A ball of mass 3 kg strikes a wall with a ve | 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 comp

Vedclass · Accepted Answer

$30 \ kg \cdot m/s$

Vedclass · Answer

(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 normal to the wall be along the $x$-axis. The initial velocity is $\vec{v}_i = (v \cos 30^\circ) \hat{i} - (v \sin 30^\circ) \hat{j}$.The final velocity is $\vec{v}_f = -(v \cos 30^\circ) \hat{i} - (v \sin 30^\circ) \hat{j}$.The change in momentum is $\Delta \vec{p} = m(\vec{v}_f - \vec{v}_i) = m(-2v \cos 30^\circ) \hat{i}$.The magnitude of the change in momentum is $|\Delta p| = 2mv \cos 30^\circ$.Given $m = 3 \ kg$, $v = 10 \ m/s$, and $\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$.$|\Delta p| = 2 	imes 3 	imes 10 	imes \frac{\sqrt{3}}{2} = 30\sqrt{3} \approx 51.96 \ kg \cdot m/s$. Note: If the angle with the wall is $60^\circ$, the angle with the normal is $30^\circ$. If the angle with the normal is $60^\circ$, $|\Delta p| = 2 	imes 3 	imes 10 	imes 0.5 = 30 \ kg \cdot m/s$. Given the options, the angle with the normal is $60^\circ$.

$A$ ball of mass $3 \ kg$ strikes a wall with a velocity of $10 \ m/s$ at an angle of $30^\circ$ with the normal to the wall and rebounds at the same angle and speed. What is the change in momentum of the ball in $MKS$ units?

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