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Restitution

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    See attachment for question. I need help with part three, where I thought that sqrt(0.05) was the answer. The answer is 0.5, by the way.
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    (Original post by bbrain)
    See attachment for question. I need help with part three, where I thought that sqrt(0.05) was the answer. The answer is 0.5, by the way.
    By restitution, since the ground isn't moving:

    "velocity after impact" = -e ("velocity before impact")

    Post some working if it's not coming out.
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    The speed of B immediately before impact = 2 ms^-1
    The speed of B immediately after impact = sqrt(2^2-2*g*0.05) = sqrt(3)
    Then e is not 0.5

    Further help please.
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    (Original post by bbrain)
    The speed of B immediately after impact = sqrt(2^2-2*g*0.05) = sqrt(3)
    That's not correct, but I don't know what you're doing to get there.
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    So what should I do?
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    Knowing it reaches a height of 0.05m, you can either use conservation of energy, or v^2=u^2+2as.

    Since you're not telling me what you did use, I can't tell why it's incorrect.
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    Got it. Since v=2gh=2*10*0.05 =1, and the speed before impact =2, 1/2=0.5
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    (Original post by bbrain)
    Got it. Since v=2gh=2*10*0.05 =1, and the speed before impact =2, 1/2=0.5
    v^2=2gh

    Hence v^2=1 and hence v=1.

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Updated: May 23, 2012
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