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A particle moves in the x-y plane under ...

A particle moves in the `x-y` plane under the action of a force `vecF` such that the value of its linear momentum `vecP` at any time `t is P_(x) = 2 cos t, P_(y) = 2 sin t`.
The angle `theta` between vecF and vecP` at a given time `t` will be:

A

`theta=0^(@)`

B

`theta=30^(@)`

C

`theta=90^(@)`

D

`theta=180^(@)`

Text Solution

Verified by Experts

The correct Answer is:
C
`vecp=2 cos thati+2sinthatj`
`vecF=(dvecp)/(dt)=-2sint hati+2 cos hattj`
`vecp.vecF=0 rArr vecp_|_ "to" vecF`
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