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.