A ball of mass 3 kg strikes a wall at an ang | vedclass

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(A) The change in momentum $\Delta \vec{p}$ is given by $\vec{p}_f - \vec{p}_i$.Since the ball rebounds at the same angle and speed,the velocity compo

Vedclass · Accepted Answer

$30 \ kg \cdot m/s$

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(A) The change in momentum $\Delta \vec{p}$ is given by $\vec{p}_f - \vec{p}_i$.Since the ball rebounds at the same angle and speed,the velocity component parallel to the wall remains unchanged,while the component perpendicular to the wall reverses direction.Let the wall be along the $y$-axis. The initial velocity is $\vec{v}_i = (v \cos 60^\circ) \hat{i} - (v \sin 60^\circ) \hat{j}$.The final velocity is $\vec{v}_f = -(v \cos 60^\circ) \hat{i} - (v \sin 60^\circ) \hat{j}$.The change in momentum is $\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = m(-2v \cos 60^\circ) \hat{i}$.The magnitude of the change in momentum is $|\Delta p| = 2mv \cos 60^\circ$.Given $m = 3 \ kg$,$v = 10 \ m/s$,and $\cos 60^\circ = 0.5$.$|\Delta p| = 2 	imes 3 	imes 10 	imes 0.5 = 30 \ kg \cdot m/s$.

$A$ ball of mass $3 \ kg$ strikes a wall at an angle of $60^\circ$ with a velocity of $10 \ m/s$ and rebounds at the same angle and speed. What is the change in momentum of the ball in $MKS$ units?

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