Kvant Physics Problem 26

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Verdicts: SKIP + SKIP
Solve time: 1m02s
Source on kvant.digital

Problem

Two horizontal half-planes located at a height $h$ one above the other smoothly transition into each other, as shown in Figure 2. On the upper half-plane, at an angle $\alpha$ to the direction of descent, a small block moves with speed $v$. How will it move on the lower half-plane? Assume that the block does not bounce, i.e. it moves while remaining in contact with the surface of the slope. Friction is neglected.

Figure 2

Setup and Assumptions

A small block of mass $m$ moves without friction on a rigid surface consisting of two horizontal half-planes connected by a smooth spatial transition. The upper half-plane lies at height $h$ above the lower half-plane. The motion occurs under uniform gravity with acceleration $g$, directed vertically downward.

At a point on the upper plane the block has speed $v$ and its velocity makes an angle $\alpha$ with the direction of descent, defined as the horizontal axis along which the surface transitions. The component of motion perpendicular to this direction lies entirely within the horizontal plane.

The surface is smooth, so the reaction force is always normal to the surface and does no work. The block remains in continuous contact with the surface throughout the motion. Air resistance is neglected. The motion is analyzed in an inertial reference frame fixed to the Earth.

The unknowns are the speed and direction of motion of the block on the lower horizontal half-plane.

Physical Principles

Mechanical energy is conserved because the only force performing work is gravity, which is conservative, and the normal reaction force is always perpendicular to the instantaneous displacement.

The work energy theorem applies in the form

$\frac{mv^2}{2} + mgy = \text{const}.$

In directions where no external force component acts, the corresponding velocity component remains constant. In particular, if the surface has translational symmetry along a horizontal direction, the corresponding component of velocity is conserved.

The final motion on the lower plane is uniform rectilinear motion because the surface is horizontal and frictionless, so no tangential force acts there.

Derivation

Choose a horizontal coordinate system with the $x$ axis directed along the descent direction of the surface transition and the $y$ axis perpendicular to it in the horizontal plane. The initial velocity components are

$v_x = v\cos\alpha, \qquad v_y = v\sin\alpha.$

The surface varies only in the $x$ and vertical $z$ directions, so it is invariant under translations along $y$. The reaction force from the surface lies in the plane normal to the surface and therefore has no net component that can change the velocity in the $y$ direction. The equation of motion in the $y$ direction reduces to

$\frac{dv_y}{dt} = 0,$

so

$v_y = v\sin\alpha$

throughout the motion.

Gravity performs work only through the change in vertical height from $h$ to $0$. The gain in kinetic energy equals the loss in gravitational potential energy,

$\frac{m v_f^2}{2} = \frac{m v^2}{2} + mgh,$

which gives the final speed magnitude

$v_f^2 = v^2 + 2gh.$

At the same time, the final velocity can be decomposed into components on the lower plane as

$v_f^2 = v_{x,f}^2 + v_{y,f}^2.$

Since $v_{y,f} = v\sin\alpha$, the horizontal component along the descent direction follows from energy balance,

$v_{x,f}^2 = v_f^2 - v_{y,f}^2 = (v^2 + 2gh) - v^2\sin^2\alpha,$

which simplifies to

$v_{x,f}^2 = v^2\cos^2\alpha + 2gh.$

Taking the positive root consistent with motion in the direction of descent,

$v_{x,f} = \sqrt{v^2\cos^2\alpha + 2gh}.$

The direction of motion on the lower plane is determined by the angle $\beta$ that the velocity makes with the descent direction, defined by

$\tan\beta = \frac{v_{y,f}}{v_{x,f}} = \frac{v\sin\alpha}{\sqrt{v^2\cos^2\alpha + 2gh}}.$

Result

The speed on the lower half-plane is

$v_f = \sqrt{v^2 + 2gh}.$

The direction is given by

$\tan\beta = \frac{v\sin\alpha}{\sqrt{v^2\cos^2\alpha + 2gh}}.$

If needed in component form,

$v_{x,f} = \sqrt{v^2\cos^2\alpha + 2gh}, \qquad v_{y,f} = v\sin\alpha.$

Substituting the physical quantities gives the final expressions with units:

$v_f = \sqrt{v^2 + 2gh}\ \text{m/s}.$

Sanity Checks

Dimensional consistency follows because both terms inside the square root for $v_f^2$ have units of $\text{m}^2/\text{s}^2$, since $gh$ has units $\text{m}^2/\text{s}^2$. The expression for $v_{x,f}^2$ also combines $v^2\cos^2\alpha$ and $2gh$, which are consistent in units.

In the limit $h \to 0$, the speed reduces to $v_f \to v$, and the direction satisfies $\tan\beta \to \tan\alpha$, preserving the initial motion on a single horizontal plane. In the limit $\alpha \to 0$, motion is entirely along the descent direction, giving $v_f^2 = v^2 + 2gh$ with no transverse component, consistent with one-dimensional acceleration under gravity-induced height drop.

The most sensitive step is the separation of velocity components using symmetry along the transverse horizontal direction; introducing any fictitious force component in that direction would incorrectly alter $v_y$ and spoil the angular relation.