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(C) Let the velocity of the particle at the base of the inclined plane (point $A$) be $v$. Using the work-energy theorem: $W_g = \Delta K.E.$$mgh = \f

Vedclass · Accepted Answer

$2m \sqrt{2gh} \sin(	heta/2)$

Vedclass · Answer

(C) Let the velocity of the particle at the base of the inclined plane (point $A$) be $v$. Using the work-energy theorem: $W_g = \Delta K.E.$$mgh = \frac{1}{2}mv^2 - 0$$\implies v = \sqrt{2gh}$.At point $C$ (on the horizontal surface),the velocity is $v$ in the horizontal direction: $\vec{P}_C = mv \hat{i}$.At point $A$ (at the bottom of the incline),the velocity is $v$ directed along the incline at an angle $	heta$ below the horizontal: $\vec{P}_A = mv \cos 	heta \hat{i} - mv \sin 	heta \hat{j}$.The change in momentum between $A$ and $C$ is $\Delta \vec{P} = \vec{P}_C - \vec{P}_A = (mv - mv \cos 	heta) \hat{i} + mv \sin 	heta \hat{j}$.The magnitude is $|\Delta \vec{P}| = \sqrt{(mv(1 - \cos 	heta))^2 + (mv \sin 	heta)^2}$.$|\Delta \vec{P}| = mv \sqrt{1 + \cos^2 	heta - 2 \cos 	heta + \sin^2 	heta} = mv \sqrt{2 - 2 \cos 	heta}$.Using the identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$:$|\Delta \vec{P}| = mv \sqrt{4 \sin^2(	heta/2)} = 2mv \sin(	heta/2)$.Substituting $v = \sqrt{2gh}$,we get $|\Delta \vec{P}| = 2m \sqrt{2gh} \sin(	heta/2)$.

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