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Collision of point mass and sphere in particular fashion

  1. Apr 5, 2017 #1
    41ce2ea4b61b1f2f1f832b12f4509907fdb01de2.jpg


    1. The problem statement, all variables and given/known data

    a point mass is projected at an angle of 60° from the horizontal. it collides with the sphere at a maximum height of trajectory with a sphere of radius of 72.5cm such that it is projected off the sphere once again at an angle of 60° from the horizontal.The sphere, due to the collision, is bounced off the floor. The point mass, in its descent, collides with the sphere again, through the same horizontal of the previous point-mass-sphere collision, and is projected at an angle of 30° below the horizontal . if the density of the sphere is 50kg/m3 and uniform mass distribution then the mass of point mass,m and the initial velocity of the point mass is u. find floor(m + u) . Assume all the collisions are elastic, and the sphere is rigid.

    2. Relevant equations
    mvi = mvfinal
    xf = xo + vot + 0.5 a t2

    3. The attempt at a solution
    couldn't try because i dont know why does the sphere bounces

    someone please tell me why does the sphere bounce then i could probably start as in i know it is because the mass hit it but what exactly is the reason?
     
  2. jcsd
  3. Apr 5, 2017 #2
    Imagine that the sphere is actually a tiny height (0mm) above the ground - and at the time of the collision, motionless. Then the collision with the point will cause the sphere to move with a downward component, strike the surface, and bounce elastically.
     
  4. Apr 5, 2017 #3

    BvU

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    I'll do as you ask and then you can start as you promised.
    Drat -- Scotty was faster! Anyway: point mass m acquires vertical momentum upwards, so sphere gets downward momentum. F (change in momentum per unit time) downward gets inverteed by the ground and results in upward bounce.
     
  5. Apr 5, 2017 #4
    so say mass goes up with momentum x then sphere goes down with momentum x then rebound with momentum x also ? is it correct
     
  6. Apr 5, 2017 #5

    BvU

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    Assume it is.
     
  7. Apr 6, 2017 #6
    wait based on conservation of momentum at point of first collision i am getting mass of point mass is same as mass of ball is that correct?

    so let ucos60 be initial velocity of point mass before collision
    then P be momentum of ball after collision
    and h be velocity of ball after collision
    m be mass of point mass and M be mass of ball
    so
    m ucos60 = mhcos60 + Px
    mhsin60 = Py
    time of flight for point mass and ball must be same after collision
    hence
    for point mass: 2hsin60/g = t
    for ball : Vy = mhsin60/M
    so for the two times to be same mustn't m/M be 1

    am i right if wrong where did i go wrong
     
  8. Apr 6, 2017 #7

    BvU

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    Don't know until you post your work
    [edit] ah, the work is following after this question mark. Will look, but I think we need an expert opinion (@haruspex for example....)

    My hunch is that the point mass should be lighter than the big ball, but... who knows what evil lurks in the hearts of exercise composers (*)


    (*) The Shadow knows
     
    Last edited: Apr 6, 2017
  9. Apr 6, 2017 #8
    that was my work until now just what want to know if i am on right track
     
  10. Apr 6, 2017 #9

    BvU

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    Is there a basis for your assumption that point mass hits the ball at the very moment the ball hits the ground again ? (In other words: did you render the complete problem statement ?)
     
  11. Apr 6, 2017 #10
    ooh so that isn't the case?because if it didn't the ball would keep bouncing until it hits the mass
     
  12. Apr 6, 2017 #11
    ok then this is how i want to approach the problem tell me if anything wrong with it

    upon first collision momentum initial in x = momentum final in x
    momentum initial in y = momentum final in y (momentum initial in y = 0 )
    kinetic energy of the mass initially = kinetic energy of the mass finally + kinetic energy of the ball
    do i need anymore equations
     
  13. Apr 6, 2017 #12
    i did that because the sphere only bounced one time according to the picture
     
  14. Apr 6, 2017 #13

    BvU

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    I don't know if that is the case or not -- still trying to formulate a complete problem statement.
    Is THAT picture in post #1 an original part of the problem statement ? Or did you sketch it yourself ?
     
  15. Apr 6, 2017 #14
  16. Apr 6, 2017 #15

    BvU

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    Goodness ! They target a brilliant audience, but their illustrating skills are rather limited -- and adding a mass and a velocity is close to complete heresy to ordinary physicists like me ?:) .

    But now I understand your assumption and I agree.
     
  17. Apr 6, 2017 #16
    so it is correct ? and by the way what does the last part about the mass bouncing off at 30 degrees got to do with the question
    on a side note i regard you way better than me at physics
     
  18. Apr 6, 2017 #17

    BvU

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    unclear to me too. probably Nanayaranaraknas has a purpose for it, but it eludes me ! Must be some smart insight that reveals the path to the solution !

    make no mistake there: a PhD (which I do have) or a professorship (which I don't) are no guarantee whatsoever against stupidities of all kinds, including the worst. In this exercise, to me you are the leading problemsolver -- I just try to help by asking some questions...
     
  19. Apr 6, 2017 #18
    ok this is all i got after this i am stuck i need help!!!! this problem is driving me crazy

    mu2 = mh2 + MV2 ---1
    mu = mh + 2MVx ---2
    hsin60 = Vy ----3
    V2 = Vx2 + Vy2
    hence since M = m
    u2 = h2 + V2
    u2 = h2 + (u2 + 4h2 - 2uh)/4M

    EDIT: i need one more equation

    wait you need to include rotational motion that is going to make the problem so much more harder !!!

    and when i mean total momentum in y to be zero i mean right after the collision before hitting the ground. after hitting the ground the velocity in y will just become negative as before and x-velocity remains unchanged
     
  20. Apr 6, 2017 #19

    haruspex

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    I am inclined to dismiss this question.

    The m+u in the target makes no sense unless it is specified what units m and u are to be expressed in. Asking for m+u is anyway bizarre. If it is possible to find that then it is possible to find m and u separately. (The same need not be true if asked for mu.). So why not ask for m and u?

    In terms of m and u, we know everything about the initial impact: the velocity, location and angle of incidence. (You can use u and the launch angle to find the height of the impact.). Thereafter, we can know the position of each body in terms of m, u and t. That they should come together in that specific relationship at the same moment that the sphere hits the ground again gives three equations, so in principle we can find m, u and t.

    It is not possible to use the final piece of information, the angle of m's path after second impact, without making some assumption about whether the ball bounces just before or just after the second impact with m. Either way, it now makes the problem overspecified. It almost certainly will lead to a contradiction.
     
    Last edited: Apr 6, 2017
  21. Apr 6, 2017 #20
    sir @haruspex me and @BvU believe mass of point mass and ball are the same if you may please take a look at my calculations in post #6 and tell us if we are correct thanks

    and i so far only managed to get 3 equations namely momentum in x and momentum in y after first collision and finally conservation energy all these three collectively could only eliminate one variable please tell me what is the other equation i need to use
    thanks!!
     
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