Question 165: Laws of Motion

Question

Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is $$t(\sqrt{2}+1)$$ s. The value of t is ___________.
(use $$\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$$ )

Accepted Answer

$$AB = 10\sqrt 2 $$ m
$${v_A} = \sqrt {2 	imes 10 	imes 10}  = 10\sqrt 2 $$ m/s
$${v_C} = 10\sqrt 2 $$ m/s
$${a_{BC}} = g\sin (30^\circ ) = 5$$ m/s
2
$${t_{BC}} = 2\sqrt 2 \,s\,\left( {{{{v_c}} \over {{a_{BC}}}}} ight)$$
$${t_{AB}} = {{{v_A}} \over {5\sqrt 2 }} = 2\,s$$
$${t_{AB}} + {t_{BC}} = 2(\sqrt 2  + 1)$$
$$ \Rightarrow t = 2$$

Laws of Motion

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