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Collision problem cant figure it out

  1. Mar 23, 2007 #1
    Question: A ball of mass m is suspended from a string of length l. A lump of clay of mass 3m is thrown horizontally at the ball w/ a speed of v0 making a direct hit and sticking to it. Find the speed of the combined object right after the collision, and how high it rises.

    Answer: 3/4vo 9/32(v0*v0/g)

    the answer is given but i need to know how this problem is actually solved. Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 23, 2007 #2
    Have you tried to solve using conservation laws?
     
  4. Mar 23, 2007 #3
    the book i was given for this class shows barely anything at all in it. I dont really know where to begin. The only equation given in this entire chapter relating to a inelastic equation (which im assuming this collision is) would be>>>>>> m1v1+m2v2=(m1+m2)v' however...where does a distance come into this ie. lenth of the height? And how woud i actually utalize this equation when almost no information is given?
     
  5. Mar 23, 2007 #4
    First, use that equation (which is a special case of the law of conservation of momentum) to find the velocity the combined mass would be having immediately after collision. Once you have figured out the velocity, use the law of conservation of energy to find the height.
     
  6. Mar 23, 2007 #5
    hmm...how do i do that if no actual masses or velocities are given?

    v'= m1v1+m2v2/ total mass.....but i have no numbers, am i missing something?
     
  7. Mar 23, 2007 #6
    Not every problem comes with numbers involved with every quantity. As you can see from the answer, it is a multiple of v0. You know one mass is 3 times the other, and the ball was at rest, initially. Substitute these details into the equation.
     
  8. Mar 23, 2007 #7
    awesome i see what u mean. The momentum of the first object is 0 because its at rest. the total mass is obviously 4m therefore: 3m vo/ 4m... the ms cancel leaving 3/4 vo. Ok i get that now. However... what is the conservation of energy formula that i need to solve for the distance?
     
  9. Mar 23, 2007 #8
    Have you come across the terms 'kinetic energy' and 'potential energy' before?
     
  10. Mar 23, 2007 #9
    Yes but im confused as to how to actually set this up to work...how do i use the velocity that i got, or do i need to. Am i suppossed to use, delta K + delta Ug(ptoential gravitational force) = 0? And if so...how would i actually set this up?

    also... K= mgh
    U = -mgh
     
  11. Mar 23, 2007 #10
    Exactly.

    Kinetic energy is (1/2)mv2!!! and U = mgh.

    Delta {something} refers to change in the something, usually at two different times. You know what the velocity of the combined mass is after colission. Consider the point to be at the 'zero' height. Now can you solve for the height to which the mass will go to?
     
  12. Mar 23, 2007 #11
    Sry this is frustrating for me...wouldnt the velocity i solved for actually be the initial velocity, right after the collosion occured, and the final velocity be 0(because it will come to a stop briefly when it reaches its max height)? So im still confused as to how this is actually set up.

    1/2mv2'-1/2mv2 + mgh'-mgh = 0
    0 - 1/2m(3/4vo)2 + mgh' - mg(0)= 0
    -1/2m(3/4vo)2 + mgh'
    -1/2m(3/4vo)2/mg = -h



    this doesnt seem right...do u have any suggestions as to how to set the equation up?
     
  13. Mar 23, 2007 #12
    That is correct.

    Why doesn't it seem right? It's perfect*. You just have to simplify it to get it to look like it is given in your book.


    *-1/2m(3/4vo)2/mg = -h (If that 2 in bold means "squared," then it's correct). It is customary to denote powers by ^. Example: (3/4vo)^2
     
  14. Mar 23, 2007 #13
    2m(9/16)/mg( this is where im confused...g would be 9.8m/s^2?) if i multiply 4m(9.8) i get 39.2 m^2s^2

    none of this seems to reduce down to the answer in my book of 9/32 v0^2/g
     
    Last edited: Mar 23, 2007
  15. Mar 23, 2007 #14
    [tex]\frac{\frac{1}{2}m\left(\frac{9v_{o}^{2}}{16}\right)}{mg}[/tex]

    Does that make more sense?
     
    Last edited: Mar 23, 2007
  16. Mar 23, 2007 #15
    hmm...so i dont actually have to plug a mass(inthe denominator) or 9.8 m/s^2 into the formula?
     
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