1. Limited time only! Sign up for a free 30min personal 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!

Potential energy and momentum question

  1. Apr 10, 2010 #1
    I included what i did under the pictures to make it easier to read.
    but im not sure, with the ski questoin, if v= root(2gd) thats somthing i pulled out of my text book, and if d is 20 or the distance of the slope.
    but if thats all right my next worry is that i shouldnt of solved it as a projectile, i found the range of the jump, found the time it took to reach the range, then halved the time to find when the skier was at max height. and got 4.4m or somthing

    is there any other way to convert kinetic / potential energy to a velocity?
    my tutor did somthing with the KE formula, he differentiated it, then v turned into a and he intergrated it, but im not 100% what he did.

    I'm pretty confident with the neutron question because it seems right having the particle with the smaller angle having a higher velocity.
    But thanks to anyone who has a look


    [PLAIN]http://img199.imageshack.us/img199/8657/eeebb.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 10, 2010 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    For the skier problem, you can take a shortcut and write the energy conservation equation saying that the mechanical energy at the starting point is equal to the mechanical energy at maximum height. Note that at max height, the speed is just (1/2)mvx2. If you write down the equation, you will see that the masses cancel on each side, so the answer is independent of the skier's mass.

    For the neutron problem you need to write the momentum conservation equations correctly in the x and in the y direction. You need to say that total momentum in the x-direction is the same before and after the collision and likewise for the y-direction. Momentum is a vector, so be sure to calculate its components correctly. Both of your equations in this problem are incorrect in that regard.
     
    Last edited: Apr 10, 2010
  4. Apr 10, 2010 #3
    Pxi = Px1 + Px2
    Pyi = 0 = mVy1 + mVy2
    mVy1 = - mVy2
    Vy1 = -Vy2

    Vx1 = V1cos30, Vx2 = V2 cos 60

    so

    300m = m(V1 cos30) + m (V2 cos 60)

    300 = (V1 cos 30) + (V2 cos 60)

    is that right?
    it mustn't be because when i try to get values for the seperate v's i get 300
     
    Last edited: Apr 10, 2010
  5. Apr 10, 2010 #4

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    What is this in terms of v1 and v2?
    This is correct.
    Can you show me exactly how you get 300 m/s?
     
  6. Apr 10, 2010 #5
    I tried just then going like
    In terms of the y components
    V1 sin 30 = - V2 sin (-60)
    and i got V1 = root(3) * V2

    Then i tried subbing that into the equation for the x components and found V2 = 150
    I've just been trying to do that again to find V1 with V2 = V1/root(3)
    but it isnt working out

    and with the other thing i did before and kept getting 300 i did

    300 = (V1 cos 30) + (V2 cos 60)

    changed them to exact values and rearranged to find V1
    found V1= (-0.5V2 + 300)/ (root(3)/2) then pluged it into the formula and got

    300 = -V2/2 + 300 + V2/2
    i did it afew different ways but its all a mess

    was doing the y component thing the right way?

    could i do 300 = V1 + V2 and sub in V1= root3 V2
     
  7. Apr 10, 2010 #6

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    You have the correct ingredients, but you don't seem able to assemble them properly. One step at a time. You have

    V1 = root(3) * V2 ( Eq.1)

    300 = (V1 cos 30) + (V2 cos 60) (Eq. 2)

    What do you get if you take the value for V1 in equation (1) and put it in equation (2)?
     
  8. Apr 10, 2010 #7
    are you sure thats right?
    when i do that i get V2= 150 and v1 = 259 or somthing
    if i plug it into
    300 = v1 + v2 i get v1 = 190 and v2 = 110 which seems pretty realistic
    because wouldnt the particle with the lower angle have a higher velocity
    and also isnt equation 2 the equation for the x components? V1cos30 = Vx1 ?
     
  9. Apr 10, 2010 #8

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    That's correct.
    Why do you think the above equation is correct? What is it an expression of?
    It ain't necessarily so. The numbers speak for themselves. If v1 = sqrt(3)*v2 = 1.732*v2, which one is larger, v1 or v2?
     
  10. Apr 10, 2010 #9
    If Pi = P1+P2

    mVi = mV1 + mV2 m is the same
    Vi = v1 + v2
    300 = v1 + v2 ?
    dont 150 and 259 have to add to give 300 for conservation of momentum to be kept?
     
  11. Apr 10, 2010 #10

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    You are forgetting that the equation

    vi=v1+v2

    is a vector equation. Because the angle between v1 and v2 is 90o, when you draw the vector addition diagram, you get a right triangle where 300 is the hypotenuse and 150 and 259 are the right sides. Check if the Pythagorean theorem is obeyed.
     
  12. Apr 10, 2010 #11
    ohhhh
    that makes sense
    thanks for all your help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Potential energy and momentum question
Loading...