What Is the Impulse Imparted to a Soccer Ball When 'Heading' It?

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SUMMARY

The discussion focuses on calculating the impulse imparted to a soccer ball when 'heading' it, specifically a size 5 ball with a mass of 0.446 kg, deflected at an angle of 31.0° and maintaining a speed of 10.40 m/s. The impulse is defined as the change in momentum, which requires understanding vector components due to the change in direction. The initial classroom calculations yielded incorrect results, prompting a reevaluation of the method to find the change in velocity using vector subtraction, leading to a clearer understanding of the impulse calculation.

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Becca93
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Homework Statement

Recent studies have raised concern about 'heading' in youth soccer (i.e., hitting the ball with the head). A soccer player 'heads' a size 5 ball deflecting it by 31.0°, and keeps its speed of 10.40 m/s constant. A size 5 ball has a mass of approximately 0.446 kg. What is the magnitude of the impulse which the player must impart to the ball?

After getting this incorrect, I was given the hint: "The impulse is the momentum imparted to the ball, and it changes the ball's momentum. Remember that momentum is a vector."The attempt at a solution

My prof has done this question in class a number of times, however the answers he came to in class are not being accepted as correct answers.

The way we did it in class is as follows:

I = fΔt = Δp
I = (Ʃp)f - (Ʃp)i
I = (Ʃmv)f - (Ʃmv)i
I = m(mf-mi) = mΔv

"Since the velocity magnitude is the same before and after collision, only the change in direction contributes to the change in velocity."

Δv = v

10.4cos31 = 8.91 m/s

Original classroom answer:

I = (.446)(8.91)
I = 30972 N*s

Revised classroom answer:

|Δv| = (8.9 - 10.4)

I = Δp = mΔv = m(v2 - v1) = m(10.4mgcosθ - 10.4)
I = m(8.9 - 10.4)
I = -0.669As I said, both of these answers are incorrect. Can anyone help explain what is wrong and what process needs to be followed to get the right answer? The question feels like it should be simple, but I just can't seem to spot what where my prof went wrong explaining it to the class.
 
Last edited:
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You must find the change in velocity by subtracting the velocity vectors. Take the original velocity direction as the +x axis. Now find the components of the final velocity vector. Subtract to find the components of the change in velocity vector, then find its magnitude. (You can subtract the vectors any way you like; using components is just one way.)
 
Hi Becca93! :smile:

Since your speed before equals your speed after, the corresponding vectors span an isosceles triangle with an angle of 31 degrees in between.
Do you know how to calculate the length of the third side?
 
Doc Al said:
You must find the change in velocity by subtracting the velocity vectors. Take the original velocity direction as the +x axis. Now find the components of the final velocity vector. Subtract to find the components of the change in velocity vector, then find its magnitude. (You can subtract the vectors any way you like; using components is just one way.)

Okay, I think I understand how to do that now. My prof and I must have misunderstood the question. Thank you!

I like Serena said:
Hi Becca93! :smile:

Since your speed before equals your speed after, the corresponding vectors span an isosceles triangle with an angle of 31 degrees in between.
Do you know how to calculate the length of the third side?

Got it! Thanks! I don't know why it didn't register to me that since it was an isosceles triangle I could use another method to find the base. Thank you.
 

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