① After the wooden block m2 rebounds, it drives m1 to move when the spring returns to its original length for the first time. Assume that the speed of the wooden block m2 at this time is v0. According to the law of conservation of mechanical energy:
< p>v0=3m/sWhen the spring stretches the longest, the speed of wooden blocks m1 and m2 is the same, set to v, and the left direction is specified as the positive direction. According to the law of conservation of momentum, we get: m2v0= ( m1+m2) v
Solution: v=1m/s
According to the law of conservation of energy, the maximum elastic potential energy of the spring Ep=12m2v02-12 (m1+m2)v2= 3J
②When the spring returns to its original length again, m1 obtains the maximum speed v1, and the speed of m2 at this time is v2
It is stipulated that the right direction is the positive direction, according to the law of conservation of momentum Obtain: m2v0=m1v1+m2v2
According to the law of conservation of mechanical energy: 12m2v02=12m1v12+12m2v22
Solution: v1=2m/s
Answer: ① When the spring is stretched to its longest length, the elastic potential energy of the spring is 3J.
② In the subsequent movement process, the maximum speed of the wooden block m1 is 2m/s.