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A particle sides down from rest on an in...

A particle sides down from rest on an inclined plane of angle `theta` with horizontal. The distances are as shown. The particle slides down to the position `A`, where it velocity is `v`.
.

A

`(v^(2) - 2 gh)` will remain zero

B

`(v^(2) - 2gs sin theta)` will remain zero

C

`[(v^(2)-2gs (H-h))/((p-s))]` will remain zero

D

`[v^(2) -(2gsH)/(p)]` will remain zero.

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`h = s sin theta`
`(1)/(2) mv^(2)+ mg (H-h) =mgH`
`(1)/(2) mv^(2) -mgh =zero`
`v^(2) -2gh = zero , v^(2) -2gs(sin theta) =zero`
But `sin theta = (H)/(P) =((H-h))/((p-s))`
So, `v^(2)-2gs ((H-h))/((p-s)) = zero` and `v^(2)-2gs (H)/(p) = zero`.
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Knowledge Check

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