A boat heads into shallow water and gets stuck in sand. The rower stands up and uses an oar to push the boat off the sandbank. The resulting change of momentum involves which of the following: the boat, the water, the earth, the rower

The rower and boat both change velocity after the push, and as $P(momentum)=m(mass)*v(velocity)$, the momentum changes.

A force is applied to the earth. $F(force)=m(mass)*a(acceleration)$. So there is an acceleration. $V=u+at$, so if there is an acceleration, the velocity changes. $P=m*v$, so the momentum of the earth changes.

But why not the water?

Surely when the boat moves through the water there is a force being applied on the water by the front of the boat. So wouldn't the momentum of the water change too?

Answer in my textbook is:

While there may be more friction if the boat was on the sand and out of the water entirely, the momentum cange does not involve the water but occurs on the rower and the boast as a result of a force along the earth. Answers are the rower, the boat and the earth only.

  • $\begingroup$ What math are you expected to be able to do with the water? I would not have expected fluid dynamics before intro to momentum. Generally things you can't guess are skipped; I notice air didn't even make the list. $\endgroup$ – user118047 Apr 25 '17 at 20:54
  • $\begingroup$ Ah fair enough. A minimal (say end of high school or first year uni) level of math. The only fluid mechanics that is part of the course is Bernoulli (if that counts). This is not a structured course as such, rather a uni entrance exam. That must have been what they have done. Thank you. $\endgroup$ – K-Feldspar Apr 25 '17 at 21:06
  • $\begingroup$ Any change in momentum of the water would be dissipated through the whole lake and negligible, I would imagine. $\endgroup$ – koldrakan Apr 26 '17 at 6:59

I would argue that the book is wrong.

There is a boat-shaped "hole" in the water. If the boat moves, the "hole" also moves.

Ignoring all the local details of how the water moves, the net result is that if the boat is moving with (vector) velocity $\vec v$, a hole-sized volume of water is moving with velocity $-\vec v$ to create a new hole in one place and fill up the old one. That movement creates a constant non-zero momentum in the total volume of the water.

However, each individual particle of water only moves for a fairly short amount of time (and in a complicated way) and then stops, once the boat has moved past it.

It may be easier to visualize what is happening to the water if you consider a solid cylinder sinking down a tube full of water, where the cylinder is almost the same size as the tube. If the length of the cylinder is $L$, as it falls from the top to the bottom, the water is raised up the cylinder through a distance $L$. That clearly involves an average upwards velocity of the water while the cylinder is falling, and therefore an average upwards momentum in the water.

As @notstoreboughtdirt said in a comment, there is also a boat-shaped hole in the air, and the same argument holds true - but the mass of air involved is very small compared with the masses of the water and the boat, so it might be reasonable to ignore the momentum change of the air caused by this effect.

| cite | improve this answer | |
  • $\begingroup$ Thank you that was very useful, especially the cylinder analogy. Yes they must have ignored the water part. Cheers. $\endgroup$ – K-Feldspar Apr 25 '17 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.