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A particle strikes a horizontal frictionless floor with a speed u at an angle $θ$ with the vertical and rebounds with a speed v at an angle $ϕ$ with the vertical. The co-efficient of restitution between the particle and floor is e. The angle $ϕ$ is equal to

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$θ$

$\tan^{- 1} (etan θ)$

$\tan^{- 1} (\frac{1}{e} \tan θ)$

$(1 + e) θ$

answer is C.

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Detailed Solution

$ϕ =$ $\tan^{- 1} (\frac{1}{e} \tan θ)$
$∴$ vertical component after collision = eu cos $θ$
& Horizontal component u sin $θ$

But V cos $ϕ$ = e u cos $θ$ and

u sin $θ$ = v sin $ϕ$
$\Rightarrow \tan ϕ = \frac{usin θ}{eucos θ} \Rightarrow ϕ = \tan^{- 1} (\frac{1}{e} \tan θ)$

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