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Collision/Projectile Motion Problem -NEED HELP-

  1. Dec 4, 2011 #1
    A Ball with mass Mo froms H1(Total Height),before it hits the ground it bounces off a stationary inclined plane at height (H2) with angle (Ω), it falls a distance of (H2) and travels a distance (L) over a time (∆t)
    FIND L and ∆t

    Image of Problem: http://s9.postimage.org/4xn9jpklb/Velocity.png

    (this is my verbal expression of the question, there will be a picture attached) unfortunately there was no written part to this question.

    am only having problem with one part specifically.... Which is the part i assume everyone has problems on..

    The collision between the inclined plane, or how exactly to find the velocity after it has been reflected by the inclined plane. (I know the angle of reflection is = to angle of incidence) so therefore angle Ω of the plane is equal to angle Ω of the velocity after collision...

    I know that Px=Px' momentum is conserved in the X direction...

    My Attempt:

    Et1= Eg1=mgh1

    Et2= Eg2 + Ek2 = mg(h1-H2) + 1/2MoV2^2

    So Et1=Et2

    mogh1=mog(h1-h2) +1/2MoV2^2 (masses cancel)

    V2 = Sqrt 2gh1-2g(h1-h2)

    (V2 is the velocity just before it collides with the plane)

    Now i've Drawn a FBD (my frame of reference/axis is along the surface horizontally and vertically the positive direction is towards the normal force) of when the ball is on the inclined plane (and it hits the plane so the normal force is perpindicular to the surface)

    so when i do my net forces i get

    fnetx= fgsinΩ
    fnety= fn -fgcosΩ=0 (therefore fn = fgcosΩ)

    Now i'm curious am i confusing this part, am i supposed to take the Normal force as the one which has components?

    Also when i do... this is where i get really stuck..

    MOMENTUM!!!

    p=p'

    Since the inclined plane remains stationary during collison that means its vertical and horizontal compenents are both zero...

    let v2x =v2cosΩ
    v2y = v2sinΩ
    so im left with

    0 + mov2x = 0 + mov2x'
    and
    0 + mov2y = 0 + mov2y'


    so now components are equal but when i continued with these steps and showed my teacher the answer she said it was incorrect.... could someone please help me with Collision between an inclined plane and the ball and finding the velocity after the collision.. i have been working on this question for weeks, my semester for physics gr12 finished 3 weeks ago.. but i really want to answer this question.
     
    Last edited: Dec 4, 2011
  2. jcsd
  3. Dec 4, 2011 #2
    Ah also, The angle Ω is 30° and this question is to be solved in terms of variables.
     
    Last edited: Dec 4, 2011
  4. Dec 4, 2011 #3

    jedishrfu

    Staff: Mentor

    okay I think your omega angle s/b (pi/2 - 2*omega) right?

    omega defines the plane incline relative to horizontal

    if the vertical drop continues thru to the ground state then the complementary angle of the
    small triangle formed is pi/2-omega

    and the angle between the vertical drop and incline is (pi/2-omega)

    since the ball reflects off the incline then the reflection angle relative to the incline is also (pi/2 - omega)

    consequently the angle relative to the horizontal is ((pi/2 - omega) - omega)

    hence (pi/2 - 2* omega)

    now you should have the velocity of the ball the moment it hits the incline and the angle it takes as it bounces off the incline
     
  5. Dec 4, 2011 #4

    jedishrfu

    Staff: Mentor

    so I get the following eqns:

    vx = V2 * cos( pi/2 - 2 * omega)
    vy = V2 * sin( pi/2 - 2 * omega)

    y = H2 + (vx * t) - 1/2 * g * t^2

    x = V2 * cos( pi/2 - 2 * omega) * t

    then set y = 0 and solve for t

    plug t into x and find the distance traveled
     
  6. Dec 4, 2011 #5
    Anyway you know how to do this without radiants, i haven't quite learned them yet.. or could you possibly post a drawing so i could learn from your drawing how to use radiants?
     
  7. Dec 4, 2011 #6

    Doc Al

    User Avatar

    Staff: Mentor

    Assuming it's a perfectly elastic collision, the speed will not change during the collision.
     
  8. Dec 4, 2011 #7
    So assuming it's not an elastic collision, what can i do to calculate the velocity after the collision, and also how to calculate the angle assuming my strategy of angle of incidence is = to angle of reflection is incorrect.
     
  9. Dec 4, 2011 #8

    Doc Al

    User Avatar

    Staff: Mentor

    Why would you assume it's not an elastic collision? (If it isn't, then I think you do not have enough information to solve it.)
     
  10. Dec 4, 2011 #9

    Doc Al

    User Avatar

    Staff: Mentor

    So what angle does the ball make with respect to the horizontal when it bounces off the incline?
     
  11. Dec 4, 2011 #10
    The angle the ball makes after collision is 30 degrees from the horizontal

    it says when it hits the surface of the plane it is perpendicular to the surface


    wait i'm wrong.. the geometry says it must be the opposing angle of 60 degrees... derrr
     
    Last edited: Dec 4, 2011
  12. Dec 4, 2011 #11

    jedishrfu

    Staff: Mentor

    just replace the pi/2 with 90 degrees and compute the sin / cos values from that.

    wrt radians, they are just another method of describing angle measure that is more natural to use in math problems especially in the mathematical series that define the values of sin and cos functions.

    2π = 360
    π = 180
    π/2 = 90

    wikipedia has a good writeup on radian measure for angles at:

    http://en.wikipedia.org/wiki/Radians
     
  13. Dec 4, 2011 #12

    jedishrfu

    Staff: Mentor

    I was trying to say that your geometry is wrong wrt to the angle.

    Just because the incline is omega doesn't mean the angle of the ball bouncing off is also omega

    I think its actually 90 degrees - 2*omega

    As an example say the incline was at 80 degrees then by your analysis it bounces off at 80 degrees to the horizontal right. But if you try the experiment it will be 70 degrees below the horizontal ie

    angle = 90 degrees - 2 * 80 degrees = - 70 degrees to the horizontal

    As another example say the incline is 45 degrees then by your analysis the ball bounces off at 45 degrees instead of zero degrees to the horizontal.

    angle = 90 degrees - 2 * 45 degrees = 0 degrees to the horizontal
     
  14. Dec 4, 2011 #13


    thanks a lot bro this is very clear. I'm going to start practicing in radiant and get familiar with them. i appreciate this.
     
  15. Dec 4, 2011 #14

    Doc Al

    User Avatar

    Staff: Mentor

    That's correct.

    :confused:
     
  16. Dec 4, 2011 #15
    c.c. i'm confused so bad.
     
  17. Dec 4, 2011 #16

    Doc Al

    User Avatar

    Staff: Mentor

    Why is that?

    The collision with the incline is elastic. The speed doesn't change and the angle of incidence equals the angle of reflection. From that point on, it's a projectile motion problem.
     
  18. Dec 4, 2011 #17
    Okay can you just go through the angle part with me... Cause i confused myself with this geometry...


    So if it bounces so that the object is perpindicular to the surface...

    How can i prove that the angle on the plane = the angle of the velocity.
     
  19. Dec 4, 2011 #18
    okay so...

    Energy:

    et1=et2
    mgh1=1/2mv2^2 + mg(h1-h2)

    v2 = √2gh2

    Momentum:

    p=p'

    0 + mv2 = 0 + mv2'

    v2=v2'

    Kinematics:
    v2'x = v2'cos30
    v2'y= v2'sin30

    Uniform accelerated motion
    H2 = v3^2 - V2'y^2
    --------------
    2g

    V3 = √2GH2 + v2^2


    H2 = v3 + v2
    --------- (ΔT)
    2

    Δt = 2h2
    ------------------
    √2gh2 + √(2gh2)sin30


    Am i correct up to this ponit?? :O
     
  20. Dec 4, 2011 #19

    jedishrfu

    Staff: Mentor

    see diagram and compare to your angles
     

    Attached Files:

  21. Dec 5, 2011 #20

    okay so in order to get the angle, you compared your x axis, with the two forces of normal..

    and you subtracted the two 2Alpha from 90 degrees, because that is that quadrant...


    parallel lines are referring to the bottom of the plane being paralell with the axis?

    And thanks a lot for spending your time man. i really appreciate this. I'm just asking a lot of questions, so that in the future if i ever encounter a similar problem i under stand things better, and geometry has never been my strong suit.
     
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