Solving Impulse & Momentum Problems: Troubleshooting an Error

In summary, the conversation is about someone asking for help understanding a problem involving a hockey puck on a frictionless surface with applied forces. They are specifically struggling with part c, where a force of 12.2 N is applied to the left for a certain amount of time, and they are trying to find the magnitude of the final velocity of the puck. They share their solution and ask for help in understanding where they went wrong. After some discussion, they realize that they calculated the velocity incorrectly and the correct answer should be 0.68 m/s instead of -0.68 m/s. They also mention using a program called "Mastering Physics" for their studies.
  • #1
verd
146
0
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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  • #2
Initial momentum: mv = 0.16*2.98 kgm/s
Impulse: Ft = -12.2*0.048 kgm/s

The impulse is the change in momentum.
 
  • #3
...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.
 
  • #4
Nevermind, I'm getting the same answer you did.
 
  • #5
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.
 
  • #6
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...
 
  • #7
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!
 
  • #8
Yeah I justn oticed that and was abuot to say. I used masteringphysics for kinematics an E&M, i hated it.
 

FAQ: Solving Impulse & Momentum Problems: Troubleshooting an Error

1. What is impulse and momentum?

Impulse is the change in momentum of an object over a period of time. Momentum is the product of an object's mass and velocity and is a measure of its motion.

2. What are common errors when solving impulse and momentum problems?

Some common errors include not properly identifying the initial and final states of the system, not considering all external forces acting on the system, and not using the correct equations for momentum and impulse.

3. How do I troubleshoot an error in my solution to an impulse and momentum problem?

First, double check all of your calculations to ensure they are correct. Then, review your problem-solving steps to see if you made any mistakes or missed any important factors. If you are still unable to find the error, try approaching the problem from a different angle or seeking guidance from a peer or instructor.

4. Can I use the same equations for all impulse and momentum problems?

No, the equations used will vary depending on the specific scenario and what is known about the system. It is important to carefully read and analyze the problem before choosing the appropriate equations.

5. How does solving impulse and momentum problems relate to real-world applications?

Solving impulse and momentum problems is essential in understanding and predicting the motion of objects in real-world scenarios. This knowledge is used in fields such as engineering, physics, and sports to design and improve systems and equipment.

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