvsin θ=mv_0 sin 45° ⇒ vsin θ=(v_0 /( (√2))) ....(i) v_(separation) =e(v_(approach) ) ⇒ vcos θ=(1/( (√2)))×v_0 cos 45° ⇒ vcos θ=(v_0 /2) ...(ii) (i)÷(ii) gives tan θ =(√2) ⇒ 𝛉=tan^(−1) (√2) v^2 sin^2 θ+v^2 cos^2 θ =(v_0 ^2 /2)+(v_0 ^2 /4) ⇒ v = ((v_0 (√3))/2) .
$${parallel}\:{to}\:{ground}\:\left({if}\:{friction}\:{be}\right. \\ $$$$\left.{zero}\:{then}\right)\:{we}\:{conserve}\:{linear} \\ $$$${momentum}. \\ $$$${let}\:{initial}\:{speed}\:{be}\:\boldsymbol{{v}}_{\mathrm{0}} \:,\:{and}\:{final} \\ $$$${speed}\:{be}\:\boldsymbol{{v}}. \\ $$$${mv}\mathrm{sin}\:\theta={mv}_{\mathrm{0}} \mathrm{sin}\:\mathrm{45}°\: \\ $$$$\Rightarrow\:\:\:\:{v}\mathrm{sin}\:\theta=\frac{{v}_{\mathrm{0}} }{\:\sqrt{\mathrm{2}}}\:\:\:\:\:\:….\left({i}\right) \\ $$$$\:\:\:{v}_{{separation}} \:={e}\left({v}_{{approach}} \right) \\ $$$$\Rightarrow\:\:\:\:{v}\mathrm{cos}\:\theta=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×{v}_{\mathrm{0}} \mathrm{cos}\:\mathrm{45}° \\ $$$$\:\:\:\:\:\Rightarrow\:\:{v}\mathrm{cos}\:\theta=\frac{{v}_{\mathrm{0}} }{\mathrm{2}}\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)\boldsymbol{\div}\left({ii}\right)\:{gives} \\ $$$$\:\:\:\mathrm{tan}\:\theta\:=\sqrt{\mathrm{2}}\:\:\:\:\Rightarrow\:\:\:\boldsymbol{\theta}=\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{2}} \\ $$$$\:\:{v}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+{v}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta\:=\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}}+\frac{{v}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:\:\boldsymbol{{v}}\:=\:\frac{\boldsymbol{{v}}_{\mathrm{0}} \sqrt{\mathrm{3}}}{\mathrm{2}}\:. \\ $$$$ \\ $$
Commented by Tinkutara last updated on 25/Oct/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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