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(C) The kinetic energy of the particle is given by $K = (1/2)MV^2$. Since the speed $V$ is uniform,the kinetic energy remains constant throughout the

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Momentum changes by $2MV$

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(C) The kinetic energy of the particle is given by $K = (1/2)MV^2$. Since the speed $V$ is uniform,the kinetic energy remains constant throughout the motion. Therefore,the change in kinetic energy is $0$.Let the initial velocity at one point be $\vec{v}_1 = V\hat{i}$.At the diametrically opposite point,the velocity will 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,$\Delta \vec{p} = M(-V\hat{i}) - M(V\hat{i}) = -2MV\hat{i}$.The magnitude of the change in momentum is $|\Delta \vec{p}| = 2MV$.

$A$ 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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