Solving Impulse & Momentum Problems: Troubleshooting an Error

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Homework Help Overview

The discussion revolves around a physics problem related to impulse and momentum, specifically involving a hockey puck moving on a frictionless surface. The problem includes calculating the final velocity of the puck after applying forces in different directions over a specified time interval.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculations for impulse and momentum, questioning the correctness of the original poster's approach, particularly for part c of the problem. There is an exploration of the definitions of impulse and momentum, as well as the handling of signs in the calculations.

Discussion Status

Some participants have provided insights into the calculations, with one noting the importance of considering the magnitude of the final velocity as an absolute value. There is an acknowledgment of potential issues with the software used for the problem, suggesting that it may not have flagged errors in the calculations.

Contextual Notes

Participants mention the use of a specific educational program, 'Mastering Physics', which may influence their understanding and the feedback they receive on their answers. There is also a repeated emphasis on the need to clarify the problem statement and ensure all details are correctly interpreted.

verd
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Okay, So... I'm having trouble understanding what I did wrong here. I can't seem to understand how I got this problem wrong.
Here's the problem statement:

A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck is moving to the right at 2.98 m/s.

a.) Calculate the magnitude of the velocity of the puck after a force of 24.9 N directed to the right has been applied for 4.80×10−2 s.
b.) What is the direction of the velocity of the puck after a force of 24.9 N directed to the right has been applied for 4.80×10−2 s.
c.) If instead, a force of 12.2 N directed to the left is applied from t=0 to t= 4.80×10−2 s, what is the magnitude of the final velocity of the puck?
d.) What is the direction of the final velocity of the puck in this case?


So I've got everything right-- except for part c. ...I would imagine that I'd do it the exact same way I did part a. This is what I did for part c:


[tex]J_{x} = impulse[/tex]
[tex]J_{x} = p_{2x} - p_{1x}[/tex]
[tex]J_{x} = F_{x}(t_{2}-t_{2}) = -12.2(.048) = -.5856[/tex]
[tex]p_{2x} = J_{x} + p_{1x} = -.5856 + (.16)(2.98) = -.1088[/tex]
[tex]v_{2x} = \displaystyle{\frac{p_{2x}}{m}} = \displaystyle{\frac{-.1088}{.16}} = -.68[/tex]m/s

Why is this wrong? What did I do wrong?


Thanks!
 
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Initial momentum: mv = 0.16*2.98 kgm/s
Impulse: Ft = -12.2*0.048 kgm/s

The impulse is the change in momentum.
 
...Okay, so Ft stands for... impulse?

If so, I indicated that I already found the value for that one, which would be approximatley -.5856


I still don't know what I did wrong.
 
Nevermind, I'm getting the same answer you did.
 
Haha, weird. ...We use the program, 'Mastering Physics' at the University I attend... And the damn thing has been kind of quirky already. Perhaps it could be messing things up, because I keep doing the problem, different ways, and I still get the same answer.

Who knows.
 
I keep rereading it, thinking I missed out on something. This is EXACTLY what I was given:

A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck is moving to the right at 2.98 m/s.
If instead, a force of 12.2 N directed to the left is applied from t=0 to t= 4.80×10−2 s, what is the magnitude of the final velocity of the puck?

And this is how I came about my solution... That apppears to be incorrect:

verd said:
Okay, So... I'm having trouble understanding what I did wrong here. I can't seem to understand how I got this problem wrong.
Here's the problem statement:

A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck is moving to the right at 2.98 m/s.

a.) Calculate the magnitude of the velocity of the puck after a force of 24.9 N directed to the right has been applied for 4.80×10−2 s.
b.) What is the direction of the velocity of the puck after a force of 24.9 N directed to the right has been applied for 4.80×10−2 s.
c.) If instead, a force of 12.2 N directed to the left is applied from t=0 to t= 4.80×10−2 s, what is the magnitude of the final velocity of the puck?
d.) What is the direction of the final velocity of the puck in this case?


So I've got everything right-- except for part c. ...I would imagine that I'd do it the exact same way I did part a. This is what I did for part c:


[tex]J_{x} = impulse[/tex]
[tex]J_{x} = p_{2x} - p_{1x}[/tex]
[tex]J_{x} = F_{x}(t_{2}-t_{2}) = -12.2(.048) = -.5856[/tex]
[tex]p_{2x} = J_{x} + p_{1x} = -.5856 + (.16)(2.98) = -.1088[/tex]
[tex]v_{2x} = \displaystyle{\frac{p_{2x}}{m}} = \displaystyle{\frac{-.1088}{.16}} = -.68[/tex]m/s

Why is this wrong? What did I do wrong?


Thanks!


Can anyone else seem to figure how this is wrong?... There has to be something...
 
Hah. I got it.

what is the magnitude of the final velocity of the puck? As we know, magnitudes are absolute values. So our answer would be 0.68 instead of -0.68.

Although, usually the program will tell you that you got your negatives and positives mixed up. Strange it didn't do it this time.

Thanks, by the way!
 
Yeah I justn oticed that and was abuot to say. I used masteringphysics for kinematics an E&M, i hated it.
 

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