Question

A small block of mass $M$ moves on a frictionless surface of an inclined plane, as shown in the figure. The angle of the incline suddenly changes from ${60}^{\circ }$ and ${30}^{\circ }$ at point $B$.
The block is initially at rest at $A$. Assume that collisions between the block and the incline are totally inelastic.The speed of the block at point $C$, immediately before it leaves the second incline is:

• A. $\sqrt{120}m/s$
• B. $\sqrt{105}m/s$
• C. $\sqrt{90}m/s$
• D. $\sqrt{75}m/s$

Answer 1

Answer: (B)

$\sqrt{105}m/s$

As the block comes down from B to C, the change in height can be obtained as:

$H=3\sqrt{3}\times tan{60}^{\circ }$

$H=3m$

Since the energy of the system will remain conserved, the kinetic energy and potential energy at $B$ will be equal to the kinetic energy at $C$

Applying mechanical energy conservation between points $B$ and $C$ gives:

$\frac{1}{2}m{v}_C^2=\frac{1}{2}m{v}_B^2+mgh$

It can be simplified as:

$v_C^2=v_B^2+2gh$

or $v_C^2=45+2\times 10\times 3$

$\Rightarrow v_C=\sqrt{105}m/s$.