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…

Question

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: 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

Aicharya · Accepted Answer

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.

$ \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). !

Paragraph: A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in…

Solution