Impulse momentum method for linear motion

In summary, the conversation is discussing a problem where the formula FT=mVf-mVi is used to find the magnitude of a force. The question arises about whether the sign of the force matters, and it is explained that for this problem, only the magnitude of the force is needed. Another question is asked about why the package and flatcar are considered separately, and it is explained that this is necessary in order to analyze the effect of the friction force between them. It is also clarified that whether to apply the impulse momentum to them separately or together depends on the specific problem.
  • #1
freshbox
290
0

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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  • #2
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.
 
  • #3
So -ve or +ve doesn't matter?
 
  • #4
freshbox said:
So -ve or +ve doesn't matter?
Not in presenting your final answer.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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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  • #8
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.
 
  • #9
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.
 

1. What is the impulse momentum method for linear motion?

The impulse momentum method for linear motion is a mathematical approach used to analyze and predict the motion of objects in a straight line. It is based on the principles of Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.

2. How does the impulse momentum method differ from other methods of analyzing linear motion?

The impulse momentum method differs from other methods, such as the kinematic equations or energy methods, in that it takes into account the time during which a force acts on an object. It also considers the change in momentum of an object, rather than just its position or velocity.

3. What is the equation used in the impulse momentum method?

The equation used in the impulse momentum method is FΔt = mΔv, where F is the net force acting on the object, Δt is the time interval during which the force acts, m is the mass of the object, and Δv is the change in velocity of the object. This equation is derived from Newton's second law of motion.

4. How is the impulse momentum method used in real-world situations?

The impulse momentum method is used in various real-world situations, such as analyzing the impact of collisions, calculating the force needed to change the velocity of an object, and designing safety measures in sports and transportation. It is also used in industries such as aerospace and automotive engineering to predict and optimize the motion of objects.

5. What are the limitations of the impulse momentum method?

While the impulse momentum method is a useful tool for analyzing linear motion, it has some limitations. It assumes that the object is moving in a straight line and that the net force acting on the object is constant. It also does not take into account external factors such as air resistance, which can affect the motion of an object in the real world.

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