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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.

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  • $\begingroup$ The inflow and outflow processes are not symmetrical, but it is not easy to understand. Take a look at the reverse sprinkler problem. $\endgroup$
    – rodrigo
    Dec 8, 2015 at 11:44
  • $\begingroup$ 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. $\endgroup$
    – pwf
    May 23, 2017 at 21:58

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The boat propulsion has 2 phases:

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.

Now about your question: in my opinion you're probably right (I've never seen a pop pop boat moving, but I guess the boat should not move continuously) and the initial reaction force should not be completely cancelled. This happends also because phases are not stationary, in particular water push and pull is related to the quantity of water in the tank, which affects weight (so affect drags) and flux pressure

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  • $\begingroup$ "I've never seen a pop pop boat moving" !! It works very well indeed. It is sold as a toy. I used it in my childhood and perhaps many children would have used the same. Just search on youtube, youtube.com/watch?v=g3OLhFx8KZY Could you elaborate more about your last section. I didn't get what phases are not stationary mean. I guess when we push water out(phase 1), it happens so fast that the drag is lesser and when we pull in the water, it happens slowly making drag more so overall the boat moves forward. $\endgroup$
    – user31782
    Dec 9, 2015 at 15:31
  • $\begingroup$ Also can we somehow make a mathematical model of phase 2 to know how much moment does water transfer in phase 2 when it strikes the tank and how much did the boat gained in backward direction before the striking of water? $\endgroup$
    – user31782
    Dec 9, 2015 at 15:33

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