A ball of mass 50 g moving at a speed of 2.0 m/s strikes a plane surface at an angle of incidence ${45}^{\circ }$.The ball is reflected by the plane at equal angle of reflection with the same speed.Calculate (a) the magnitude of the change in momentum of the ball (b) the change in the magnitude of the momentum of the ball.

Mass=50 g=0.05 kg

$v=\left(2cos{45}^{\circ }\right)\overrightarrow{i}-2sin{45}^{\circ }\overrightarrow{j}$

$v_1=-2cos{45}^{\circ }\overrightarrow{i}-2sin{45}^{\circ }\overrightarrow{j}$

(a) Change in momentum=$m\overrightarrow{v}-m\overrightarrow{v_1}$

$=0.05(2cos{45}^{\circ }\overrightarrow{i}-2sin{45}^{\circ }\overrightarrow{j})-0.05(-2cos{45}^{\circ }\overrightarrow{i}-2sin{45}^{\circ }\overrightarrow{j})$

$=0.1cos{45}^{\circ }\overrightarrow{j}-0.1sin{45}^{\circ }\overrightarrow{i}+0.1sin{45}^{\circ }\overrightarrow{j}+0.1sin{45}^{\circ }\overrightarrow{j})$

$=0.2cos{45}^{\circ }\overrightarrow{j}$

$\therefore Magnitude=\frac{0.2}{\sqrt{2}}=\frac{0.2828}{2}$

(b) The change in the magnitude of the momentum of the ball

$|\overrightarrow{P_2}|-|\overrightarrow{P_1}|=2\times 0.5-2\times 0.5=0$