A sphere of mass m comes down on a smooth inc | vedclass

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Question

Let velocity of the particle at the base of the inclined plane be $v$ Work-energy theorem: $\quad W_{g}=\Delta K . E$ $m g h=\frac{1}{2} m v^{2}-0$

Vedclass · Accepted Answer

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

Vedclass · Answer

Let velocity of the particle at the base of the inclined plane be $v$ Work-energy theorem: $\quad W_{g}=\Delta K . E$ $m g h=\frac{1}{2} m v^{2}-0$

$\Longrightarrow v=\sqrt{2 g h}$

Momentum of the particle at point $\mathrm{C}, \quad \vec{P}_{c}=m v \hat{i}$ Momentum of the particle at point $A,$ $\quad \vec{P}_{A}=m v \cos 	heta \hat{i}-m v \sin 	heta \hat{j}$

Change in momentum between $A$ and $C,$ $\quad \Delta \vec{P}=\vec{P}_{c}-\vec{P}_{A}$ $\Delta \vec{P}=m v(1-\cos 	heta) \hat{i}-m v \sin 	heta \hat{j}$

$|{\Delta \vec{P}}|=\sqrt{m^{2} v^{2}(1-\cos 	heta)^{2}+m^{2} v^{2} \sin ^{2} 	heta}$

$|\Delta \vec{P}|=m v \sqrt{1+\cos ^{2} 	heta-2 \cos 	heta+\sin ^{2} 	heta}$

$|\Delta \vec{P}|=m v \sqrt{2(1-\cos 	heta)} \quad\left(\because 1-\cos 	heta=2 \sin ^{2} \frac{	heta}{2}ight)$

$\Longrightarrow|\Delta \vec{P}|=2 m v \sin \left(\frac{	heta}{2}ight)=2 m \sqrt{2 g h} \sin \left(\frac{	heta}{2}ight)$

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