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Question

A ball falls on an inclined plane of inclination θ from a height h above the point of impact and makes a perfectly elastic collision.Where will it hit the plane again ?

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Solution

The ball strikes the inclined plane at origin with velocity v0=2gh.As the ball elastically rebounds,it recalls with same velocity v0,at the same angle θ from the normal or y-axis.Let the ball strikes the incline second time at any point P,which is at a distance l from the origin along the incline.From the equation,

y=v0yt+12wyt2

0=v0 cos θ t12g cos θ t2

Which t is the same time of motion of ball in air which moving from origin to P.

As t0,so t=2v0g

Now from the equation x=v0xt+12wxt2

l=v0sin θt+12g sin θ t2

So, l=v0sin θ(2v0g)+12g sin θ (2v0g)2

=4v20sinθg

Hence the plane will hit again at a distance l=8 h sin θ


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