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3-2.Motion in Plane
hard
A particle is projected from a point $A$ with velocity $u\sqrt 2$ at an angle of $45^o$ with horizontal as shown in fig. It strikes the plane $BC$ at right angles. The velocity of the particle at the time of collision is
A
$\frac{{\sqrt 3 u}}{2}$
B
$\frac{u}{2}$
C
$\frac{{2u}}{{\sqrt 3 }}$
D
$u$
Solution
Let $v$ be the velocity at the time of collision.
Then. $u \sqrt{2} \cos 45^{\circ}=v \sin 60^{\circ}$
$\Rightarrow \quad(u \sqrt{2})\left[\frac{1}{\sqrt{2}}\right]=\frac{\sqrt{3} v}{2}$
$\therefore \quad v=\frac{2}{\sqrt{3}} u$
Standard 11
Physics
