A ball of mass 50 g travelling at a speed of 2.0 m/s strikes a...

Mass of the ball, $m = 50 \text{ g} = 0.05 \text{ kg}$

Speed of the ball, $v = 2.0 \text{ m/s}$

Angle of incidence, \theta = 45^{}^{{}^{\circ}}

The component of momentum parallel to the surface remains unchanged during the collision.

The component of momentum perpendicular to the surface is reversed.

Initial momentum perpendicular to the surface, $p_{i} = -mv \cos \theta$

Final momentum perpendicular to the surface, $p_{f} = mv \cos \theta$

Magnitude of the change in momentum, $|\Delta p| = |p_{f} - p_{i}| = 2mv \cos \theta$

Substituting the given values:

|\Delta p| = 2 \times 0.05 \times 2.0 \times \cos 45^{}^{{}^{\circ}}

$|\Delta p| = 0.2 \times \dfrac{1}{\sqrt{2}}$

$|\Delta p| = 0.1 \sqrt{2} \approx 0.1 \times 1.414 = 0.1414 \text{ kg-m/s}$

Rounding off, the change in momentum is $0.14 \text{ kg-m/s}$.