A particle is moving in xy-plane and at any instant of time its position vector is given by r→=acosθi^+asinθj^. Where θ is the angle between r→ and positive x-axis. If the path of the particle is circular, then unit vector of linear momentum of the particle at that instant is

If the path followed by the particle is circular, then linear momentum vector p→ and position vector r→ must always be perpendicular. If p^=li^+mj^, then p^.r→=0⇒alcosθ+amsinθ=0⇒lcosθ+msinθ=0. Also l2+m2=1 solving, l=±sinθ and m=∓cosθ.∴ p^=sinθi^−cosθj^ or p^=−sinθi^+cosθj^.

A particle is moving in xy-plane and at any instant of time its position vector is given by r→=acosθi^+asinθj^. Where θ is the angle between r→ and positive x-axis. If the path of the particle is circular, then unit vector of linear momentum of the particle at that instant is

If the path followed by the particle is circular, then linear momentum vector p→ and position vector r→ must always be perpendicular. If p^=li^+mj^, then p^.r→=0⇒alcosθ+amsinθ=0⇒lcosθ+msinθ=0. Also l2+m2=1 solving, l=±sinθ and m=∓cosθ.∴ p^=sinθi^−cosθj^ or p^=−sinθi^+cosθj^.

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A particle is moving in xy-plane and at any instant of time its position vector is given by $\vec{r} = (a \cos θ \hat{i} + a \sin θ \hat{j})$ . Where $θ$ is the angle between $\vec{r}$ and positive x-axis. If the path of the particle is circular, then unit vector of linear momentum of the particle at that instant is

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$(- \sin θ \hat{i} + \cos θ \hat{j})$

$(\cos θ \hat{i} - \sin θ \hat{j})$

both (2) and (3) are right

$(\sin θ \hat{i} - \cos θ \hat{j})$

answer is D.

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Detailed Solution

If the path followed by the particle is circular, then linear momentum vector $\vec{p}$ and position vector $\vec{r}$ must always be perpendicular. If $\hat{p} = (l \hat{i} + m \hat{j})$ , then $\hat{p} . \vec{r} = 0 \Rightarrow a l \cos θ + a m \sin θ = 0 \Rightarrow l \cos θ + m \sin θ = 0$ . Also $l^{2} + m^{2} = 1$ solving, $l = \pm \sin θ$ and $m = \mp \cos θ$ .