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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
`P_x=2 cos t, P_y = 2 sin t`
`therefore vecP=2costveci + 2 sinvecj`
`vecF=(dvecP)/(dt)=-2 sint veci + 2cos vecj`
`vecF.vecP=0 therefore theta=90^@`
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Knowledge Check

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