Q. A body is projected at an angle 60° with the horizontal with kinetic energy K. When the velocity makes an angle 30° with the horizontal, the kinetic energy of the body will be
(a) K/2
(b) K/3
(c) 2K/3
(d) 3K/4
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Ans: (b)
Sol: According to question ,
$\frac{1}{2}m u^2 = K $
As horizontal component remains unchanged ,
$u cos60^o = v cos30^o $
$ \frac{u}{2} = v \frac{\sqrt{3}}{2} $
$v = \frac{u}{\sqrt{3}}$
$K.E = \frac{1}{2}m v^2 $
$K.E = \frac{1}{2}m (\frac{u}{\sqrt{3}})^2 $
$K.E = \frac{1}{3}(\frac{1}{2}m u^2)$
$K.E = \frac{1}{3}K $