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Elastic and un-elastic collisions

  1. Nov 29, 2005 #1
    dear god i just typed a full page and it decided to give me a 404, i hate computers.

    First time i had to use this forum in a long time but im very stuck today and cant find a way out of it after over an hour and with reading my notes so here goes.

    1)a proton of mass m travelling at 300m/s collides with a stationary carbon nucleus of mass 12m, the collision is elastic and head-on.

    (a) find the velocity of the centre of mass of the system, for this i took an imaginary particle of mass 13m and used conservation of momentum to find a velocity of 23.07m/s ....not sure if that is right
    (b) find the velocity of the proton after the collision in the centre of mass reference frame ive tried playing around with conservation of momentum for a while and with conservation of kinetic energy but my answer seems to always come out at either 0 or 300, not got a clue what im doing wrong
    (c)find the velocity of the proton after the collision in the lab reference frame i really am stuck on how to do this with conservation of momentum giving me one equation but 2 un-knowns

    2)a bullet of mass m1 is fired with a speed v into the bob of a ballistic pendulum of mass m2. find the maximum height h attained by the bob if the bullet passes through the bob and emerges with a speed v/2 unfortunately i was hungover during this lecture and ended up falling asleep...... whoooops, but it was a birthday night before so how could i not?

    if anyone could point me in the right direction with these it would be greately appreciated

  2. jcsd
  3. Nov 30, 2005 #2
    been looking at these some more (at a more reasonable hour) but im very confused, logically i would attempt to find the velocity of the proton in the lab reference frame possibly using m1/m2 = (v4-v2)/(v3-v1) however the order in which the questions are ordered makes me think that i am missing something very simple any advice pls?
  4. Nov 30, 2005 #3
    Are you given the equation for the velocity of centre of mass, or do you have to derive it yourself?
  5. Nov 30, 2005 #4
    i have typed everything i am given but as far as i am aware to get the velocity of the centre of mass all you do is have your "before" diagram as normal then on your "after" diagram you just draw a single imaginary particle with the complete mass of the system and work out the velocity of that, which is how i came to the conclusion of 23.07m/s
  6. Nov 30, 2005 #5
    (a) Your value is correct. However it is unclear whether you are supposed to be finding the velocity of the centre of mass before or after the collision. I will assume before.
    (b) You can now use sum of velocities to determine the initial velocities of the proton and carbon atom in the centre of mass frame (i.e. subtract 23.7 m/s from each velocity). The determine the resultant velocity of the proton in that frame. (You have the relevant equation?)
    (c) Then add the velocity of the centre of mass back to the photon's resultant velocity.
  7. Nov 30, 2005 #6
    first of all thanks for your answer however, if i do subtract 23.7 from each velocity i will get a velocity of -23.7 for the mass m2, which is wrong as it will be travelling in the same direction as the proton was originally

    to calculate the resultant velocity of the proton in that frame, assuming that 23.7 is positive and the 276.3 i would get from subtracting 23.7 from 300 is infact negative and the proton travells back the other way, as it should? then would it not just be 23.7m/s x 2 ? as i believe this is the relative velocity of the centre of mass of the system to the proton? and if not, no i dont have the equation and just looked through the motion of centre of mass section in my textbook and cannot find anything that seems to fit this problem exept the equation to calculate the velocity of the centre of mass

    thanks again
  8. Nov 30, 2005 #7
    ??? As the distance between the two particles decreases, both particles will be moving towards their common centre of mass: i.e. in opposite directions. Please note I said to subtract Vcm from both velocities to find their initial velocities. Then you can determine the resultant velocities.

    If I follow you correctly, then yes you are right. If you have not been given this equation (and are not expected to derive it):

    [tex]v_{1} = \frac{u_{1}(m_{1} - m_{2}) + 2m_{2}u_{2}}{m_{1} + m_{2}}[/tex]

    then you can approximate by saying that the change in m2's velocity is negligible therefore:

    [tex]v_{1} = -u_{1} + 2v_{2}[/tex]
  9. Nov 30, 2005 #8
    now im gettin somewhere, didnt really want to approximate so i used this equation to give me v1 = -253.84m/s in the CM reference frame

    then adding back on my 23.07m/s (Vcm) i get the final velocity of the proton to be -276.9 in the lab reference frame

    could you just check those over for me

    thanks again for your help

    sorry re - read and re-ran the equations realising i was using actual velocities as my inputs and found that v1=-276.916 in CM and then -300 in lab reference frame

    which proves your approximation equation that it is roughly equal to the negative of the original velocity in the lab reference frame
    Last edited: Nov 30, 2005
  10. Nov 30, 2005 #9
    Show me your working. I think you've gone wrong somewhere. Having compared the propert equation with the approximation myself, I would say the approximation isn't justified here. (I got a difference of 50 m/s!!!)
  11. Nov 30, 2005 #10
    actually re-running it i get a v1 in the CM reference of -276.93

    and then adding on the Vcm i get 253.86 this what you get?

    the error was because i forgot about the minus sign
  12. Dec 1, 2005 #11
    That looks more like it, though I get -277.5 (but I'm not using a decent calculator):

    [tex]v_{1} = \frac{276.3(1 - 12) + 2 * 12 * -23.7}{1 + 12} = -277.5[/tex]?
  13. Dec 1, 2005 #12
    yep thats what i got, thanks alot :)

    sorted the second question aswell, thanks alot
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