The conservation of momentum in the process of direct collision between A and B, with the initial velocity direction of A as the positive direction, is obtained by the law of conservation of momentum:: mav0=mdvd 1…②,
It is found that the initial velocity of slider A is v0=2m/s…
According to the law of conservation of energy, the mechanical energy lost in the frontal collision between A and B can be obtained: △ E =12mav02-12mdvd12 … ④.
Substituting the data, the result is: △ e =1j;
(2) As shown in Figure B, the speed of D is: VD2 =-0.5m/s...⑤ At the moment when the spring first recovers its deformation.
In the system composed of D, C and spring, the conservation of momentum and mechanical energy with the initial velocity direction of D and C systems as the positive direction is obtained by the law of conservation of momentum:
mdvd 1=mdvd2+mcvc2…⑥
According to the law of conservation of energy:12mdv2d1=12mdv2d2+mcv2c2 … ⑦.
The mass of the slider C obtained by substituting the data is: MC = 6 kg … 8;
(3) Let the speed of d be vd3 at the moment before slamming the slider c, and then:
mdvd 1=mdvd3+mcvx，vd3= 1-3vx…
After that, when the sliders C and d*** accelerate instantaneously, the elastic potential energy of the spring is the largest, and the initial velocity direction of D is positive, which is obtained by the law of conservation of momentum:
mdvd3+mcvx=(md+mc)v '，
d:V′= 1？ 6 vx4……
According to the law of conservation of energy, the maximum elastic potential energy: EP = 12 mdv2d 1? 12(MD+MC)v′2，
Solution: EP= 1? ( 1? 6vx)24，
When VX = 16m/s, EP can get the maximum value, and the maximum value is EPM =1j;
Answer: (1) The initial velocity of slider A is 2m/s, and the mechanical energy △E lost by the frontal collision between A and B is1j; ;
(2) The mass of the slider c is 6 kg;
(3) After that, the expression of the maximum elastic potential energy Ep of the spring is: EP= 1? ( 1? 6vx)24. When VX = 16m/s, EP can get the maximum value, and the maximum value is EPM =1j.