Paragraph: A small block of mass M moves on a frictionless surface...

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

B

immediately after it strikes the second incline is

Options

B. $\sqrt{45} \mathrm{~m} / \mathrm{s}$

Between A and B

, height fallen by block

h_1=\sqrt{3} \tan 60^{\circ}=3

\mathrm{ m} .

\thereforespeed 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

Bimmediately after it strikes the

incline is,

\begin{aligned} v_2 & =\text{ component\text{ 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)$