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Homework Help: Finding Velocity and Impulse using Energy Considerations.

  1. Apr 16, 2013 #1
    My problem/issue has come up with part C of this question but it wont make sense unless I post parts A and B first and I would be grateful if you could check them whilst they are here :) .

    1. The problem statement, all variables and given/known data
    an object that has a mass of 800g drops from a height 2.5m. Once it has hit the ground it rises to a height of 1.8m .

    Part A: Find the speed at which the object hits the ground .
    Part B: Find the speed with which the objectrebounds from the ground
    Part C: Find the impulse the object receives from the ground.

    2. Relevant equations

    3. The attempt at a solution
    Part A:
    \frac{1}{2} mv^2 &=mgh \\
    \frac{1}{2} 0.4v^2 &=0.4 \times 0.4 \times 9.8 \times 2.5 \\
    0.2v^2 &=9.8 \\
    ∴v &=\sqrt{\frac{9.8}{0.2}}=7m/s

    Part B:
    \frac{1}{2} mv^2&=mgh \\
    \frac{1}{2} 0.4v^2&=0.4 \times 9.8 \times 1.8 \\
    0.2v^2&=7.056 \\
    ∴v&=\sqrt{\frac{7.056}{0.2}} =5.94m/s

    Part C:
    I=Δp&=mv-mu \\
    Δp&=(0.4 \times 7) - (0.4 \times 5.94) \\

    I think I got the final and initial velocities either the wrong way around or completely wrong altogether! A friend of mine has put the final velocity to be zero as he reckons it is when the object has bounced back into the air the highest it can go again. therefore being zero before speeding back down to earth.

    Any help or advice is appreciated :) .
    Last edited: Apr 16, 2013
  2. jcsd
  3. Apr 16, 2013 #2
    Your friend is thinking of the impulse on the ball as it rebounds, of which there is none. The impulse the problem is asking for occurs as the ball comes into contact with the ground surface.

    How did you get the final and initial velocities the wrong way? Positive impulse means it occurs in the object's direction of motion, and negative impulse means it acts against it.

    Btw in your calculations, it's ##5.94 m/s##, correct?
  4. Apr 16, 2013 #3
    You've also got both velocities as positive - one will be negative depending on the direction of your vertical axis.
  5. Apr 16, 2013 #4
    You found the magnitudes velocities correctly.

    When computing the impulse, though, you ignored their directions.
  6. Apr 16, 2013 #5
    Ah yeah, it does state at the beginning of the problem sheet that only the magnitude of the velocities are required. I should have mentioned that.

    If I do try and include the direction though, I cannot square root the fraction because it contains a negative, or even if I do the fraction and try to square root the answer is the same problem, is there a way around that?

    God knows, I probably went through them too fast think it was too easy and just made a simple mistake in determining the variables. So the magnitude of the impulse is correct its just that its in the opposite, minus, direction then?

    Ah yeah, that was just a typo, I have corrected it.

    Thanks peeps.
  7. Apr 16, 2013 #6
    The value v in the kinetic energy is speed not velocity, so it must be positive.
    When calculating the change in momentum, you are working with the velocity so you have to include the sign.
    Just choose either up or down as your positive direction (it doesn't matter which) and redo part (c)
  8. Apr 16, 2013 #7
    Assuming down to be positive is the below correct?

    Δp=mv-mu \\
    Δp=(0.4 \times 7)-(0.4 \times -5.94) = 5.176Ns \\

  9. Apr 16, 2013 #8
    Did you just edit this?
    I think the answer before was correct - you have the velocities the wrong way round now.
    (What sign do you expect the impulse to have?)
  10. Apr 16, 2013 #9
    Yeah I did, I thought I did but got confused, so the below was right?

    Δp=mv-mu \\
    Δp=(0.4 \times -5.94) - (0.4 \times 7)=-5.176Ns
  11. Apr 16, 2013 #10
    Looks good :)
    The impulse is negative as you should expect, i.e. it acts in the upward direction.
  12. Apr 16, 2013 #11
    The final velocity (immediately after the ball rebounds) is in the opposite direction to the initial velocity (before the ball rebounds). Taking up to be the positive sense and down to be negative, you just needed to include the direction of the final velocity, that's all.

    No problem.
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