Impulse momentum method for linear motion

[freshbox]

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?

 

[Doc Al]

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.

 

[freshbox]

So -ve or +ve doesn't matter?

 

[Doc Al]

So -ve or +ve doesn't matter?

Not in presenting your final answer.

 

[freshbox]

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.

 

[Doc Al]

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.

 

[freshbox]

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...

 

[Doc Al]

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.

 

[freshbox]

why did the question consider the package and flatcar individually?

 

[Doc Al]

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.

 

[freshbox]

So anything with friction between them I must treat them separately if not I can treat them as a whole?

 

[Doc Al]

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.

 

[freshbox]

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?

 

[Doc Al]

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.

 

[freshbox]

Roger.