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    Ok...I'm at a complete loss with this question...there are no diagrams supplied or anything..this is all ur given.....

    "A 5 kg ball is dropped from a height of 12m above one end of a uniform bar that pivots at its centre. The bar has mass 8kg and is 4m in length. At the other end of the bar sits another 5kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision. How high will the other ball go after the collision?"

    The answer given is 1.87 m so if any1 can get this answer and explain where it comes from it would b greatly appreciated.

    Thanks in advance for any help.
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    Find

    (1) the moment of inertia of the bar with the two balls,
    (2) the speed with which the first ball hits the bar,
    (3) the angular impulse that the first ball exerts on the bar,
    (4) the angular speed of the bar just after the first ball hits,
    (5) the speed of the second ball just after the first ball hits,
    (6) the answer!
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    (Original post by Jonny W)
    Find

    (1) the moment of inertia of the bar with the two balls,
    (2) the speed with which the first ball hits the bar,
    (3) the angular impulse that the first ball exerts on the bar,
    (4) the angular speed of the bar just after the first ball hits,
    (5) the speed of the second ball just after the first ball hits,
    (6) the answer!
    Thanks a lot....I was actually very nearly there.... I had just forgotten when calculating the moment of inertia of the rod to add the moment of inertia of the 2 balls.

    Neway thanks again for your help.
 
 
 
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