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Force exerted via collision

  1. May 23, 2010 #1
    1. The problem statement, all variables and given/known data

    A 100 g snooker ball with speed 50 cm/s strikes a cushion at an angle of 45° and bounces back off the same speed and angle. If the ball is in contact with the cushion for 0.05 s, what is the average force exerted by the cushion on the ball?

    3. The attempt at a solution

    I get final momentum=initial momentum because the angle, mass and the speed are the same before and after.

    [tex]p_i=(0.1)(0.5 cos 45) = p_f[/tex]

    The change in momentum is 0.

    So, the impulse is equal to [tex]I=\Delta \vec{p}[/tex]. To find the force I think we can use the formula

    [tex]\vec{I}=\sum \vec{F} \Delta t[/tex]

    But how can I find force when the change in momentum is zero? Is there something wrong with my strategy or did I make some mistakes?
     
  2. jcsd
  3. May 23, 2010 #2
    You need to approach these questions systematically
    ie.
    Vi=50cm/s* cos45i+50cm/s* sin45j
    vf=-50cm/s*cos45+50cm/s* sin45j
    j=pf-pi
    =0.5cos45i*2*100i+0.5cos45*2*100j as cos 45=sin 45
    =0.0707i+0.0707j
    J=0.1=Ft
    F=0.05/0.1
    F=0.5N




    Did you understand the other momentum problem from before??
     
  4. May 23, 2010 #3
    I think you made a mistake when subtracting the initial and final momenta
     
  5. May 23, 2010 #4
    whys that what answer did you get??? the plus is there as its a double negative
     
  6. May 23, 2010 #5
    you made a mistake when subtracting the j-components for the impulse. If you notice, it's equal for both the final and initial momenta so the change in the j-direction is zero.
     
  7. May 23, 2010 #6
    yep your right,,,, . thats my bad sorry but the basic principals are still right. agree??
     
  8. May 23, 2010 #7
    Yes, that's right.
     
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