In the horizontal surface of a large table, there is a small round hole into which a vertical cylindrical tube is inserted. Inside the tube, a piston moves upward with a constant velocity V 0 = 6 m

Question

In the horizontal surface of a large table, there is a small round hole into which a vertical cylindrical tube is inserted. Inside the tube, a piston moves upward with a constant velocity $V_{0} = 6 m / s$ , whose upper surface is tilted at an angle $α = 1 5^{\circ}$ to the horizontal. From a certain height above, a small ball falls (see the figure). At the moment immediately before an elastic collision of the ball with the piston, the velocity of the ball is directed downward and is equal to $V_{0}$ . The point at which the collision occurs is exactly at the level of the table surface. After the collision with the piston, the ball falls onto the surface of the table at a distance $ℓ$ from the hole in the table. Air resistance can be neglected. Let the angle of inclination of the top surface of the piston to the horizontal be adjustable. At what value of the angle $α$ will the maximum possible flight distance $ℓ_{max}$ of the ball be achieved?

TestHub · Accepted Answer

In the piston frame the ball approaches at $2 V_{0}$ downward and reflects elastically off the surface tilted at $α$ , leaving at angle $2 α$ from the vertical. In the lab frame:

$v_{h} = 2 V_{0} sin 2 α$

$v_{v} = 2 V_{0} cos 2 α + V_{0}$

The flight ends at table level, so

$ℓ = v_{h} \cdot \frac{2 v _{v}}{g}$

$ℓ \propto sin 2 α (2 cos 2 α + 1)$

Maximising $f (x) = sin 2 x + sin x$ with $x = 2 α$ :

$f^{'} (x) = 2 cos 2 x + cos x = 0$

$4 cos^{2} x + cos x - 2 = 0$

$cos x = \frac{- 1 + 33}{8}$

so $α = \frac{1}{2} cos^{- 1} (\frac{- 1 + 33}{8})$ .

Correct option: (A)

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