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.

 

[Nickie]

I would first clarify the scenario and the variables involved. From the given information, it seems that there are three boats with the same mass P traveling in a line with the same speed v. Two identical loads of mass P1 are thrown simultaneously from the middle boat to the front and rear boats with the same speed u relative to the boat. The question is asking for the speeds of the boats (v1, v2, and v3) after the loads have been thrown across.

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force. In this case, the system consists of the three boats and the two loads.

Initially, the total momentum of the system is given by 3Pv, since all three boats have the same mass and velocity. After the loads are thrown, the total momentum of the system remains the same. This can be expressed as:

3Pv = (P+P1)v1 + (P+P1)v3 + (P-2P1)v2

Where v1, v2, and v3 are the velocities of the three boats after the loads have been thrown.

To solve for these velocities, we can use the fact that the speed of the masses relative to the ground will be v+u and v-u. This means that the velocity of the second boat, v2, will remain the same as the initial velocity v, since the masses were thrown from it with equal and opposite velocities.

Substituting this into the equation, we get:

3Pv = (P+P1)v1 + (P+P1)v3 + (P-2P1)v

Simplifying further, we get:

3Pv = Pv1 + Pv3 + 2Pv - 2P1v

Rearranging the terms, we get:

2P1v = Pv1 + Pv3

Substituting the values of v1 and v3 in terms of v and u, we get:

2P1v = P(v+u) + P(v-u)

Simplifying, we get:

2P1v = 2Pv + Pu - Pu

Therefore, the velocity of the boats after the loads have been thrown can be expressed as:

v1 = v + u

v2 = v

v3 = v