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Easy kinematics

  1. Aug 1, 2008 #1
    Dunno if its a school level qn for u all , but its certainly not for me :D

    A ball freely falls from a height h onto an inclined plane forming an ange alpha with the horizon. Find the ratio of the distance between the points at which the jumping ball strikes the inclined plane first , second and third times after rebounding from falling from the given height.Consider the impacts betweeen the ball and the plane to be absolutely elastic.

    The answer is 1:2:3 and it kinda shocked me !!
  2. jcsd
  3. Aug 1, 2008 #2


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    It sure looks quite complicated. It seems as though I'd have to calculate the velocity vectors each time it bounces off the plane under the assumption that the ball rebounds the same angle from the normal to the incline plane and then re-calculate the next point of impact. There has to be a trick we can use somewhere, perhaps by energy considerations?
  4. Aug 1, 2008 #3
    i indeed did calculate all those velocity vectors , and what i got was a good messup of sin and cos alpha :D
  5. Aug 2, 2008 #4
    LOL i saw a similar questions in one of the mit ocw assignments. Its got to do with projectile motion.
    This is what i think but i might be wrong:
    When the ball strikes the inclined plane, it leaves the inclined plane at angle alpha from the horizon. Since the collision is elastic, K.E. is conserved and the velocity when it strikes is the velocity when it leaves.(correct me if i'm wrong i forgot my momentum stuff).
    Ok this means that mgh = K.E. = (mv^2)/r
    Solve for v and you'll get sqrt(rgh) = v.
    This is the speed at which it leaves the inclined plane.
    Now, what distance x and y does it move. From this, you can calculate how much does it move, diagonally on the inclined plane.

    I think that's the idea.
  6. Aug 2, 2008 #5
    question* 10char.
  7. Aug 2, 2008 #6


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    Well actually the problem isn't with how to do it It's just that it is very tedious to solve it in this manner. Is there an easier way?
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