Solution: (1) At point G, the athlete and the skateboard do circular motion together, let centripetal acceleration be a n, and the speed be, and the athlete is affected by the gravity Mg and the skateboard support force n, then:

Solution: = 6.5m/s

(2) Let the skateboard slide down from point A to BC orbit at the speed of v 1. The law of conservation of mechanical energy is as follows:

Solution:

After the athletes and skateboards slide down the BC track from point A, the speed is also v 1. Athletes jump from skateboard B to skateboard A, assuming that the horizontal speed when kicking skateboard B is v 2 and the horizontal displacement when flying in the air is S, then: S = V2T2.

Let the horizontal distance between skateboard A and skateboard B during take-off be s 0, then: S 0 = V 1T 1.

Let the displacement of the skateboard in t 2 time be s 1, then: s 1 = v 1t2.

S = S 0+S 1, that is, V2T2 = V 1 (T 1+T 2).

After the athlete falls on the skateboard A, the speed of moving with the skateboard A is V, and the law of conservation of momentum is MV1+MV2 = (m+m) V.

It can be solved from the above equation:

Substituting the data, the result is: v = 6.9 m/s.

(3) Suppose that after the athlete leaves skateboard B, the speed of skateboard B is v 3, and there are

Mv 2 +mv 3 =(M+m)v 1

V3 =-3 m/s can be calculated as follows: │ V3 │ = 3 m/s < V 1 = 6 m/s, and the B plate will move back and forth between the two platforms with the same mechanical energy.

The mechanical energy of the system becomes:

△E=88.75 J