According to the law of conservation of mechanical energy: (m1+m) GH =12 (m1+m) v12,
It is known that h = 2v02g, and the solution is: v1= 2v0; = 2v0
Let the horizontal speed (relative to the ground) of people jumping out of armored vehicles be V.
The momentum of each system is conserved when people jump out of car A and then into car B,
Suppose a person jumps off the A car and jumps on the B car, and the speeds of the two cars are v 1' and v2' respectively.
According to the law of conservation of momentum: (m1+m) v1= mv+m1v1',
When people jump on the B car: Mv-m2v 0 =(M+m2)v2',
Solution: v 1'= 6v 0-2v? ①，v2′= 12v- 12v 0②，
The critical condition that two cars cannot collide again is: v 1'= V2',
When v 1'= v2', it is obtained from ① ②: v= 135v0,
When v 1'=-v2', it is obtained from ① ② that v= 1 13v0,
Therefore, the range of v is:135v0 ≤ v ≤113v0;