Impulse momentum method for linear motion

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Homework Help Overview

The discussion revolves around the impulse-momentum method in linear motion, specifically addressing the application of the formula FT = mVf - mVi in various scenarios. Participants explore the implications of sign conventions in force calculations and the treatment of objects in motion, particularly when friction is involved.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the sign of the force calculated using the impulse-momentum formula and whether the sign matters when reporting the final answer. There is also exploration of how to determine the sign of values used in calculations. Additionally, questions arise regarding the treatment of objects (package and flatcar) individually versus as a whole in the context of friction.

Discussion Status

The conversation is ongoing, with participants providing insights into the importance of sign conventions and the necessity of treating objects separately when friction is a factor. Some guidance has been offered regarding the interpretation of results and the conditions under which impulse momentum can be applied.

Contextual Notes

Participants note that the problem does not specify the direction of motion, leading to ambiguity in sign conventions. The discussion also highlights the need for clarity on the assumptions made when applying the impulse-momentum method, particularly in relation to frictional forces.

freshbox
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Homework Statement


I was told by a friend that M1+I1-2=M2 is actually FT=mVf-mVi.

So using FT=mVf-mVi to solve the following question...

The Attempt at a Solution


FT=mVf-mVi
(F)(0.15)=800(0)-800(5000/3.6)
F=-7407.4N

Ans is 7407.4N positive. Can I ask did I get my sign wrong?
 

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freshbox said:
Ans is 7407.4N positive. Can I ask did I get my sign wrong?
All they want is the magnitude of the force.

If you choose to make the initial velocity positive, then the force will indeed be negative.
 
So -ve or +ve doesn't matter?
 
freshbox said:
So -ve or +ve doesn't matter?
Not in presenting your final answer.
 
So how do i determine where the values that i am going to put in is going to be +ve or -ve?

The formula is FT=mVf-mVi, i am just subbing in the values in, but in the end i got negative.
 
freshbox said:
So how do i determine where the values that i am going to put in is going to be +ve or -ve?

The formula is FT=mVf-mVi, i am just subbing in the values in, but in the end i got negative.
There's nothing wrong with your solution. I'm just pointing out that often when they ask for some quantity they just want the magnitude of that quantity.

If you had assumed the car was initially moving to the left, then using your sign convention you would have found that the force was positive. The direction of motion (or sign convention) wasn't specified so presumably they just want the magnitude of that force.
 
Ok, sorry I would like to ask another question attached below. Actually I tried this question:

FT=mVf-mVi
-441.45t=150(0)-150(1.2)
-441.45t=-180
t=0.407s

I don't understand this question why do they consider them individually? For work energy method is about setting of datum, is there anything special for impulse momentum that I need to know?


Thanks...
 

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freshbox said:
Ok, sorry I would like to ask another question attached below. Actually I tried this question:

FT=mVf-mVi
-441.45t=150(0)-150(1.2)
-441.45t=-180
t=0.407s
Your error is thinking that the package ends up with zero final velocity. It just stops moving with respect to the flatcar. After the package stops sliding, the package and flatcar end up moving with the same nonzero speed.
 
why did the question consider the package and flatcar individually?
 
  • #10
freshbox said:
why did the question consider the package and flatcar individually?
In order to analyze the effect of the friction force between package and flatcar, you need to treat them separately so that the friction is an external force.
 
  • #11
So anything with friction between them I must treat them separately if not I can treat them as a whole?
 
  • #12
freshbox said:
So anything with friction between them I must treat them separately if not I can treat them as a whole?
It all depends on the particular problem. The key thing is to realize that you can apply the impulse momentum to them separately or together, as needed.
 
  • #13
So for this question I have to apply impulse momentum separately because there is a friction force between them if not I won't be able to solve them. Correct?
 
  • #14
freshbox said:
So for this question I have to apply impulse momentum separately because there is a friction force between them if not I won't be able to solve them. Correct?
Right.
 
  • #15
Roger.
 

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