Because of the large reversal of F1, F2, etc., the combined force of the system is zero, the momentum is conserved, and the center of mass of the system is still, so A and B both do simple harmonic motion with respect to the center of mass. Taking the center of mass O as the reference system (mass center), the spring connecting A and B can be regarded as two segments. The original length of the left side is lBO=mm+m0l0, the stiffness coefficient is m+m0mk; the original length of the right side is lAO=m0m +m0l0, the stiffness coefficient is m+m0m0k; it is known from T=2mk that the vibration periods of A and B are both T=2mm0k(m+m0). Because both A and B do simple harmonic motion with respect to the center of mass, they are each at the highest speed in the equilibrium position. Let the constant force F1=F2=F, when the A and B reach the equilibrium position, the elongation of the spring is x, there is F=kx, and the direction to the right is defined as the positive direction. According to the conservation of momentum and the conservation of energy, mvAm+m0vBm can be obtained: mvAm+m0vBm =0, Fx=12mv2Am+12m0v2Bm+12kx2. Solving vAm=m0F2km(m0+m), vBm=mF2km0(m0+m). From the above analysis, we can smoothly obtain the velocity time images of A and B (as shown), so as to obtain the motion laws of A and B. System centroid uniform motion example 2 As shown, two objects A and B of mass m and m0 respectively are connected by a light spring with a kinematic coefficient k on a smooth horizontal surface, and now A is horizontally to the right initial velocity v0. Let it start to exercise, try to analyze the movement laws of A and B. Since the combined force of the system is zero, the momentum of the system is conserved, the mechanical energy is conserved, and the center of mass of the system moves to the right at a speed of mm+m0v0. In the same way as in Example 1, A and B both do T=2mm0k(m+m0) relative to the center of mass. Harmonic movement. In the process of motion, when the spring returns to the original length, the speeds of A and B are vA and vB respectively, according to the conservation of momentum and the conservation of energy: mv0=mvA+m0vB, 12mv20=12mv2A+12m0v2B. Solving vA=v0, vB=0, or vA=m-m0m+m0v0, vB=2mm+m0v0. Discussion (1) When m=m0, the solution is vA=v0, vB=0, or vA=0, vB=v0. Thus, the speed time images of A and B can be obtained. (2) When m>m0, the solution is vA=v0, vB=0, or vA=m-m0m+m0v0>0, vB=2mm+m0v0>v0. Thus, the speed time images of A and B can be obtained. (3) when m It can be seen from the above analysis that although the physical process of the spring double oscillator model is complex and the motion scene is difficult to imagine, as long as the characteristics of the model are recognized, the general method of analyzing the model is mastered, and such problems can be appropriately modified, Can improve their ability to solve problems, and thus continuously improve the quality of scientific thinking. Bathroom Door,Shower Doors ,Glass Shower Doors ,Bathroom Glass Door Wooden & Timber Door Co., Ltd. , http://www.china-steeldoors.com