A smooth sphere moves on a horizontal surface | vedclass

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(B) The wall is parallel to the $\hat{j}$ axis,meaning the wall lies in the $y$-direction. The normal to the wall is along the $x$-direction $(\hat{i}

Vedclass · Accepted Answer

$-\hat{i} + 2\hat{j}$

Vedclass · Answer

(B) The wall is parallel to the $\hat{j}$ axis,meaning the wall lies in the $y$-direction. The normal to the wall is along the $x$-direction $(\hat{i})$.Before collision,the velocity components are $v_x = 2 	ext{ m/s}$ and $v_y = 2 	ext{ m/s}$.During the collision,the velocity component parallel to the wall $(v_y)$ remains unchanged because the wall is smooth.Thus,$v_y' = v_y = 2 	ext{ m/s}$.The velocity component perpendicular to the wall $(v_x)$ changes according to the coefficient of restitution $e$:$v_x' = -e \cdot v_x = -(1/2) \cdot 2 = -1 	ext{ m/s}$.Therefore,the final velocity vector is $\vec{v} = v_x'\hat{i} + v_y'\hat{j} = -\hat{i} + 2\hat{j}$.

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