Calculating the magnitude of an impulse

  • Thread starter Nkele
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  • #1
Nkele
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Homework Statement


Aplayer bounces a 0.43-kg soccer ball off her head, changing
the velocity of the ball from
Vi = (8.8 m/s)x + (-2.3 m/s)y
to
Vf = (5.2 m/s)xN + (3.7 m/s)y
[/B]
. If the ball is in contact with the
player’s head for 6.7 ms, what are (a) the direction and (b) the
magnitude of the impulse delivered to the ball?

Will I be wrong if I use just change in velocities(ie excluding) in x-comp and y-comp to calculate direction of the impulse.

I did calculation which gave me theta of -59.03, and when using coordinates the theta falls in second quadrant.





Homework Equations


1. delta P(x comp)= Pf(x)-Pi(x)= -1.548
2. delta P(y comp)=Pf(y)-Pi(y)=2.58


The Attempt at a Solution


Using the solutions I calculated magnitude of impulse as 3.03 kg.m/s
 

Answers and Replies

  • #2
kuruman
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That looks about right.
 
  • #3
Nkele
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So is using just velocities o calculate magnitude acceptable
 
  • #4
kuruman
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So is using just velocities o calculate magnitude acceptable
If I understand correctly, you used the equation for the impulse, ##\vec J=m\Delta \vec v## to find the impulse in component form and obtained the magnitude using the Pythagorean theorem. Then you found the angle using the expression ##\theta = \arctan(J_y/J_x)##. That's how you should think of it and it is the correct method. Simplifying it to "using just velocities" is dangerous and could cause trouble next time you encounter something like this.
 
  • #5
Nkele
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Ok, thank you very much for clarification
 

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