Kvant Physics Problem 26
A small block moves without friction on a smooth surface consisting of two horizontal half-planes separated in height by a vertical distance $h$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 13m24s
Source on kvant.digital
Problem
Two horizontal half-planes, located at a height $h$ one above the other, merge smoothly into each other as shown in Figure 2. A small block moves along the upper half-plane with speed $v$ at an angle $\alpha$ to the direction of descent. How will it move on the lower half-plane? Assume that the block does not bounce, i.e., it moves without losing contact with the surface of the descent. Neglect friction.
Fig. 2
Setup and Assumptions
A small block moves without friction on a smooth surface consisting of two horizontal half-planes separated in height by a vertical distance $h$. The planes are connected by a smooth descending transition. Before entering the transition, the block moves on the upper half-plane with speed $v$ and its velocity makes an angle $\alpha$ with the direction of descent shown in the figure.
Let the $x$ axis be directed along the line of greatest descent of the transition, and let the $y$ axis lie in the horizontal plane, perpendicular to $x$. The initial velocity components are
$$v_x=v\cos\alpha, \qquad v_y=v\sin\alpha.$$
The unknown is the direction and speed of motion after the block reaches the lower horizontal half-plane.
The surface is assumed perfectly smooth, friction is neglected, the block remains in contact with the surface at all times, air resistance is neglected, and the Earth is taken as an inertial reference frame. The transition between the planes is smooth, so the normal reaction is finite and no impact occurs.
Physical Principles
The solution rests on conservation of mechanical energy and on conservation of the momentum component associated with a cyclic coordinate.
Since friction is absent, the normal reaction does no work. Hence the mechanical energy of the block is conserved:
$$\frac{mv^2}{2}+mgh = \frac{mV^2}{2},$$
where $V$ is the speed on the lower half-plane.
The shape of the surface is translationally invariant in the $y$ direction. Consequently no external force acts in the $y$ direction. The component of momentum parallel to the $y$ axis is conserved:
$$mv_y=\text{const}.$$
Therefore
$$V_y=v_y=v\sin\alpha.$$
Derivation
The conservation of mechanical energy gives the speed on the lower plane:
$$\frac{mv^2}{2}+mgh = \frac{mV^2}{2}.$$
After canceling the mass and multiplying by $2$,
$$V^2=v^2+2gh.$$
Hence
$$V=\sqrt{v^2+2gh}.$$
Next, consider the component of velocity perpendicular to the direction of descent. Since the surface does not depend on the coordinate $y$, the force component along $y$ vanishes. Thus
$$V_y=v\sin\alpha.$$
The total speed on the lower plane is already known. The component along the direction of descent on the lower plane is then
$$V_x=\sqrt{V^2-V_y^2}.$$
Substituting the expressions for $V$ and $V_y$,
$$V_x = \sqrt{(v^2+2gh)-v^2\sin^2\alpha}.$$
Using $1-\sin^2\alpha=\cos^2\alpha$,
$$V_x = \sqrt{v^2\cos^2\alpha+2gh}.$$
Let $\beta$ be the angle between the final velocity and the direction of descent. Then
$$\tan\beta = \frac{V_y}{V_x}.$$
Substituting the velocity components,
$$\tan\beta = \frac{v\sin\alpha} {\sqrt{v^2\cos^2\alpha+2gh}}.$$
This expression determines the direction of motion on the lower half-plane.
Result
The speed on the lower half-plane is
$$V=\sqrt{v^2+2gh}.$$
Its components are
$$V_y=v\sin\alpha,$$
and
$$V_x=\sqrt{v^2\cos^2\alpha+2gh}.$$
The angle $\beta$ that the trajectory on the lower half-plane makes with the direction of descent satisfies
$$\boxed{ \tan\beta= \frac{v\sin\alpha} {\sqrt{v^2\cos^2\alpha+2gh}} }.$$
Equivalently, the block emerges with speed
$$\boxed{ V=\sqrt{v^2+2gh} }$$
and continues in a straight line on the lower horizontal plane at the angle $\beta$ given above.
No numerical values are provided in the statement, so a numerical substitution cannot be performed.
Sanity Checks
The quantity $2gh$ has dimensions of velocity squared, so the expression
$$V^2=v^2+2gh$$
is dimensionally consistent.
If $h=0$, the two planes coincide. Then
$$\tan\beta = \frac{v\sin\alpha}{v\cos\alpha} = \tan\alpha,$$
which gives $\beta=\alpha$, while $V=v$. The motion is unchanged, as expected.
If $\alpha=0$, the initial motion is directed along the descent. Then
$$V_y=0,$$
and
$$\beta=0.$$
The block remains aligned with the descent direction and merely gains speed from the loss of potential energy.
If $\alpha=\frac{\pi}{2}$, the initial velocity is perpendicular to the descent direction. Then
$$V_y=v, \qquad V_x=\sqrt{2gh},$$
so the descent generates a new velocity component only in the $x$ direction.
The most common algebraic error occurs when determining the final direction. Using
$$V_x^2=V^2-V_y^2$$
with an incorrect sign would lead to an unphysical result and an incorrect value of $\beta$. The conservation of the transverse component $V_y=v\sin\alpha$ is the crucial step.