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thushananth01

  • 2 years ago

A ball of mass 50g falls from a height of 2.0m and rebounds to a height of 1.2m. How much kinetic energy is lost on impact?

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  1. thushananth01
    • 2 years ago
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    I need your help :) @Callisto @Vincent-Lyon.Fr @ganeshie8

  2. Callisto
    • 2 years ago
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    |dw:1354449116877:dw|

  3. thushananth01
    • 2 years ago
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    yes, in this case the loss of kinetic energy will be equal to the gain in potential energy after impact

  4. Callisto
    • 2 years ago
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    At position 1: Total energy = PE = \(mgh_1\) At position 2: Total energy = (KE + energy loss) = total energy at position 1 So far so good?

  5. thushananth01
    • 2 years ago
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    yes i got my answer as 0.5886 which is equal to PE after impact

  6. thushananth01
    • 2 years ago
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    (Ep+Ek) at 1 = (Ep + Ek) at 2 mgh = 1/2m(V,)^2 Kinetic energy at the beginning = mgh (Ep+Ek) at 2 = (Ep + Ek) at 3 mgh = mgh1 + 1/2m(v,,)^2 so Loss in kinetic energy = mgh1?

  7. Callisto
    • 2 years ago
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    Hmmm.. At position 3: Total energy = PE = \(mgh_2\) = KE at position 2 So, KE at position 2 = 0.05 x 9.81x 1.2 And from (KE + energy loss) = total energy at position 1 0.05 x 9.81x 1.2 + Energy loss = 0.05 x 9.81 x 2.0 So, you can get the energy loss

  8. thushananth01
    • 2 years ago
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    wht is the energy loss?

  9. Callisto
    • 2 years ago
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    |dw:1354449520295:dw|

  10. Callisto
    • 2 years ago
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    What do you think?

  11. thushananth01
    • 2 years ago
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    how is KE at position3 = 0?

  12. Callisto
    • 2 years ago
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    Velocity at the max point is 0, so KE=0 :|

  13. Shadowys
    • 2 years ago
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    note: in perfect bounce, the height of rebound is equal to the original height of drop. So, the loss in potential energy =loss in kinetic energy.(For quick answer)

  14. Callisto
    • 2 years ago
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    Oh... Is 1.2m the max. height of the rebound?

  15. thushananth01
    • 2 years ago
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    tht wsnt given in the question?

  16. thushananth01
    • 2 years ago
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    @Shadowys can u work it out using equations?

  17. Shadowys
    • 2 years ago
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    eh? Since it isn't a perfect rebound, loss of kinetic energy = loss of potential energy =\(mg(h-h_o)\) This is the potential energy is turned completely to kinetic energy during the drop, and vice versa during the rebound after hitting the ground.

  18. thushananth01
    • 2 years ago
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    @Shadowys what ws your answer? can u calculate and let me know

  19. thushananth01
    • 2 years ago
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    @Callisto Kinetic energy at lowest point = Loss in potential energy 1/2mv^2 = mgh Ek = 0.05*9.81*2=0.981 Ep + Ek at 2 = Ep + Ek at 3 0.981 = 0.05*9.81*1.2 + Ek Ek at 3 = 0.3924J there loss in kinetic energy = K.E before impact - K.E after impact 0.981-0.3924= 0.5886

  20. Shadowys
    • 2 years ago
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    .392J

  21. Shadowys
    • 2 years ago
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    why would the Kinetic energy at lowest point = Loss in potential energy?

  22. Callisto
    • 2 years ago
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    Kinetic energy at lowest point =/= Loss in potential energy Kinetic energy at lowest point = Loss in potential energy - energy loss due to impact

  23. thushananth01
    • 2 years ago
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    Ep + Ek at 1 = Ep + Ek at 2 mgh + 0 = 0 + 1.2mv^2 mgh = Ek

  24. Shadowys
    • 2 years ago
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    mechanical energy is not really conserved since you've loss of rebound height, right?

  25. thushananth01
    • 2 years ago
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    ok i understood....is it some thing similar like this. Ep + EK at A = Ep + Ek at B + W.d against friction?

  26. Shadowys
    • 2 years ago
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    yup! so you see i don't use that reasoning because of that work against friction/etc thingy. Very complicated.

  27. thushananth01
    • 2 years ago
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    one more question K.E at position3 = 0?

  28. Shadowys
    • 2 years ago
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    yeah. As Callisto said, that's when the velocity becomes 0, i.e. it stops.

  29. thushananth01
    • 2 years ago
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    thanks:)

  30. Shadowys
    • 2 years ago
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    you're welcome :)

  31. Vincent-Lyon.Fr
    • 2 years ago
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    Final mechanical energy = Initial mechanical energy - energy loss. Hence Eloss = mg(h1-h2) = 0.40 J with g = 10 m/s²

  32. thushananth01
    • 2 years ago
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    correct!

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