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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

`40^@`

D

`90^@`

Text Solution

Verified by Experts

The correct Answer is:
D
`vecF = (dvecP)/(dt) = -kA sin (kt) hati - kA cos (kt)hatj`
`vecP = A cos (kt) hati - A sin (kt) hatj`
Now `vecF cdot vecP = 0`
Therefore, angle between `vecF and vecP` should be `90^@`.
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