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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 isA. `0^(@)`B. `30^(@)`C. `45^(@)`D. `90^(@)` |
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Answer» Correct Answer - D `vecF=(vec(dP))/(dt)=kAsin(kt)hati-kAcos(kt)hatj` `vecP=cos(kt)hati-Asin(kt)hatj` Now `vecF.vecP=0` Therefore angle betweenn `vecF` and `vecP `should be `90^(@)` |
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