A particle of mass m , kg moving with a veloc | vedclass

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(B) Let the initial velocity be $\vec{v}_1$ and the final velocity be $\vec{v}_2$. From the geometry,the initial momentum is $\vec{P}_1 = m\vec{v}_1 =

Vedclass · Accepted Answer

$2mv \sin 	heta$

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(B) Let the initial velocity be $\vec{v}_1$ and the final velocity be $\vec{v}_2$. From the geometry,the initial momentum is $\vec{P}_1 = m\vec{v}_1 = -mv \sin 	heta \hat{i} - mv \cos 	heta \hat{j}$. The final momentum is $\vec{P}_2 = m\vec{v}_2 = mv \sin 	heta \hat{i} - mv \cos 	heta \hat{j}$. The change in momentum is $\Delta \vec{P} = \vec{P}_2 - \vec{P}_1$. $\Delta \vec{P} = (mv \sin 	heta \hat{i} - mv \cos 	heta \hat{j}) - (-mv \sin 	heta \hat{i} - mv \cos 	heta \hat{j})$. $\Delta \vec{P} = 2mv \sin 	heta \hat{i}$. The magnitude of the change in momentum is $|\Delta \vec{P}| = 2mv \sin 	heta$.

$A$ particle of mass $m \, kg$ moving with a velocity $v \, m/s$ strikes a surface as shown in the figure and rebounds with the same speed. The magnitude of the change in momentum is:

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