Question

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 the magnitude of the change in momentum of the ball.

• A. 0 zero
• B. 0.14 kg m/s
• C. 0.14 kg m/s
• D. zero

Answer 1

The correct option is C

0.14 kg m/s

The situation in the question can be represented diagrammatically as - shown in figure.

As clear from the figure ,
$\overrightarrow{v_i}=2cos450\hat{i}-2sin450\hat{j}$
$\overrightarrow{v_i}=2(\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j})$
Similarly, $\overrightarrow{v_f}=2(\frac{-1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j})$
Change in momentum of the ball is
${\overrightarrow{P}}_f-{\overrightarrow{P}}_i=0.05(-\sqrt{2}\hat{i}-\sqrt{2}\hat{j})-0.05(\sqrt{2}\hat{i}-\sqrt{2}\hat{j})$
$=-\left(0.05\right)\sqrt{2}\hat{i}-0.05\sqrt{2}\hat{j}-0.05\sqrt{2}\hat{i}+0.05\sqrt{2}\hat{j}$
$=-2\sqrt{2}\left(0.05\right)\hat{i}$
$=-0.14\hat{i}kgm/s$
$\therefore |{\overrightarrow{P}}_f-{\overrightarrow{P}}_i|=0.14kgm/s.$