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  • A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle \theta with speed v_{o} and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle \theta with speed v_{o} and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.



 

Answers (1)

The particle rebounds from P. when it strikes plane inclined at v0 speed. Hence the speed of particle after it rebounds from P will be v0 We assume the new axis X’OX and YOY’ axis at P as origin ‘O’.  The components of g and v0 in the new OX and OY axis are:

Focusing on the motion of the particle from O to A,

s=ut+\frac{1}{2}gt^{2}

Here, t=T which is the time of flight

0=T[v\; \cos\; \theta -\frac{1}{2}g\; \sin\; \theta \; T]

So, either T=0\; or \left [ v\; \cos\; \theta -\frac{1}{2}g\; \sin\; \theta \; T \right ]=0

S_{x}=u_{x}t+\frac{1}{2}a_{x}t^{2}

L=[\frac{2v}{g}]v\; \sin \theta +\frac{1}{2}g\; \sin\; \theta [\frac{2v}{g}]^{2}

L=\frac{4v^{2}}{g}.\sin \theta

 

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infoexpert23

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