An object of mass m , kg with speed v , m/s s | vedclass

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(A) Let the normal to the wall be along the $y$-axis and the wall be along the $x$-axis.The initial velocity is $\vec{v}_1 = v \sin 	heta \hat{i} - v

Vedclass · Accepted Answer

$2m v \cos 	heta$

Vedclass · Answer

(A) Let the normal to the wall be along the $y$-axis and the wall be along the $x$-axis.The initial velocity is $\vec{v}_1 = v \sin 	heta \hat{i} - v \cos 	heta \hat{j}$.The final velocity is $\vec{v}_2 = v \sin 	heta \hat{i} + v \cos 	heta \hat{j}$.The initial momentum is $\vec{P}_1 = m\vec{v}_1 = m v \sin 	heta \hat{i} - m v \cos 	heta \hat{j}$.The final momentum is $\vec{P}_2 = m\vec{v}_2 = m v \sin 	heta \hat{i} + m v \cos 	heta \hat{j}$.The change in momentum is $\Delta \vec{P} = \vec{P}_2 - \vec{P}_1$.$\Delta \vec{P} = (m v \sin 	heta \hat{i} + m v \cos 	heta \hat{j}) - (m v \sin 	heta \hat{i} - m v \cos 	heta \hat{j}) = 2 m v \cos 	heta \hat{j}$.The magnitude of the change in momentum is $|\Delta \vec{P}| = 2 m v \cos 	heta$.

An object of mass $m \, kg$ with speed $v \, m/s$ strikes a wall at an angle $\theta$ with the normal and rebounds at the same speed and same angle. The magnitude of the change in momentum of the object will be

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