The figure shows a body of mass m moving with | vedclass

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(A) The velocity at point $A$ is directed along the tangent,which is $\vec{v}_A = v\hat{j}$.The velocity at point $B$ is directed along the tangent,wh

Vedclass · Accepted Answer

$v\sqrt{2}$

Vedclass · Answer

(A) The velocity at point $A$ is directed along the tangent,which is $\vec{v}_A = v\hat{j}$.The velocity at point $B$ is directed along the tangent,which is $\vec{v}_B = -v\hat{i}$.The change in velocity is $\Delta \vec{v} = \vec{v}_B - \vec{v}_A = -v\hat{i} - v\hat{j}$.The magnitude of the change in velocity is $|\Delta \vec{v}| = \sqrt{(-v)^2 + (-v)^2} = \sqrt{2v^2} = v\sqrt{2}$.Alternatively,using the formula for change in velocity for an angle $	heta$ between two velocity vectors: $|\Delta \vec{v}| = 2v \sin(	heta/2)$.Here,the angle $	heta$ between the radii to $A$ and $B$ is $90^\circ$.$|\Delta \vec{v}| = 2v \sin(90^\circ/2) = 2v \sin(45^\circ) = 2v 	imes (1/\sqrt{2}) = v\sqrt{2}$.

The figure shows a body of mass $m$ moving with a uniform speed $v$ along a circle of radius $r$. The change in velocity in going from $A$ to $B$ is

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