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A paarticle moves in the xy-plane under ...

A paarticle moves in the xy-plane under the action of a force F such that the componentes of its linear momentum p at any time t and `p_(x)=2cos` t, `p_(y)=2sin` t. the eangle between F and p at time l is

A

`90^(@)`

B

`0^(@)`

C

`180^(@)`

D

`30^(@)`

Text Solution

Verified by Experts

The correct Answer is:
A

Given that `vecp=p_(x)hati+p_(y)hatj=2costhati+2sinthatj`
`therefore vecF=(dvecp)/(dt)=-2sinthati+2costhatj`
Now. `vecF.vecp=0` i.e., angle between `vecF and vecp` is `90^(@)`.
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