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Momentum of soccer ball

  1. Aug 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball with mass 100 gr is kicked with angle -30o to the ground and has speed 20 ms-1. Then, the ball is kicked and reversed with angle 30o and speed 25 ms-1. The player touches the ball for 0.1 s. Find:
    (i) Change in momentum
    (ii) Force by the player
    (iii) Work by the player
    (iv) Power exerted on the ball


    2. Relevant equations
    p = momentum = mv = F.t
    P = power = W/t
    Work = change in kinetic energy

    3. The attempt at a solution
    I'm not very sure about the question at part "the ball is kicked and reversed with angle 30o". I tried to solve this problem with assumption that the ball is kicked back in a straight line.

    (i) change in momentum = mv-mu = 0.1 (25 - (-20) ) = 4.5 Ns

    (ii) [tex]F=\frac{p}{t}=\frac{4.5}{0.1}=45N[/tex]

    (iii) W = 1/2 mv2 - 1/2 mu2 = 11.25 J

    (iv) P = W/t = 11.25 / 0.1 = 112.5 Watt

    Do I get it right ?

    Thanks
     
  2. jcsd
  3. Aug 28, 2009 #2

    Hootenanny

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    Re: Momentum

    I would read this statement as "the ball is initially travelling with speed 20m/s at an angle of -30 degrees. The ball is then kicked and travels at an angle of 30 degrees and speed 25m/s".

    In this case, the motion is not rectilinear.
     
  4. Aug 28, 2009 #3
    Re: Momentum

    Hi Hootenanny

    So you mean that the motion of the ball is like "V" letter?

    Thanks
     
  5. Aug 28, 2009 #4

    Hootenanny

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    Re: Momentum

    Technically, a sidewards "V" since the angles are given with respect to the ground, but yes I would say that the trajectory of the ball is V-shaped.
     
  6. Aug 28, 2009 #5
    Re: Momentum

    Hi Hootenanny

    Ok so I've tried new solution.

    (i) Because the trajectory of the ball is V-shaped, I broke the velocity onto two components.

    a. When the ball comes:
    *Vx = 20 cos 30o to the right
    *Vy = 20 sin 30o downward

    b. When the ball kicked:
    *Vx = 25 cos 30o to the right
    *Vy = 25 sin 30o upward

    Px = 25 cos 30o-20 cos 30o=5 cos 30o
    Py = 25 sin 30o - (-20 sin 30o) = 45 sin 30o

    Change in momentum = [tex]\sqrt{(P_x)^2+(P_y)^2} = 5\sqrt{21}\;Ns[/tex]


    (ii) [tex]F=\frac{p}{t}=\frac{5\sqrt{21}}{0.1}=50\sqrt{21} \;N[/tex]

    (iii) W = 1/2 mv2 - 1/2 mu2 = 11.25 J

    (iv) P = W/t = 11.25 / 0.1 = 112.5 Watt

    Do I get it right?

    Thanks
     
    Last edited: Aug 28, 2009
  7. Aug 28, 2009 #6

    Hootenanny

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    Re: Momentum

    I haven't checked your arithmetic, but your method looks fine two.

    One, somewhat minor point: since we are dealing with planar motion, both the change in momentum and the force should be vector quantities.
     
  8. Aug 28, 2009 #7
    Re: Momentum

    Hi Hootenanny

    I've re-checked my answer and I have doubt at (ii). The question is asking about the force by the player, so should we only consider the momentum when the ball kicked with speed 25 ms-1 because the player doesn't give any forces when the ball comes?

    Thanks
     
  9. Aug 29, 2009 #8

    Hootenanny

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    Re: Momentum

    Newton's second law states that the net external force acting on a body is equal to the rate of change of the momentum. Therefore, you need to consider the change in momentum, not just the out-going momentum.
     
  10. Aug 29, 2009 #9
    Re: Momentum

    Hi Hootenanny

    Ok now I get it.

    Thanks a lot for your help :)
     
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