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)

Paragraph: A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle

Question: If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point

B

, immediately after it strikes the second incline is

Options

A $\sqrt{30} \mathrm{~m} / \mathrm{s}$
B $\sqrt{15} \mathrm{~m} / \mathrm{s}$
Czero
D $-\sqrt{15} \mathrm{~m} / \mathrm{s}$

Correct answer

C. zero

Step-by-step solution

In elastic collision, component of

v_1

parallel to

B C

will remain unchanged, while component perpendicular to

B C

will remain unchanged in magnitude but its direction will be reversed. Paragraph: A small block of mass M moves on a frictionless surface of an inclined plane, as shown in figure. The angle

\begin{aligned}& v_{\|}=v_1 \cos 30^{{}^{\circ}}=(\sqrt{60})\left(\frac{\sqrt{3}}{2}\right)=\sqrt{45} \mathrm{~ms}^{-1}\\ & v_{\perp}=v_1 \sin 30^{{}^{\circ}}=(\sqrt{60})\left(\frac{1}{2}\right)=\sqrt{15} \mathrm{~ms}^{-1} \end{aligned}

Now vertical component of velocity of block :

\begin{aligned}v & =v_{\perp} \cos 30^{{}^{\circ}}-v_{\|} \cos 60^{{}^{\circ}} \\ & =(\sqrt{15})\left(\frac{\sqrt{3}}{2}\right)-(\sqrt{45})\left(\frac{1}{2}\right)=0 \end{aligned}

\therefore$ correct option is (c).

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