An almost identical problem is done on page 101. Let v represent the little fish's velocity before lunch. Set the momentum before lunch equal to the momentum after. $$(5 kg)(1 m/s) + (1 kg) v = 0$$ $$5 m/s + v = 0$$ $$v = -5 m/s$$ So the little fish was approaching the big fish at at speed of 5 m/s. Its initial momentum was equal and opposite to that of the big fish, because the system's momentum was zero.