A steel ball falls from a height nbsp- h on a floor for which the coefficient of restitution is e- nbsp- The height attained by the ball after two rebounds is nbsp-eh- e-2h- e-3h- e-4h-

The correct option is D e^4h Ist collisionVelocity of strike =√(2gh) ⇒ Velocity of rebound =e×√(2gh)=u ⇒ height attained = ...

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Standard XII

Physics

Oblique Collision

A steel ball ...

Question

A steel ball falls from a height h on a floor for which the coefficient of restitution is e. The height attained by the ball after two rebounds is

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e^{2} h

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e^{3} h

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e^{4} h

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Solution

The correct option is D $e^{4} h$
$I s t$ collision
Velocity of strike $= \sqrt{2 g h}$
$\Rightarrow$ Velocity of rebound $= e \times \sqrt{2 g h} = u$
$\Rightarrow$ height attained $= \frac{u^{2}}{2 g} = \frac{2^{2} \times 2 g h}{2 g}$
$= e^{2} h$
$I I n d$ collision
Velocity of strike $= \sqrt{29 \times e^{2} h}$
$= e \sqrt{2 g h}$
Velocity of rebound $= e \times (e \sqrt{2 g h})$
$= e^{2} \sqrt{2 g h} = v$
$\Rightarrow$ height attained $= \frac{v^{2}}{2 g} = \frac{e^{4} \times 2 g h}{2 g}$ $8 e^{4} h$ .

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A steel ball falls from a height h on a floor for which the coefficient of restitution is e. The height attained by the ball after two rebounds is

The correct option is D e4hIst collisionVelocity of strike =√2gh⇒ Velocity of rebound =e×√2gh=u⇒ height attained =u22g=22×2gh2g=e2hIInd collisionVelocity of strike =√29×e2h=e√2ghVelocity of rebound =e×(e√2gh)=e2√2gh=v⇒ height attained =v22g=e4×2gh2g8e4h.