A ball is moving with speed 20 , m/s collides | vedclass

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Question

(B) From the figure,the ball approaches the surface at an angle of $30^{\circ}$ with the normal and rebounds at the same angle.Initial velocity vector

Vedclass · Accepted Answer

$20 \sqrt{3}$

Vedclass · Answer

(B) From the figure,the ball approaches the surface at an angle of $30^{\circ}$ with the normal and rebounds at the same angle.Initial velocity vector: $\vec{v}_i = (20 \sin 30^{\circ} \hat{i} - 20 \cos 30^{\circ} \hat{j}) \, m/s$Final velocity vector: $\vec{v}_f = (-20 \sin 30^{\circ} \hat{i} - 20 \cos 30^{\circ} \hat{j}) \, m/s$Change in velocity: $\Delta \vec{v} = \vec{v}_f - \vec{v}_i$$\Delta \vec{v} = (-20 \sin 30^{\circ} \hat{i} - 20 \cos 30^{\circ} \hat{j}) - (20 \sin 30^{\circ} \hat{i} - 20 \cos 30^{\circ} \hat{j})$$\Delta \vec{v} = -40 \sin 30^{\circ} \hat{i}$Magnitude of change in velocity: $|\Delta \vec{v}| = 40 	imes \sin 30^{\circ} = 40 	imes 0.5 = 20 \, m/s$.Wait,re-evaluating based on the standard reflection geometry: The component parallel to the surface remains unchanged,while the component perpendicular to the surface reverses. Let the normal be along the $y$-axis. $\vec{v}_i = v \sin 30^{\circ} \hat{i} - v \cos 30^{\circ} \hat{j}$$\vec{v}_f = -v \sin 30^{\circ} \hat{i} - v \cos 30^{\circ} \hat{j}$$\Delta \vec{v} = \vec{v}_f - \vec{v}_i = -2v \sin 30^{\circ} \hat{i} = -2(20)(0.5) \hat{i} = -20 \hat{i}$.If the normal is along the $x$-axis:$\vec{v}_i = -v \cos 30^{\circ} \hat{i} - v \sin 30^{\circ} \hat{j}$$\vec{v}_f = v \cos 30^{\circ} \hat{i} - v \sin 30^{\circ} \hat{j}$$\Delta \vec{v} = \vec{v}_f - \vec{v}_i = 2v \cos 30^{\circ} \hat{i} = 2(20)(\frac{\sqrt{3}}{2}) \hat{i} = 20\sqrt{3} \hat{i}$.Given the options,the magnitude is $20\sqrt{3}$.

$A$ ball is moving with speed $20 \, m/s$ collides with a smooth surface as shown in the figure. The magnitude of the change in velocity of the ball will be ........... $m/s$.

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