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A particle moves in the xy plane under t...

A particle moves in the `xy` plane under the influence of a force such that its linear momentum is `vecP(t) = A [haticos(kt)-hatjsin(kt)]`, where `A` and `k` are constants. The angle between the force and momentum is

A

`0^(@)`

B

`30^(@)`

C

`45^(@)`

D

`90^(@)`

Text Solution

Verified by Experts

The correct Answer is:
D
`vecP(t) = A[hati cos (kt)- hatj sin (kt)]`
`vecF =(dvecp)/(dt) =Ak[-hati sin (kt) -hatj cos(kt)]`
Here, `vecP.vecF =0` therefore
The angle between `vecP` and `vecF` is `90^(@)`
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