Why is vec{v} times vec{p} = 0 for a particle | vedclass

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(B) The linear momentum $\vec{p}$ of a particle is defined as $\vec{p} = m\vec{v}$,where $m$ is the mass and $\vec{v}$ is the velocity vector.Since $m

Vedclass · Accepted Answer

Because the cross product of two parallel vectors is zero.

Vedclass · Answer

(B) The linear momentum $\vec{p}$ of a particle is defined as $\vec{p} = m\vec{v}$,where $m$ is the mass and $\vec{v}$ is the velocity vector.Since $m$ is a scalar,$\vec{p}$ is always in the same direction as $\vec{v}$.Two vectors are parallel if the angle between them is $0^\circ$.The cross product of two vectors $\vec{A}$ and $\vec{B}$ is given by $\vec{A} 	imes \vec{B} = |A||B| \sin(	heta) \hat{n}$.For $\vec{v}$ and $\vec{p}$,the angle $	heta = 0^\circ$,so $\sin(0^\circ) = 0$.Therefore,$\vec{v} 	imes \vec{p} = |v||p| \sin(0^\circ) \hat{n} = 0$.This confirms that for any particle,the cross product of its velocity and momentum is zero because they are collinear.

Why is $\vec{v} \times \vec{p} = 0$ for a particle moving in a straight line,and how does this relate to the angular momentum of a rotating particle?

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