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    What did everyone get for question 2 ( I think) about the speed at which the box falls?
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    Omg I spent so long on that one but I still couldn't do it :/
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    I got the square root of 30 - 5. something I can't remember right now
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    (Original post by Leah2706)
    Omg I spent so long on that one but I still couldn't do it :/
    It was really confusing, I ended up working out velocity through KE = 1/2 mv^2 and got 5.5m/s but i doubt it was right.
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    (Original post by bonnie_x)
    I got the square root of 30 - 5. something I can't remember right now
    Same!!
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    ke = gpe lost
    ke = 0.5*m*v^2
    gpe lost = 11J (from last question)
    mass = 3.2kg
    11J = 0.5*3.2*v^2
    11 = 1.6*v^2
    11/1.6 = v^2
    √(11/1.6) = v
    2.62 = v

    I think that's right :/

    (sorry had to re-do all working to remember it)
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    Didn't the height change from 0.5m to 1.5m? I got the kinetic energy (GPE) as 48j
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    The answer was 5.48m/s. This is, of course, the same as the above poster: 5.5m/s. And, also after socializing after the exam to everybody I know -- I can certainly confirm the answer is 5.48m/s
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    TSR Support Team
    (Original post by Torticci)
    ke = gpe lost
    ke = 0.5*m*v^2
    gpe lost = 11J (from last question)
    mass = 3.2kg
    11J = 0.5*3.2*v^2
    11 = 1.6*v^2
    11/1.6 = v^2
    √(11/1.6) = v
    2.62 = v

    I think that's right :/

    (sorry had to re-do all working to remember it)
    You made exactly the same mistake I did (first other person to see do this haha)
    It gave a new mass and weight, and you had to calculate the GPE from these - it wasn't the same as from the previous question!
    However all your working is right so you should get a method mark (well I'm praying haha)
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    You had to work out the GPE and then the GPE is the same as kinetic energy and so you used the kinetic energy formula to work out the velocity squared and square root it. I completely blanked on this question though
 
 
 
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