2m/s = 120m/min，5m/s = 300m/min。 When the first puppy catches up with Xiaoming, the distance between the two puppies is 6 minutes. Suppose the first puppy doesn't run, and it takes 10 minutes for the second puppy to catch up with Xiao Ming, that is, the second puppy will run 65438+. At this time, the first puppy ran back to a place for 10 minutes, adding up to 20 minutes, and subtracting 6 minutes from the middle is 14 minutes.

Solution: suppose that after Xiaoming left for x minutes, the first puppy began to chase Xiaoming:

It takes [120 x/(300-120)] × 2 = 4x/3 minutes for the first puppy to catch up with Xiao Ming.

Six minutes later, another puppy also set out from place A and caught up with Xiaoming.

Time to catch up with Xiao Ming and return to A: [120 (x+6)/(300-120)] × 2 = (4x/3+8) minutes.

So:

The time interval between two puppies returning to a place is 6+(4x/3+8)-4x/3= 14 minutes.