Paragraph: A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in…
Paragraph:
A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in figure. The angle of the incline suddenly changes from $60^{\circ}$ to $30^{\circ}$ at point $B$. The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic $\left(g=10 \mathrm{~m} / \mathrm{s}^2\right)$ Question:
The speed of the block at point $C$, immediately before it leaves the second incline is
$\sqrt{120} \mathrm{~m} / \mathrm{s}$
$\sqrt{105} \mathrm{~m} / \mathrm{s}$
$\sqrt{90} \mathrm{~m} / \mathrm{s}$
$\sqrt{75} \mathrm{~m} / \mathrm{s}$
Solution
Height fallen by the block from $B$ to $C$. $h_2=3 \sqrt{3} \tan 30^{\circ}=3 \mathrm{~m}$
Let $v_3$ be the speed of block, at point $C$, just before it leaves the second incline, then
$
\begin{aligned}
v_3 & =\sqrt{v_2^2+2 g h_2} \\
& =\sqrt{45+2 \times 10 \times 3} \\
& =\sqrt{105} \mathrm{~ms}^{-1}
\end{aligned}
$
$\therefore$ correct option is (b).
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