Please turn on javascript in your browser to play chess.
Posers and Puzzles

Posers and Puzzles

  1. 11 Oct '08 23:20
    http://img2.tapuz.co.il/forums/1_122429217.pdf
  2. Standard member PBE6
    Bananarama
    16 Oct '08 17:57
    Originally posted by David113
    http://img2.tapuz.co.il/forums/1_122429217.pdf
    Interesting! From the link:

    Two bodies of masses M and m are placed on a slippery surface next to a vertical wall. At some point m is given some velocity v towards M. Find an expression for the number of times the two bodies will collide.

    All collisions in the problem are completely elastic (both the collisions between m and M and the collisions between M and the wall). Note that the two bodies may have any masses (i.e. not necessarily m < M).

    Sketch of figure:

    |
    |___M__m_
  3. Standard member PBE6
    Bananarama
    17 Oct '08 17:57
    Going to redo the sketch to make it a bit easier to work with:

    ___1___2___|

    Since all collisions are elastic, the following velocity formulas apply:

    v1 = (u1(m1-m2) + 2m2u2)/(m1+m2)
    v2 = (u2(m2-m1) + 2m1u1)/(m1+m2)

    where: m1,m2 are the masses, u1,u2 are the initial velocities, and v1,v2 are the final velocities.

    The initial velocity of mass 2 is 0, so we can simplify the above expressions to get the following:

    v1 = u1(m1-m2)/(m1+m2)
    v2 = 2m1u1/(m1+m2)

    There are 3 cases here:

    Case 1: m1=m2

    If both masses are the same, the velocities after the first collision are:

    v1 = 0
    v2 = u1

    In other words, there is a complete trade of momentum between the two masses. After rebounding off the wall, the velocity of m2 gets reversed so we have:

    v1 = 0
    v2 = -u1

    After the second collision, we have:

    v1 = -u1
    v2 = 0

    And m2 is stopped dead in its tracks again. Therefore, when both masses are the same there will only be 2 collisions.

    Case 2: m2>m1

    If m2 is greater than m1, the velocities after the first collision are:

    v1 = u1(m1-m2)/(m1+m2)
    v2 = 2m1u1/(m1+m2)

    Since m2>m1, (m1-m2) is negative which means m1 reverses direction. Comparing the magnitude of these velocities, we have:

    |v2|/|v1| = |2m1u1|/|u1(m1-m2)| = |2m1/(m1-m2)| = 2m1/(m2-m1)

    There are two sub-cases here. Letting |v2|/|v1| = 1 to find the point where both velocities are the same, we find:

    2m1/(m2-m1) = 1
    2m1 = m2-m1
    3m1 = m2
    m2/m1 = 3

    Case 2(a)

    m2 is at least 3 times as massive as m1. Plugging this back into the equation above, we see that v2 is less than (or equal to) v1. Therefore, m2 will never catch up with m1, so there is only 1 collision.

    Case 2(b)

    m2 is greater than m1, but less than 3 times as massive. Plugging this back into the equation above, we see that v2 is greater than v1. Therefore, m2 will eventually catch up with m1. My first guess is to say that there will only be 2 collisions total, but I'll have to double-check (more work on this in the next post!).

    Case 3: m1>m2

    If m1 is greater than m2, the velocities after the first collision are:

    v1 = u1(m1-m2)/(m1+m2)
    v2 = 2m1u1/(m1+m2)

    Since m1>m2, (m1-m2) is positive which means m1 continues in the same direction. Comparing the magnitude of these velocities, we have:

    |v2|/|v1| = |2m1u1|/|u1(m1-m2)| = |2m1/(m1-m2)| = 2m1/(m1-m2)

    This is strictly greater than 1, which means m2 will increase in speed after each collision (I think...need to double check that). There will be a finite number of collision until eventually the momentum from the quick moving m2 reverses the direction of m1, after which m2 should start slowing down resulting in a finite number of collisions akin to Case 2 (small object hitting a large object).

    As in Case 2, when I get more time I'll double check the math and try to find an expression for the number of collisions in each unanswered case.
  4. 18 Oct '08 01:18 / 5 edits
    The singular variable of interest in this puzzle is the ratio of the inside still object to the outside moving object. I refer to this ratio of still ball to moving ball as k.

    And I have determined in my examination of this little puzzler that if k is at least 1, but less than 3, there are 2 collisions, one coming, one going.

    At k=3, the moving ball reflects off the still ball the same velocity that the still ball gains, and thus the two balls end up staying a constant distance apart, once the still ball reflects off the wall.

    However, I was interested in finding the point where the follwing occurs.

    The outside moving ball is bigger than the inside static one (k < 1).

    1) Big ball hits little ball, sending it flying, but still moving forward.
    2) Little ball reflects back at big ball.
    3) Both hit and reflect off each other, and end up traveling at the same speed in opposite directions.

    In other words, where is the 2 collision border for k?

    The answer seems to be the solution to k^2 - 10k + 5 = 0, with the only solution in applicable range being 5 - 2*sqrt(5), a bit more than 1/2. (0.527864...)

    Between this value and less than 3, you have 2 collisions.

    Below this value, you will see a third collision as the static ball catches up to its bigger outer cousin again.

    If you were to look for where the fourth collision starts to occur, I suspect you would have to solve a cubic equation.

    Or perhaps the answer is a bit cleaner than it seems right now..
  5. 18 Oct '08 02:05
    Also found the point where the second collision stops the moving ball, assuming it is bigger.

    k = 3 - 2*sqrt(2) or about 0.172

    Been pondering what condition must be met for the inside ball not to be able to catch the outside ball, and as near as I can figure, it's some relationship between percentage of mass in each ball and overall momentum of the system...