Finding the Speed of a head hit by a soccer ball

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SUMMARY

The discussion focuses on calculating the difference in speed acquired by a soccer player's head when heading a ball, comparing a bare head to a helmeted head. The measured accelerations are 200 m/s² for the bare head and 80 m/s² for the helmeted head, both over a duration of 0.006 seconds. The calculated speeds are 1.2 m/s for the bare head and 0.48 m/s for the helmeted head, resulting in a speed difference of 0.57 m/s, which rounds to 0.56 m/s. The discussion emphasizes the importance of impulse duration in helmet design, as it spreads the impact over a longer time frame.

PREREQUISITES
  • Understanding of basic physics concepts such as acceleration and impulse
  • Familiarity with calculus, specifically integration for calculating area under curves
  • Knowledge of kinematic equations related to motion
  • Experience with interpreting graphical data in physics
NEXT STEPS
  • Study the principles of impulse and momentum in physics
  • Learn how to calculate the area under curves using integration techniques
  • Explore the physics of sports equipment design, particularly helmets
  • Investigate the effects of headgear on impact forces in sports
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Physics students, sports safety researchers, and engineers involved in sports equipment design will benefit from this discussion, particularly those interested in the dynamics of head impacts in soccer.

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Homework Statement


When a soccer ball is kicked toward a player and the player deflects the ball by "heading" it, the acceleration of the head during the collision can be significant. Figure 2-38 gives the measured acceleration a(t) of a soccer player's head for a bare head and a helmeted head, starting from rest. At time t = 7 .0 ms, what is the difference in the speed acquired by the bare head and the speed acquired by the helmetedhead?

Homework Equations

The Attempt at a Solution


I uploaded the picture of the graph and question also I was able to get an answer. I found the peak of the helmet graph to be 80 m/s^2 and .006s. I found the other peak from the bare graph to be 200 m/s^2 and .006s.
I multiplied 80 m/s^2 * .006s=.48 m/s and 200 m/s^2 *.006s =1.2 m/s
1.2 m/s- .48 m/s =.57 m/s
the answer of the problem is .56 m/s, I was wondering if I did the problem the right way?
 

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The final velocity is the area under each curve v = ∫a dt. You have a whole bunch of triangle and rectangle areas to calculate. Go for it.
 
As an aside, the graph completely misrepresents reality. The player intends to impart a specific impulse to the ball. If that is being achieved in both graphs then the helmet must be heavier than the head!
The real benefit of the helmet is the extent to which it spreads the impulse over a longer duration.
 

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