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A ball is moving with speed $20 \,m / s$ collides with a smooth surface as shown in figure. The magnitude of change in velocity of the ball will be ........... $m/s$
$10 \sqrt{3}$
$20 \sqrt{3}$
$\frac{40}{\sqrt{3}}$
$40$
Solution
(b)
From the above figure,
Initial velocity, $\vec{v}_{ i }=\left(-20 \sin 30^o \hat i-20 \cos 30^{\circ} \hat j \right) m / s$
Final velocity, $\overrightarrow{ v }_{ f }=\left(-20 \sin 30^{o}\hat i+20 \cos 30^o \hat j \right) m / s$
Change in velocity, $\Delta \vec{v}=\vec{v}_f-\vec{v}_i$
$\Delta \overrightarrow{ v }=\left(-20 \sin 30^{o}{ \hat{ i }}+20 \cos 30^0 j \right)-\left(-20 \sin 30^o \hat{ i }-20 \cos 30^{o}{\hat j }\right)$
$\Delta \overrightarrow{ v }=2 \times 20 \cos 30^{o}{ \hat{ j }}$
$\Delta \overrightarrow{ v }=2 \times 20 \times \frac{\sqrt{3}}{2} \hat{ j }$
$\Delta \overrightarrow{ v }=20 \sqrt{3} \hat{ j }$
