# How does a pop pop boat work?

I have been reading about the pop pop boat on wikipedea. What I don't understand is after the hot water is ejected the vacuume gets created, so when the same amount of water will be sucked in the boat should move backwards. Wikipedea says,

...while the water pushed out carries away with it momentum, which must be balanced (by Newton's third law) by an opposite momentum on the part of the boat, the water sucked in quickly impinges on the boiler tank and transfers its momentum to the boat. The initial reaction force on the boat (which would pull it backwards) is therefore cancelled by the pushing of the water when it hits the inside of the boiler. The result is that the inflow of water causes no appreciable force on the boat.

Now when the vacuum generates inside the boat feels a net backwards force force by the pressure of water present towards its head. I think that since the area of boat is larger than the nozzle so the force exerted by water on its surface must be larger as compared to the force imparted by the water being sucked. So the initial reaction force should not be completed cancelled out.

• The inflow and outflow processes are not symmetrical, but it is not easy to understand. Take a look at the reverse sprinkler problem. – rodrigo Dec 8 '15 at 11:44
• When the water is ejected it has lots of speed, so takes lots of backwards momentum with it and gives the boat forward momentum. The water that is sucked in has a much smaller speed relative to the boat before it gets sucked in (if any), so sucking it in transfers little backwards momentum to the boat. – pwf May 23 '17 at 21:58

1) WATER EJECTION: the water boils in the exaust, the pressure rises in the boiler and the water is ejected from exaust, causing a momentum transfer to the boat, which moves forward $$\text{back} \qquad \qquad \qquad \text{front}$$ $$\longrightarrow$$ $$\text{force from water expulsion}$$ Please note that the water is expulsed backwards, but the momentum is directed forward
2) WATER ASPIRATION: due to depression in the boiler after hot water expulsion, fresh water is sucked into the boiler. Since the water is sucked from the exaust (which is located on boat's poop), the water moves from the back to the front of the boat. The opposite contribution is given from aspiration from the back of the boat. So no net momentum is tranferred in this phase $$\text{back} \qquad \qquad \qquad \text{front}$$ $$\longleftarrow \qquad \qquad\longrightarrow$$ $$\text{force from pressure of inlet water} \qquad \qquad \text{force from mass transfer to boiler}$$ So the total force applied to the boat during phase 2 is zero.