Two cars,both of mass m,collide and stick tog | vedclass

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(C) Let the $x$-axis be East and the $y$-axis be North.Initial momentum of the first car (moving North): $\vec{p}_1 = m(2v)\hat{j} = 2mv\hat{j}$.Initi

Vedclass · Accepted Answer

$\frac{v}{2}\sqrt{5 - 4\sin \phi}$

Vedclass · Answer

(C) Let the $x$-axis be East and the $y$-axis be North.Initial momentum of the first car (moving North): $\vec{p}_1 = m(2v)\hat{j} = 2mv\hat{j}$.Initial momentum of the second car (moving at angle $\phi$ South of East): $\vec{p}_2 = m(v \cos \phi \hat{i} - v \sin \phi \hat{j}) = mv \cos \phi \hat{i} - mv \sin \phi \hat{j}$.Total initial momentum: $\vec{P}_{initial} = \vec{p}_1 + \vec{p}_2 = mv \cos \phi \hat{i} + (2mv - mv \sin \phi) \hat{j}$.After the collision,the cars stick together with mass $2m$ and velocity $\vec{V}' = V'_x \hat{i} + V'_y \hat{j}$.By conservation of linear momentum: $2m V'_x = mv \cos \phi \implies V'_x = \frac{v \cos \phi}{2}$.$2m V'_y = 2mv - mv \sin \phi \implies V'_y = \frac{v(2 - \sin \phi)}{2}$.The magnitude of the final velocity is $V' = \sqrt{(V'_x)^2 + (V'_y)^2} = \sqrt{\left(\frac{v \cos \phi}{2}\right)^2 + \left(\frac{v(2 - \sin \phi)}{2}\right)^2}$.$V' = \frac{v}{2} \sqrt{\cos^2 \phi + 4 - 4 \sin \phi + \sin^2 \phi} = \frac{v}{2} \sqrt{1 + 4 - 4 \sin \phi} = \frac{v}{2} \sqrt{5 - 4 \sin \phi}$.

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