According to the balance, f = mgsin30 = 12mg.
When B is instantly placed on A, the stress analysis of AB as a whole is as follows:
The external force exerted on the whole body is f = 2mgsin30-f = (2m) a.
Can you get the whole acceleration of a = 2 mg? 12? 12mg2m = G4, so A is correct;
BC. When B is instantaneously placed on A, B has an acceleration downward along the inclined plane, and the acceleration of B can be decomposed in the horizontal and vertical directions:
The acceleration of B has a horizontal component, and the gravity and supporting force are in the vertical direction, so we know that this acceleration component is provided by the friction between A and B, so B is wrong.
The acceleration of B has a vertical component, which is vertically downward. So the resultant force of A's support to B and B's gravity is vertical downward, so A's support to B is smaller than B's gravity, so C is wrong.
When D and AB slide together, the spring force increases and the acceleration of * * * decreases, so when the acceleration decreases to 0, the speed of AB is the maximum, which is known from an analysis.
F = 2mgsin 30-f' = 0
Available spring force F'=mg
So * * * and the sliding distance are the same △ x = f'? Fk=mg？ 12 mgk = mg2k, and AB has the highest speed, so D is correct.
Therefore: ad.