Question

A ball is moving with speed $20m/s$ collides with a smooth surface as shown in figure. The magnitude of change in velocity of the ball will be

• A. $10\sqrt{3}$
• B. $20\sqrt{3}$
• C. $\frac{40}{\sqrt{3}}$
• D. $40m/s$

Answer 1

The correct option is B $20\sqrt{3}$

R.E.F image

Rescue the vector into the components

Initial velocity

$\overrightarrow{V_i}=20(-cos{30}^{\circ }\hat{i}-sin{30}^{\circ }\hat{j})$

$=20(\frac{\sqrt{3}}{2}\hat{i}-\frac{1}{2}\hat{j})$

$=10(-\sqrt{3}\hat{i}-\hat{j})$

Final velocity

$\overrightarrow{V_f}=20(-sin{30}^{\circ }\hat{i}+cos{30}^{\circ }\hat{j})=10(-\hat{i}+\sqrt{3}\hat{j})$

$\therefore$ change In velocity

$\Delta \overrightarrow{v}=\overrightarrow{V_f}-\overrightarrow{V_i}=10(-\sqrt{3}\hat{i}-\hat{j}+\hat{i}-\sqrt{3}\hat{j})$

$=10\left[\right(1-\sqrt{3})\hat{i}-\sqrt{3}\hat{j}]$

So, magnitude

$|\Delta \overrightarrow{v}|=10\sqrt{(1-\sqrt{3}{)}^2(1+\sqrt{3}{)}^2}=20\sqrt{2}m/s$