1.

A particle moves in X-Y plane under the action of forces F such that the values of linear momentum ‘p’ at any time is p_(x) =2 cos t and p_(y) = 2 sin t. The angle betweenvec(F) and vec(p) at the time t will be :

Answer»

<P>0°
30°
90°
180°

Solution :Here `VEC(p)=vec(p_(x))HATI + vec(p_(y))hatj=2costhati+2sinthatj`
Now `vec(F)=(dvecp)/(t)=-2sinthati+2costhatj`. The angle will be
`costheta=(vec(F).vec(p))/(Fp)`
`costheta=(-4sintcost + 4sintcost)/(Fxxp)`
`theta=90^@`


Discussion

No Comment Found

Related InterviewSolutions