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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 $B$ immediately after it strikes the second incline is
Options:
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 $B$ immediately after it strikes the second incline is
Solution:
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Verified Answer
The correct answer is:
$\sqrt{45} \mathrm{~m} / \mathrm{s}$
$\sqrt{45} \mathrm{~m} / \mathrm{s}$
Between $A$ and $B$, height fallen by block $h_1=\sqrt{3} \tan 60^{\circ}=3 \mathrm{~m}$.
$\therefore$ speed of block just before striking the second incline,
$$
v_1=\sqrt{2 g h_1}=\sqrt{2 \times 10 \times 3}=\sqrt{60} \mathrm{~ms}^{-1}
$$
In perfectly inelastic collision, component of $v_1$ perpendicular to $B C$ will become zero, while component of $v_1$ parallel to $B C$ will remain unchanged.
$\therefore$ speed of block $B$ immediately after it strikes the incline is,
$$
\begin{aligned}
v_2 & =\text { component of } v_1 \text { along } B C \\
& =v_1 \cos 30^{\circ} \\
& =(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1}
\end{aligned}
$$
$\therefore$ correct option is (b)
$\therefore$ speed of block just before striking the second incline,
$$
v_1=\sqrt{2 g h_1}=\sqrt{2 \times 10 \times 3}=\sqrt{60} \mathrm{~ms}^{-1}
$$
In perfectly inelastic collision, component of $v_1$ perpendicular to $B C$ will become zero, while component of $v_1$ parallel to $B C$ will remain unchanged.
$\therefore$ speed of block $B$ immediately after it strikes the incline is,
$$
\begin{aligned}
v_2 & =\text { component of } v_1 \text { along } B C \\
& =v_1 \cos 30^{\circ} \\
& =(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1}
\end{aligned}
$$
$\therefore$ correct option is (b)
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