Horizontal direction: Rcos37 =v0t.

Vertical direction: RSIN 37 = 12GT2.

Solution: v0 = rcos37 2RSin37g = 433m/s.

(2) In order for the small piece to hit the baffle, the small piece must be able to move to the O point.

According to the kinetic energy theorem, FX-μ gmgs = △ ek = 0.

Solution: x = 2.5m

From Newton's second law: f-μ g = ma

Solution: a=5m/s2

From the kinematic formula: x = 12at2

Solution: t =1s.

(3) Let the coordinates of any point where a small object impacts the baffle be (x, y), then x=v0t.

y= 12gt2

From conservation of mechanical energy: Ek? = 12mv20+mgy

x2+y2=R2。

Simplified: ek = mgr24y+3mgy4.

Obtained mathematically: ekmin = 523 j

Answer: (1) The speed when leaving point O is 433 m/s;

(2) In order to make a small object hit the baffle, the shortest time for pulling force F to act is1s; ;

(3) Change the action time of the pulling force F to make small fragments hit different positions of the baffle. The minimum kinetic energy of small debris hitting the baffle is 523 J.