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Tennis ball hit by racket (momentum problem I think)

  1. Dec 22, 2011 #1
    1. The problem statement, all variables and given/known data
    A tennis ball of mass 0.022 kg is moving at 3.1 m/s at an angle of 222° to the horizontal. It is struck by a racket which exerts a force on it of 72t - 442 t^2 N for 1/10 of a second at an angle 32° to the horizontal. Find the final velocity of the tennis ball (Express answer using i and j unit vectors)

    2. Relevant equations

    So, I guess this is a momentum problem...?)

    p = mv (momentum)
    Δp = pf - pi

    3. The attempt at a solution

    My attempt here was the next:

    1.- Integrate the force exerted by racket, using tf = 0.1 s and ti = 0
    Δp = ∫ F dt = 0.218 Km m / s

    2.- Divide Δp in x and y components, by multiplying 0.218 by cos 222° and sin 222°, respectively.

    3.- Obtain ball's momentum, in components also:
    px initial = (0.022 kg)(3.1 m/s) cos 32°
    py initial = (0.022 kg)(3.1 m/s) sin 32°

    4.- Using Δp = pfinal - pinitial, solve for pfinal, which is (mass)(vel. final), then solve for
    vel. final.

    v final = (Δp + p initial) / mass = (6.05 i + 3.2 j) m/s

    That's what I got... They said integral calculus was not needed (even though it's a pretty easy integral) for this course I'm taking lol, but after some thinking this is the only way I think this can be solved.

    Any comments? Any other way to solve it w/o integral calculus?

  2. jcsd
  3. Dec 22, 2011 #2


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    hmm... the force is meant to be (72t - 442 t^2) N right? But this clearly doesn't have the physical dimensions of force. Unless t is a dimensionless parameter. (for example, time per second). What does the question say, specifically?

    edit: When I said "for example, time per second", I meant the total elapsed time divided by the time for one second to elapse. In other words, a ratio of time intervals. This is just one example, so for your equation, it may be a different ratio. I'm guessing there is more information given in the question?
    Last edited: Dec 22, 2011
  4. Dec 22, 2011 #3

    I know! I've never seen force expressed like this hehe... It has NEWTONS, so the dimensions are correct I guess. Those t's mean that the force can vary with time maybe? You are asked to find the final velocity on the ball after that force was exerted.
    All the information is there, there's nothing more :P hehe
  5. Dec 23, 2011 #4


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    The question does not give enough information. But I think I can guess what t is meant to be.
    the equation for force is: (72t - 442 t^2) N And we want it to have dimensions of Newtons, so this tells you something important about the dimensions of t. From this, what do you guess t is meant to be?
  6. Dec 23, 2011 #5
    I dont quite get it... We already have newtons, but we wanna get newtons? hehe Im guessing you could also plug in TIME where those t's are and get the total force that was exerted during that time interval ?
  7. Dec 24, 2011 #6


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    You interpreted the badly worded problem correctly, and the method of solution is correct, but there are mistakes in it.

    It is said that the force is in newtons. It would be essential to give information about t, like "t is time in seconds, starting when the ball hits the racket". As the timespan of the interaction is given as 0.1 s we can conclude that t is time in seconds, elapsed from the beginning of the interaction. The coefficients of t and t2 must have dimensions and units, the correct form of F(t) would be F(t)=72(Ns-1)t-442 (Ns-2)t2.

    The principle is correct, but the result is not. Check the unit. K (Kelvin) is the unit of temperature. Use kg for mass. And the dimension of Δp is [mass][length][time]-1. The numerical value is also inaccurate.

    Check the text of the problem. You mixed the directions.

  8. Dec 24, 2011 #7


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    If you put in time where those t's are, then we certainly won't have Newtons. You need to remember that time is another physical dimension. so time is 0.1seconds or (1/600)minutes e.t.c.

    So we know the total time is 0.1seconds, but we want a dimensionless number. I think your teacher is hoping that you use your intuition to guess what that dimensionless number is meant to be. What was your guess?
  9. Dec 27, 2011 #8

    lolll yeah, it's supposed to be Kg in there, not Kelvin. Just a typing mistake. But you also mentioned the numerical value is incorrect...

    Well, like I said, I wasn't supposed to be doing integral calculus in this subject, but here I am attempting an integral... I had the limits of the definite integral be tf = 0.1 s and ti = 0 and obtained Δp = 0.218 Kg m/s

    What did you get as an answer? Also, if there's another method that doesn't require calculus, I'd appreciate a hint hehe
  10. Dec 27, 2011 #9


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    I think you need calculus. ∫Fdt==72t2/2-442 t3/3 at t=0.1 equals to 0.213 kg m/s.

  11. Dec 27, 2011 #10
    Guess I'll have to learn those integrals before time hehe thanks a bunch for the help!
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