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(D) The impulse imparted to the particle by the wall is equal to the change in its linear momentum.Let the wall be along the $y$-axis. The velocity of

Vedclass · Accepted Answer

$\sqrt{3} m v$

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(D) The impulse imparted to the particle by the wall is equal to the change in its linear momentum.Let the wall be along the $y$-axis. The velocity of the particle before collision is $\vec{v}_i = v \cos 30^{\circ} \hat{i} - v \sin 30^{\circ} \hat{j}$.Since the collision is elastic,the component of velocity parallel to the wall remains unchanged,and the component perpendicular to the wall reverses its direction.Thus,the velocity after collision is $\vec{v}_f = -v \cos 30^{\circ} \hat{i} - v \sin 30^{\circ} \hat{j}$.The change in momentum $\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = m(-2v \cos 30^{\circ} \hat{i}) = -2mv \cos 30^{\circ} \hat{i}$.The magnitude of impulse is $|\Delta \vec{p}| = 2mv \cos 30^{\circ} = 2mv \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} mv$.

$A$ particle of mass $m$ strikes a wall with speed $v$ at an angle $30^{\circ}$ with the wall elastically as shown in the figure. The magnitude of impulse imparted to the ball by the wall is

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