A ball of mass 0.3 kg strikes a rigid vertical wall with a speed of...

Let the wall be along the y-axis and the normal to the wall be along the x-axis.

The angle made by the velocity vector with the surface of the wall is $\theta = 30^\circ$.

The component of velocity perpendicular to the wall is $v_{\perp} = v \sin 30^\circ$.
The component of velocity parallel to the wall is $v_{\parallel} = v \cos 30^\circ$.

During the collision, the parallel component remains unchanged, while the perpendicular component reverses its direction.

Initial momentum perpendicular to the wall, $p_{i} = m v \sin 30^\circ$
Final momentum perpendicular to the wall, $p_{f} = -m v \sin 30^\circ$

Magnitude of impulse = Change in momentum = $|p_{f} - p_{i}| = 2 m v \sin 30^\circ$

Impulse $= 2 \times 0.3 \times 10 \times \dfrac{1}{2} = 3 \text{ kg m s}^{-1}$.

Answer: $3 \text{ kg m s}^{-1}$