1: I think there are some doubts about this question. I wonder what the questioner means. It is mainly the state when A runs to the other end, whether there is speed V energy or stillness. I think it should be static, so x is x=0 without calculation. If it is not static, I think there is a condition missing: the length L of the flat car.
2. B. When F pulls B, B will drive A through the spring, and the spring will be elongated in the process; If F and AB are removed, the system loses external force, so there is no acceleration. However, due to the effect of the spring, there will be * * * vibration between AB, because the * * vibration without the friction spring will continue all the time, and A and D are obviously wrong. C Note that AB is not relatively static when the spring returns to its original length, but relatively static when it is the longest or shortest, so only B is correct.
3. When A is fixed, the rising height of B means that all kinetic energy is converted into gravitational potential energy. When B is in the horizontal plane, A and B are at rest.
According to mbgh = (1/2) mbv 2 and I=Ft=P2-P 1, (P 1=0).
So mbgh = l 2/2mb.
When A is not fixed and B rises to the horizontal plane, AB is relatively static and together, and the velocity in the horizontal direction is V 1.
The momentum theorem is applied to the horizontal system: mBV=(mA+mB)V 1.
Without friction, the system will have energy conservation.
(1/2) mbv2 = (1/2) (ma+MB) v12+mbgh, and mbgh = l 2/2mb, V 1=mBV/(mA+mB).
Solution: l = mbv = ((ma+MB) l 2/ma)-2