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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^(@)` |
Answer» Correct Answer - 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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