Conservation of momentum problem

[nejikun]

Homework Statement

Two boats move parallel each other inertially in opposite directions. They both have a speed of 6m/s. When they are near we move a load from one boat to the other. After this the second boat continues in the same direction with a speed of 4m/s. Find the mass of the second boat if the first boat weighs 5000N and the load weighs 600N. Water resistance should not be taken into consideration

Homework Equations

The Attempt at a Solution

I tried a few ways but none of them takes me to the same solution as the book. It should be 300kg but I keep getting other answers.

 

[tiny-tim]

Welcome to PF!

Hi nejikun! Welcome to PF!

Show us your full calculations, and then we'll see what went wrong, and we'll know how to help!

 

[nejikun]

i thought that it should be :
[m(1)+m(l)]*V(1) + m(2)*V(1)= m(1)*V + [m(2)+m(l)]*V(2)
the only problem is that i don't know the speed of the first boat after moving the load.
the other ways are logically wrong as i now understand. I have no clue for any other way to solve it.

 

[tiny-tim]

Hi nejikun!

You don't need the final speed of the first boat (though you can find it, if you want )

Hint: how much momentum does the load lose?

 

[nejikun]

Well you confused me even more with the hint :S. I read again all the theory about momentum but i am still missing something. I don't understand what you mean by the load losing momentum.

 

[tiny-tim]

The load was going at 6 m/s one way, now it's going at 4 m/s the other way …

how much was the change in its momentum? ​

 

[nejikun]

oh so all the problem was because i didn't take into consideration the direction.

 

[nejikun]

sorry to disturb you again but i still can't solve this. i can't figure out anything

 

[tiny-tim]

ok, let's do it in stages …

i] The load was going at 6 m/s one way, now it's going at 4 m/s the other way …

how much was the change in its momentum? ​

 

[nejikun]

i think it is 120 kg*m/s

 

[tiny-tim]

i think it is 120 kg*m/s

(that's with g = 10 m/s² ?)

It would help if you wrote your reasons as well …

change in momentum = change in velocity times mass …

so what is the change in velocity, and what is the mass?

No, it's not 120, that would be if the load slowed down from 6 km/s to 4km/sec, both in the same direction.

Try again! ​

 

[nejikun]

yes g=10m/s^2. So the change is velocity is 10m/s and the mass is 60 kg so the change in momentum is 600 kg*m/s

 

[tiny-tim]

yes g=10m/s^2. So the change is velocity is 10m/s and the mass is 60 kg so the change in momentum is 600 kg*m/s

(try using the X² tag just above the Reply box )

Yup!

ok, now …

ii] apply the conservation of momentum equation for the load and the second boat combined …

what do you get? ​

 

[nejikun]

i tried this : m(2)*V(1)+600=[m(2)+m(L)]*V(2) but it is not correct.

 

[tiny-tim]

i tried this : m(2)*V(1)+600=[m(2)+m(L)]*V(2) but it is not correct.

uhh?

what numbers?

 

[nejikun]

it's m(2)*6+ 600 = [m(2)+ 60]*4 and when i solve for m(2) it comes out negative

 

[tiny-tim]

it's m(2)*6+ 600 = [m(2)+ 60]*4

ah, you've counted some of the load's momentum twice …

do you see that the 60*4 is already included in the 600?

Look at it this way … the momentum gained by the load must equal the momentum lost by the second boat.

 

[nejikun]

I finally sovled it :D thank you veeery much tiny-tim for all your time ;)