1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple Conservation of Momentum Question!

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data
    What can be said about the nature of collisions (elastic/inelastic) and the conservation laws that can be applied if...?

    Collision 1:
    Total kinetic energy before collision: 0.00503 J
    Total kinetic energy after collision: 0.00245 J
    Total momentum before: 0.0322 kg m/s
    Total momentum after: 0.0228 kg m/s

    Collision 2:
    Total kinetic energy before collision: 0.00342 J
    Total kinetic energy after collision: 0.00241 J
    Total momentum before: -0.0326 kg m/s
    Total momentum after: -0.0276 kg m/s

    2. The attempt at a solution
    From what I've read, the collision of two hard objects would result in an elastic collision. But the numbers don't stay the same, so some of it must have converted to sound/energy?
    No friction because gliders on an air track were used.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 26, 2008 #2

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Firstly, no collision between two macroscopic objects is ever completely elastic. One can approximate the collision as elastic to make the calculations simpler, but that doesn't mean that the collision really is elastic. Further, how have you calculated these values? How have you measured these values? What are the errors associated with your measurements?

    It would also be helpful you could describe the experimental set-up. From what I gather, you have two gilders on an air track, but I'm not sure of much else.
     
  4. Jan 26, 2008 #3
    In Collision 1, a 192g glider collides into a 279g glider, but the 192g glider moves at a faster speed, and results in the 279g being displaced. In Collision 2, a 279g glider and a 192g glider collide into each other, and the 192g glider is displaced. (http://electron9.phys.utk.edu/video, collision_22 and collision_21 respectively)

    I calculated the total KE/momentum before and after using
    KE = 1/2mv^2
    p=mv for each cart and adding them up.
    I didn't include uncertainties in my calculations, but on the distance-time graphs of the collisions I included uncertainties of 0.005 seconds and 0.03 meters.

    Thank you so much for your help! I really appreciate it.
     
  5. Jan 26, 2008 #4

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Is it possible you could provide an example calculation, say for the first collision, perhaps we can discover what's happening there, but my initial guess would be that your errors a large compared to you actual values.
     
  6. Jan 26, 2008 #5
    I've attached my calculations in the word document. I used mathtype so it would be easier to look at =)
     

    Attached Files:

  7. Jan 26, 2008 #6

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'll take a look it as soon as it's approved. However, in the meantime I'll elaborate on my point regarding errors. When you multiply to measured quantities (both with an associated error), to find the error of the resultant quantity you much add their fractional errors in quadrature. So, if we let [itex]s,v,t[/itex] be distance, speed and time respectively; and [itex]\delta s, \delta v, \delta t[/itex] their associated errors. Then;

    [tex]v = \frac{s}{t} \Rightarrow \delta v = \pm v\sqrt{\left(\frac{\delta s}{s}\right)^2 + \left(\frac{\delta t}{t}\right)^2}[/tex]

    One has to repeat this process again with the mass, to obtain the momentum. So you see that if [itex]\delta s \approx 1/2\cdot s[/itex], then [itex]\delta v \approx \pm 1/2\cdot v[/itex]; that is, your error in speed will be ~50% your speed. This is one reason why your measurements may not agree with theory.
     
  8. Jan 26, 2008 #7
    Wow, I never thought of that. I guess because the uncertainties inherent in the experiment are so big, they make it less valid.
    Again, thank you so much! You're a life-saver :D
     
  9. Jan 26, 2008 #8

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm not saying that this is the reason, but it probably is since the distance that you measure is probably going to be < 0.6, this means that your minimum uncertainty in velocity is going to be around 50% of your nominal velocity.

    Like I said, I'll have a look at your document when it's approved as see if I can see anything wrong with your calculations.
     
  10. Jan 26, 2008 #9

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    After a brief glance through your calculations, they look okay. Look at your graphs you can see that your errors are large just by looking at the error bars, I'll leave the calculations to you, but I would imagine they are more than enough to account for the discrepancies in the momentum.
     
  11. Jan 26, 2008 #10
    Thank you =)
    Haha yeah, the error bars on the graphs are rather conspicuous.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?