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(C) In uniform circular motion,the speed $v$ of the particle remains constant,so the kinetic energy $K = \frac{1}{2} M v^{2}$ remains constant. Thus,t

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momentum changes by $2 M v$

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(C) In uniform circular motion,the speed $v$ of the particle remains constant,so the kinetic energy $K = \frac{1}{2} M v^{2}$ remains constant. Thus,the change in kinetic energy is $0$.At diametrically opposite points,the velocity vectors are in opposite directions. Let the initial velocity be $\vec{v}_1 = v \hat{i}$ and the final velocity be $\vec{v}_2 = -v \hat{i}$.The change in momentum $\Delta \vec{p}$ is given by $\Delta \vec{p} = M \vec{v}_2 - M \vec{v}_1$.Substituting the values,$\Delta \vec{p} = M(-v \hat{i}) - M(v \hat{i}) = -2 M v \hat{i}$.The magnitude of the change in momentum is $|\Delta \vec{p}| = 2 M v$.

$A$ particle of mass $M$ is moving in a horizontal circle of radius $R$ with uniform speed $v$. When it moves from one point to a diametrically opposite point,its

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