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

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    Boat Momentum
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Homework Help Overview

The discussion revolves around the application of momentum and center of mass in a boat-related physics problem. Participants are exploring the relevance of conservation of momentum and the computation of center of mass in the context of a system involving a boat and objects being thrown.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants express confusion regarding whether to use conservation of momentum and how to compute the center of mass. Questions arise about the relevance of the boat's mass in determining the system's momentum and the setup of equations for different parts of the problem.

Discussion Status

The discussion is ongoing, with some participants providing guidance on using conservation of momentum. There are multiple interpretations being explored regarding the setup of equations and the direction of momentum, particularly when an object is caught. While some clarity has been achieved, explicit consensus on the approach has not been reached.

Contextual Notes

Participants are grappling with the implications of different masses in the system and the effects of momentum transfer when objects are thrown and caught. There is an indication of confusion regarding the initial conditions and the assumptions about friction and motion.

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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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