Boat's conservation of momentum

[quark001]

Three boats with the same mass P travel in line ahead (one after the other) with the same speed v. Two identical loads of mass P1 each are thrown simultaneously from the middle boat to the front and rear boats with the same speed u relative to the boat. What are the speeds of the boats (v1, v2, and v3) after the loads have been thrown across?

My attempts:

The speed of the masses relative to the ground will be v+u and v-u. Considering only the second boat and applying conservation of momentum, you get: Pv = (P - 2P1)(v2) + P1(v + u) + P1(v - u), and v2 = v. That part is fine.

Now considering all the boats in the initial, final, and intermediate stage: 3Pv = (P+P1)v1 + (P+P1)v3 + (P-2P1)v = 2Pv + (P-2P1)(v) + P(v+u) + P1(v-u). Now the expressions for v1 and v3 as given in the solutions section should not contain v1 or v3, and they must contain u. I don't understand how that's possible since if you simplify the rightmost side of the equation, u is eliminated.

 

[Doc Al]

I don't quite understand your reasoning. Why not just find the speeds v1 and v3 just like you found the speed v2? You'll get your answers in terms of v and u.

 

[quark001]

If you simplify 2Pv + (P-2P1)(v) + P(v+u) + P1(v-u), you get 2Pv + (P-2P1)(v) + 2Pv = 5Pv - 2P1v. I could solve for v1, but my answer was independent of u.

 

[Doc Al]

If you simplify 2Pv + (P-2P1)(v) + P(v+u) + P1(v-u), you get 2Pv + (P-2P1)(v) + 2Pv = 5Pv - 2P1v. I could solve for v1, but my answer was independent of u.

Sorry, but I don't understand what you're doing.

Try this. Start with boat 1. What is its total momentum after the load is dropped into it? Use that to solve for v1.