How do momentum and center of mass play a role in solving a boat problem?

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The discussion focuses on using conservation of momentum and the center of mass to solve a boat problem involving a thrown object. Participants express confusion about whether to include the boat's mass in momentum calculations and how to set up the equations correctly. It is clarified that the momentum of the boat and the thrown object are in opposite directions, necessitating the use of negative values for the boat's velocity. The key takeaway is that the momentum before the object is thrown equals the momentum after it is caught, confirming the conservation principle. Ultimately, the original poster resolves their confusion and finds the correct solution with assistance.
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Homework Statement


physicsmidtermproblem.jpg



Homework Equations



p=mv
XCM = (m1x1 + m2x2)/(m1 + m2)

The Attempt at a Solution



I'm really confused on this problem. I'm not sure if I should be using conservation of momentum, or what. Also do I have to compute the center of mass for the boat?

Thanks
 
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You don't think Mb is relevant to determining the momentum of the system?
 
But to answer your confusion, yes, you are supposed to used conservation of momentum.
 
Ok, I'm still not really sure how to solve the problem. I don't even know where to begin really.
 
When the ball is in the air, it has m1V1.

But if the boat was at rest when it was thrown and the friction is 0, then that means that the boat and people have opposite momentum doesn't it? But they of course have a different mass than the ball, so ...
 
I'm still confused how to set up the equations for parts A and B...

would it be, for part A:

(MB+M1+M2)VBoat = MPVP
 
haydn said:
I'm still confused how to set up the equations for parts A and B...

would it be, for part A:

(MB+M1+M2)VBoat = MPVP

That looks right.

What happens then when it is caught?
 
MPVP = (M1+M2+MB)VBoat, Final

?
 
haydn said:
MPVP = (M1+M2+MB)VBoat, Final

?

No. Because the momentums are in different directions aren't they and when it is caught what happens to the total momentum?

I should have pointed out before that the V of the boat and people was opposite to the Vp.
 
  • #10
Ok, so in that first equation you said was right there should be a negative in front of VBoat?

I don't know what happens when the backpack is caught... I thought all the momentum of the backpack would transfer to the boat+person system.
 
  • #11
haydn said:
Ok, so in that first equation you said was right there should be a negative in front of VBoat?

I don't know what happens when the backpack is caught... I thought all the momentum of the backpack would transfer to the boat+person system.

It does.

But happily it must be the very same momentum that was imparted to the backpack when it was thrown.

Hence Momentum before throwing = momentum after being caught isn't it?
 
  • #12
Ok great. I figured out the right answer now. Thanks for the help!
 

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