As shown in the figure, the wood blocks a and B are side by side and fixed on the horizontal table. The length of a is l, and the length of B is 2L. A bullet enters a at the speed V1 along the horizontal direction and passes out B at the speed v2. The bullet can be regarded as a particle, and its motion can be regarded as a uniform linear motion. Then the velocity of the bullet passing through a is () A. 2(v12+v22) 3B. 2(v12+v22) 3C. 2v12+v223D. v12+2v223

As shown in the figure, the wood blocks a and B are side by side and fixed on the horizontal table. The length of a is l, and the length of B is 2L. A bullet enters a at the speed V1 along the horizontal direction and passes out B at the speed v2. The bullet can be regarded as a particle, and its motion can be regarded as a uniform linear motion. Then the velocity of the bullet passing through a is () A. 2(v12+v22) 3B. 2(v12+v22) 3C. 2v12+v223D. v12+2v223

Let the mass of the bullet be m, and apply the kinetic energy theorem to the whole process of the bullet passing through AB: 12mvb2-12mva2 = WF, because the friction force on the bullet remains unchanged, and because the length of a is l and the length of B is 2L, the work done by the friction force in the process of the bullet passing through a is 13wf, and apply the kinetic energy theorem to the process of the bullet passing through a: 12mva12-12mva2 = 13wf, and the solution is va1 = 2v12 + v223