Soccer 'Heading': Calculating Impulse of a Ball

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SUMMARY

The discussion centers on calculating the impulse imparted to a size 4 soccer ball during a heading maneuver in youth soccer. The ball, with a mass of 0.302 kg, is struck at a constant speed of 19.60 m/s and deflected by 38 degrees. The correct impulse calculated is 3.85 kg*m/s. Key insights emphasize the importance of treating momentum as a vector and the necessity of vector subtraction to determine the change in momentum.

PREREQUISITES
  • Understanding of momentum and impulse concepts
  • Familiarity with vector diagrams and vector addition/subtraction
  • Basic knowledge of physics equations, specifically p=mv
  • Ability to perform trigonometric calculations related to angles
NEXT STEPS
  • Study vector addition and subtraction in physics
  • Learn about impulse-momentum theorem applications
  • Explore the effects of angular deflection on momentum
  • Review case studies on head impacts in sports and their implications
USEFUL FOR

This discussion is beneficial for physics students, sports scientists, coaches, and anyone interested in the biomechanics of soccer, particularly regarding the safety and effects of heading the ball in youth sports.

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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 4 ball, deflecting it by 38.0 deg, but keeping its speed of 19.60 m/s constant. A size 4 ball has a mass of approximately 0.302 kg. What is the magnitude of the impulse which the player imparts to the ball?
CORRECT ANSWER: 3.85 kg*m/s

Homework Equations


p=mv
draw a vector diagram

The Attempt at a Solution


I drew a vector diagram with the vector p=mv=5.9 going in the +x direction. I then drew the deflection of 38 degrees going in the negative and positive direction.

Ok so I found the hypotenuse (the redirection) to be 7.48.

Then I thought it would be 7.48-5.9, but alas, nope.

Any help?
 
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nicknaq said:
bumpidity
The impulse on a given object is the change in momentum on that given object.

But don't forget, momentum is a vector and needs to be treated as such.
I drew a vector diagram with the vector p=mv=5.9 going in the +x direction. I then drew the deflection of 38 degrees going in the negative and positive direction.

Ok so I found the hypotenuse (the redirection) to be 7.48.
No, that's not quite right. The problem statement says, "...deflecting it by 38.0 deg, but keeping its speed of 19.60 m/s constant."

So the hypotenuse of momentum after the collision is still (0.302 [kg])(19.60 [m/s]). The hypotenuse is unchanged in magnitude.

However, what did change is the momentum vector's direction. So allow me to ask,
  • What is the momentum vector after the collision?
  • What is the momentum vector before the collision?
  • What is the difference between the two (i.e. the change in momentum)? [Hint: this involves subtracting vectors, not scalars. :wink:]
  • What is the magnitude of this change?
 

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