Five coins on a table

Let the mass of each coin be $m \text{ kg}$ .

Since all collisions are elastic and no external force is applied to the system of coins, we can apply conservation of linear momentum one by one to each of the coins.

When the first coin strikes the second one, its momentum along the side of the hexagon is: $m \cdot 24 \cos 60^{\circ} = 12m \text{ kg m s}^{-1}$ .

When the second one imparts its momentum into the third one, its momentum along the side of the hexagon is: $m \cdot 12 \cos 60^{\circ} = 6m \text{ kg m s}^{-1}$ .

When third imparts its momentum to the fourth one, its momentum along the side of the hexagon is: $m \cdot 6 \cos 60^{\circ} = 3m \text{ kg m s}^{-1}$ .

When fourth imparts its momentum to the final one, its momentum in the horizontal direction is: $m \cdot 3 \cos 0^{\circ} = 3m \text{ kg m s}^{-1}$ .

Hence, velocity with which the final coin leaves the hexagon is: $\dfrac{3m}{m} = \boxed{3 \text{ m s}^{-1}}$ .

$We\ have,$

$v_{bottom\ left\ coin}=24\ ms^{-1}$

$We\ know\ that\ exterior\ angle\ of\ regular\ hexagon\ is\ \frac{\pi }{3},$

$v_{top\ left\ coin}=v\cos \left(\frac{\pi }{3}\right)\cdot \cos \left(\frac{\pi }{3}\right)\cdot \cos \left(\frac{\pi }{3}\right)$

$v=24\cdot \frac{1}{8}$

$v=3\ ms^{-1}$