Animated Solution for Physics - Work, Energy, and Power: Comprehension Passage

Imagine you're on a roller coaster. You plunge down a steep drop, building up incredible speed, only to suddenly hit a section where the track levels out slightly. That jarring transition is exactly what our block experiences at point $B$. Let's break down the physics of this thrilling ride.

The block starts from rest at point $A$ and slides down a $60^{\circ }$ incline. To find its speed at the bottom of this first drop (point $B$), we need to know how far it has fallen vertically. Gravity only cares about the vertical drop when converting potential energy into kinetic energy.

Using basic trigonometry, the vertical height $h_1$ is related to the horizontal distance $x_1$ by the tangent of the angle:

Substituting the given horizontal distance of $\sqrt{3}\text{ m}$:

Now, we apply the principle of conservation of mechanical energy. The potential energy lost equals the kinetic energy gained:

Plugging in $g=10{\text{ m/s}}^2$ and $h_1=3\text{ m}$:

Here is where the problem gets interesting. At point $B$, the incline suddenly changes from $60^{\circ }$ to $30^{\circ }$. The block doesn't just smoothly transition; it strikes the new surface. The problem states this collision is totally inelastic.

What does a totally inelastic collision with a fixed surface mean? It means the block doesn't bounce. The component of its velocity that is perpendicular to the new surface is completely absorbed and dissipated as heat and sound. However, the component of its velocity that is parallel to the new surface remains completely unaffected.

To find this surviving parallel component, we need the angle between the incoming velocity vector and the new incline. The incoming velocity is at $60^{\circ }$ below the horizontal, and the new incline is at $30^{\circ }$ below the horizontal. The difference between them is:

The new velocity $v_2$ is simply the parallel component of the incoming velocity $v_1$:

Substituting our known values:

To simplify this, we can bring the $2$ inside the square root as a $4$:

And there we have it! The speed of the block immediately after the jarring transition at point $B$ is $\sqrt{45}\text{ m/s}$.