(20 14? As shown in the figure, Block A with mass mA=2kg is stationary on a smooth horizontal surface, and Block B with mass mB= 1kg is at an initial velocity v0=5m.

(1) The momentum conservation in the collision process between A and B takes the initial velocity direction of B as the positive direction,

According to the law of conservation of momentum, mBv0=mAvA 1+mBvB 1,

After A collides with the baffle, the speed remains unchanged, and the left direction (the speed direction of A) is the positive direction.

According to the law of conservation of momentum: mava1-mbvb1= mava2+mbvb2,

Substituting the data, we get: vA 1=2m/s, VB1=1m/s;

② In the second collision process, for B, take the left as the positive direction,

According to the momentum theorem, I=mBvB2-mBvB 1,

Substitute data, I=2.2N? s；

Answer: ① After the first collision between A and B, the speed of A is 2m/s;

② In the second collision, the impulse of A to B is 2.2N? south