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Physics Question on Motion in a plane

Figure shows a body of mass m moving with a uniform speed vv along a circle of radius rr. The change in velocity in going from AA to BB is:

A

v2v\sqrt{2}

B

v/2v/\sqrt{2}

C

vv

D

zero

Answer

v2v\sqrt{2}

Explanation

Solution

When a body rotates uniformly, then the direction of velocity changes continuously but its magnitude remains constant.
Also, direction of velocity is perpendicular to direction of motion.
∴v⃗A=j^v⃗\therefore \vec{v}_{A}=\hat{j} \vec{v}
v⃗B=−i^v⃗\vec{v}_{B}=-\hat{i} \vec{v}
Change in velocity
Δv⃗=v⃗B−v⃗A\Delta \vec{v}=\vec{v}_{B}-\vec{v}_{A}
=−i^v⃗−j^v⃗=-\hat{i} \vec{v}-\hat{j} \vec{v}
Magnitude of change in velocity is
=∣−i^v⃗−j^v⃗∣=|-\hat{i} \vec{v}-\hat{j} \vec{v}|
=v2+v2=v2=\sqrt{v^{2}+v^{2}}=v \sqrt{2}