The missing figure is attached.
Since the friction is negligible, we can apply the law of conservation of energy. The total mechanical energy at the top and at the bottom must be the same:
[tex]E_t = E_b[/tex]
At the top, the mechanical energy is sum of kinetic energy and gravitational potential energy:
[tex]E_t = K_t + U_t = \frac{1}{2} mv_t^2 + mgh[/tex] (1)
where m is the mass of the cars, [tex]v_t[/tex] is the velocity of the cars at the top (30 km/h) and h is the height at the top (18 m).
At the bottom, the mechanical energy is just kinetic energy:
[tex]E_b = K_b = \frac{1}{2}mv_b^2 [/tex] (2)
where [tex]v_b[/tex] is the velocity of the cars at the bottom of the track. By putting together (1) and (2), we find
[tex] \frac{1}{2}mv_t^2 + mgh = \frac{1}{2}mv_b^2 [/tex]
from which we can isolate [tex]v_b[/tex], the velocity of the cars at the bottom of the track:
[tex]v_b = \sqrt{v_t^2 + 2gh} [/tex]
and since [tex]v_t = 30 km/h =8.33 m/s[/tex], we find
[tex]v_b = \sqrt{(8.33 m/s)^2 + 2(9.81 m/s^2)(18 m)}=20.56 m/s =74 km/h [/tex]
