Impulse Momentum (et al.) test

In summary, the student attempted to solve a potential energy problem by using a derivative calculation and claimed that the final velocity was 5.25 m/s.
  • #1
QuarkCharmer
1,051
3

Homework Statement


I took a test today and I am pretty sure I got everything right except for this one problem that was on there which I am unsure of. If I remember correctly, it went like this:

It said there was a particle of mass 2 kg moving with a speed of 5 m/s to the right. Then there was a graph that more or less looked like this one I just made:

2dqixc8.jpg


The question asked what the final velocity was after the impulse.

Homework Equations



I believe this is correct and relevant.
[tex]\int F dt = J = ΔP = mv_{f} - mv_{i}[/tex]

The Attempt at a Solution



I wagered that since the impulse was equal to the area under the curve on a Force v. Time graph, that the problem could be done in this way:

[tex]J = \frac{1}{2}(base)(height)[/tex]
[tex]J = \frac{1}{2}(0.10)(10)[/tex]
[tex]J = 0.5[/tex]

[tex]J = mv_{f} - mv_{i}[/tex]
[tex]0.5 = (2)v_{f} - (2)(5)[/tex]
[tex]10.5 = (2)v_{f}[/tex]
[tex]5.25 = v_{f}[/tex]

So I said the velocity afterwards was 5.25 m/s (and part b asked the direction, which I claimed was to the right).

It's unlike any practice/hw/quiz problems we have had before and there was some equation she wrote on the board:
[tex]F_{avg}Δt = mv_{f} - mv_{i}[/tex]

Which I believe is basically what I did, I just didn't use that equation.

Did I mess up?
 
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  • #2
Assuming that the impulse was also to the right, you did fine.

FYI: You could have used your instructor's formula; In this case the average force is 10/2 = 5 N. Same answer, of course. (Since it's really the same thing.)
 
  • #3
Whew, thank you.

Another question that I am a little sketchy about is one where they showed some function U(x) which represented potential energy with respect to distance.The mass of the thing was m kg The question(s) were:

a.)What is force:
I simply took the derivative of U, and my final answer was U'(x). Sorry I can't remember the function, but it was something that had the general shape of y = 1/x.

b.)If the object was released from it's origin, what is it's speed at x = 2?

I claimed that mgU(0) was the total energy on the system, and so, after the thing fell or whatever, the final velocity could be found via:

[tex]mgU(0) = mgU(2) + \frac{1}{2}mv^{2}[/tex]
Solving for v obviously.

?
 
  • #4
QuarkCharmer said:
a.)What is force:
I simply took the derivative of U, and my final answer was U'(x). Sorry I can't remember the function, but it was something that had the general shape of y = 1/x.
Almost. Given a potential energy function, the associated force is given by -dU/dx. The minus sign is important.

b.)If the object was released from it's origin, what is it's speed at x = 2?

I claimed that mgU(0) was the total energy on the system, and so, after the thing fell or whatever, the final velocity could be found via:

[tex]mgU(0) = mgU(2) + \frac{1}{2}mv^{2}[/tex]
Solving for v obviously.
U(x) is the potential energy, not mgU(x). (You're confusing this with a gravitational PE problem, perhaps?)

You want:
[tex]U(0) = U(2) + \frac{1}{2}mv^{2}[/tex]
 
  • #5
Oh ok, I got confused trying to replicate the problem from memory into latex.

Yes I used [itex]F = -\frac{dU}{dx}[/itex]
and I did use U(x) for potential, not mgU(x).
 
  • #6
QuarkCharmer said:
Oh ok, I got confused trying to replicate the problem from memory into latex.

Yes I used [itex]F = -\frac{dU}{dx}[/itex]
and I did use U(x) for potential, not mgU(x).
Good! :cool:
 

What is the Impulse Momentum test?

The Impulse Momentum test is a physics experiment that measures the change in momentum of an object when it experiences an impulse, or sudden change in force, over a certain amount of time.

How is the Impulse Momentum test conducted?

The Impulse Momentum test involves using a force sensor to apply an impulse to an object and measuring the change in momentum using a motion sensor. The data is then analyzed to calculate the impulse, change in momentum, and average force.

What is the relationship between impulse and momentum?

According to Newton's second law of motion, force is equal to the change in momentum over time. This means that the greater the impulse applied to an object, the greater the change in momentum and the greater the force exerted on the object.

What are some real-life applications of the Impulse Momentum test?

The Impulse Momentum test has many practical applications, such as analyzing the impact force of a car crash, studying the force exerted on the body during sports activities, and designing safety features for amusement park rides.

How can the Impulse Momentum test be used to improve safety?

By understanding the relationship between impulse, momentum, and force, the Impulse Momentum test can be used to design and improve safety features in various industries. For example, studying the impact force of a car crash can inform the design of airbags and seatbelts to minimize injuries.

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