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(B) The initial velocity of the sphere is $\vec{u} = 2 \hat{i} + 2 \hat{j} \ m/s$.Since the wall is parallel to the $y$-axis (vector $\hat{j}$),the wa

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

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(B) The initial velocity of the sphere is $\vec{u} = 2 \hat{i} + 2 \hat{j} \ m/s$.Since the wall is parallel to the $y$-axis (vector $\hat{j}$),the wall is perpendicular to the $x$-axis.During the collision,the impulse acts only along the normal to the wall,which is the $x$-direction.Therefore,the velocity component parallel to the wall ($y$-component) remains unchanged: $v_y = u_y = 2 \hat{j} \ m/s$.For the $x$-component,we use the coefficient of restitution $e = -\frac{v_x}{u_x}$,where $u_x = 2 \ m/s$ and $e = 1/2$.$v_x = -e \cdot u_x = -\frac{1}{2} \cdot 2 = -1 \ m/s$.Thus,the velocity vector after the collision is $\vec{v} = v_x \hat{i} + v_y \hat{j} = -1 \hat{i} + 2 \hat{j} \ m/s$.

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