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Balls connected by string--momentum problem

  1. Oct 24, 2014 #1
    1. The problem statement, all variables and given/known data
    Two identical balls are interconnected with a massless and inextensible thread. The system is in gravity free space with the thread just taut. Each ball is imparted a velocity v, one towards the other ball and the other perpendicular to the first, at t=0. Then,

    (A)The thread will become taut at t=L/v
    (B)the thread will become taut at some time t<(L/v)
    (C)The thread will always remain taut for t>(L/v)
    (D)The kinetic energy of the system will always remain mv^2.

    The correct options are (A) and (C).

    2. Relevant equations
    m1u1 + m2u2 = m1v1 + m2v2
    KE= 1/2 mv^2

    3. The attempt at a solution
    There is no collision!!! How do I even start?
     
  2. jcsd
  3. Oct 24, 2014 #2

    PeroK

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    Start by using some simple numbers for L and v and drawing a diagram of where the balls are after 1, 2 and 3 secs etc. e.g L = 3m and v = 1m/s.
     
  4. Oct 24, 2014 #3
    That's not the problem. I get how the situation will look like when the string becomes taut. The string will form sort of a diagonal.
     
  5. Oct 24, 2014 #4

    PeroK

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    Maybe that's not correct. Maybe it isn't a sort of diagonal.
     
  6. Oct 24, 2014 #5
    Oh ok. Then I don't know how they'll move. How do I draw a diagram at different instants of time? I don't know what their velocities are. I figured I could use momentum conservation and velocity of centre of mass to figure out their velocities, but I can't.
     
  7. Oct 24, 2014 #6

    PeroK

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    Some basic trigonometry shows that the string becomes loose when the balls start to move. If you map the constant motion of both balls you will see when the string becomes taut again. After that conservation of momentum will be very important to see what happens.
     
  8. Oct 24, 2014 #7
    I tried mapping it out. I think the string will become taut when the two balls lie in a straight line, one above the other. The time taken to get there will be L/v, as the ball with the horizontal acceleration has velocity v. So at t=L/v, it becomes taut, and I'm pretty sure it'll stay that way. However, how do I conserve momentum in order to figure out the velocities of the two particles?
     
  9. Oct 24, 2014 #8

    PeroK

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    Remember that momentum is conserved in each direction.
     
  10. Oct 24, 2014 #9

    haruspex

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    It may help to take the initial motion of one of the balls as the frame of reference.
    When the string becomes taut again, that is a kind of collision.
     
  11. Oct 24, 2014 #10
    So I used conservation of momentum along x and y directions, assuming that after the "collision " both the balls move with the same velocities in both x and y directions. So final velocity of each ball in x and y directions is v/2 so the resultant final velocity of each ball is v/root 2 , so the final kinetic energy is 2(1/2mv^2/2) which is clearly not equal to mv^2. Hence option (d) is incorrect. Thank you PeroK and haruspex :)
     
  12. Oct 25, 2014 #11

    haruspex

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    But they won't; the system will have angular momentum.
     
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