Need help with momentum

  • Thread starter noname1
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  • #1
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The figure below gives an overhead view of the path taken by a 0.165 kg cue ball as it bounces from the rail of a pool table. The ball's initial speed is 2.00 m/s, and the angle θ1 = 30.0°. The bounce reverses the y component of the ball's velocity but does not alter the x component.

(b) What is the change in the ball's linear momentum in unit-vector notation? (The fact that the ball rolls is irrelevant to the question.)


I tried solving the one that wrong by 2*0.165*2*cos(30) = -5.72j but when i submit the answer is saying its wrong

what am i doing wrong
 

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  • #2
Delphi51
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Looks okay to me.
 
  • #3
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Check your calculation, you seem to be out by an order of magnitude.

[tex]2\cdot 0.165\cdot 2\cdot \cos \left( 30 \right)=0.5716[/tex]
 
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