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Homework Help: Impulse Baseball problem

  1. Jul 14, 2013 #1
    Hello, I am taking physics 12U through correspondence. I am having issues with this problem. I don't know why but I feel like I made a mistake in this. Can you guys check this question and verify my results.

    Thanks, very much. :smile:

    1. The problem statement, all variables and given/known data

    A baseball with a mass of 0.152 kg is moving horizontally at 32.0 m/s [E], when it is struck by a bat for 0.00200 seconds. The velocity of the ball just after the collision is 52.0 m/s [W 20° N].

    a) Find the impulse experienced by the ball. (6 marks)​

    b) Find the average net force of the ball. (2 marks)​


    [itex]m_{} = 0.152 kg[/itex]​
    [itex]v_{1} = 32.0 m/s [E][/itex]​
    [itex]t{} = 0.00200 s[/itex]​
    [itex]v_{2} = 52 m/s [W 20° N][/itex]​

    2. Relevant equations


    3. The attempt at a solution


    Break down vectors into components

    solve for components

    [itex]\vec{P}_{x}=0.152(-45.033-32)=-11.7 Ns[/itex]​
    [itex]\vec{P}_{y}=0.152(26-0)=39.52 Ns[/itex]​

    [itex]P=\sqrt{(-11.7)^{2}+(39.52)^{2}}=41.2 Ns[/itex]​

    [itex]∴\vec{ΔP}=41.2 Ns [W 73° N][/itex]​

    [itex]F_{NET}Δt=41.2 Ns[/itex]​
    [itex]\vec{F}_{NET}=\frac{41.2}{0.002}=20,609 N [W 73° N] [/itex]​
  2. jcsd
  3. Jul 14, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Check your arithmetic here.
  4. Jul 14, 2013 #3
    Thanks for pointing that out. I have corrected that error.
    Is there anything else wrong with the solution?

    solve for components

    [itex]\vec{P}_{x}=0.152(-45.033-32)=-11.7 Ns[/itex]​
    [itex]\vec{P}_{y}=0.152(26-0)=3.952 Ns[/itex]​

    [itex]P=\sqrt{(-11.7)^{2}+(3.952)^{2}}=12.4 Ns[/itex]​

    [itex]∴\vec{ΔP}=12.4 Ns [W 19° N][/itex]​

    [itex]F_{NET}Δt=12.4 Ns[/itex]​
    [itex]\vec{F}_{NET}=\frac{12.4}{0.002}=6179 N [W 19° N] [/itex]​
    Last edited: Jul 14, 2013
  5. Jul 14, 2013 #4

    Doc Al

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    Staff: Mentor

    Check your computation of those components of the final velocity.
  6. Jul 14, 2013 #5
    Wow, I can't believe I did that. I used 30 degrees instead of 20 there. hah!
    I need to pay attention when I input the values into the calculator.

    I have made the changes. Please let me know if you see anything else wrong with it. particularly part b) I don't know if that is how it is supposed to be calculated.

    solve for components

    [itex]\vec{P}_{x}=0.152(-48.86-32)=-12.29 Ns[/itex]​
    [itex]\vec{P}_{y}=0.152(17.79-0)=2.703 Ns[/itex]​

    [itex]P=\sqrt{(-12.29)^{2}+(2.703)^{2}}=12.58 Ns[/itex]​

    [itex]∴\vec{ΔP}=12.6 Ns [W 78° N][/itex]​

    [itex]F_{NET}Δt=12.6 Ns[/itex]​
    [itex]\vec{F}_{NET}=\frac{12.6}{0.002}=6300 N [W 78° N] [/itex]​
  7. Jul 14, 2013 #6

    Doc Al

    User Avatar

    Staff: Mentor

    I think you mixed up your components when calculating the angle.

    Other than that, your solution looks good. (Since the force of the bat is so much greater than the weight of the ball, it's OK to ignore the weight of ball in calculating the net force from the bat.)
  8. Jul 14, 2013 #7
    Thanks, I am going to get some sleep and then do the rest of the questions. I really botched that lol.
    I really appreciate your help.
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