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Two smooth slides lie within the same vertical plane and make angles $α$ to the horizontal (see the figure). At some moment, two small balls are released from points. A and B and they start sliding down. It took time $t_{1}$ for the first ball that started from point A to reach the ground; for the second one the time of descent was $t_{2} .$ At what time t was distance between the balls the smallest?
If $t_{1} = 5 \sec$ and $t_{2} = 3 \sec .$ Find t =?

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answer is 2.83.

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Detailed Solution

Distance between balls will be minimum when they are along same vertical line

$A B = \frac{1}{2} g \sin α (t_{1}^{2} - t_{2}^{2})$

$\frac{A B}{2} = \frac{1}{2} g \sin α t^{2}$

$t = \sqrt{\frac{t_{1}^{2} - t_{2}^{2}}{2}} = 2.82$

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