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anonymous

  • one year ago

A billiard ball moving at 2.2 m/s strikes a second billard ball at rest and continues at 45 degrees from its original direction, while the other moves at -45 degrees from the same original direction. Both billiard balls have the same mass. Find the speed of the second billiard ball. 1.6 m/s, 3.1 m/s, 0.78 m/s, or 1.1 ,/s **not sure! thank you!

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  1. perl
    • one year ago
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    can we make a diagram of the situation

  2. anonymous
    • one year ago
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    ok, i am not quite sure how!

  3. perl
    • one year ago
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    |dw:1433018245578:dw|

  4. perl
    • one year ago
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    |dw:1433018360172:dw|

  5. anonymous
    • one year ago
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    yes:) what do we do from here?

  6. perl
    • one year ago
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    note that the collision is not 'head on' , thats why it goes off an angle

  7. anonymous
    • one year ago
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    yes:) so in calculating, would it be moving at the speed of 3.1 m/s?

  8. anonymous
    • one year ago
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    ok! what do i plug in for m? :/

  9. perl
    • one year ago
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    momentum = mass *velocity In elastic collisions, momentum is conserved, therefore "initial momentum = final momentum" . Also kinetic energy is conserved. initial x momentum = 2.2*m + 0*m initial y momentum = 0*m + 0*m final x momentum = m(vf1 cos 45) + m*vf2 cos(-45) final y momentum = m (vf1 sin 45)+ m (vf2 sin(-45)) 2.2 m = m*vf1 cos 45 + m*vf2 cos(-45) 0 = m (vf1 sin 45)+ m (vf2 sin(-45)) 1/2m*2.2^2 = 1/2 m* vf1^2 + 1/2* m*vf2^2

  10. perl
    • one year ago
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    we get a system of three equations, (we can cancel m out)

  11. perl
    • one year ago
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    |dw:1433019149286:dw|

  12. anonymous
    • one year ago
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    okay! and so we get left with vf1 and vf2? we are looking for vf2 becuase of the second billiard\?

  13. perl
    • one year ago
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    right

  14. anonymous
    • one year ago
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    how can i solve that value?

  15. perl
    • one year ago
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    momentum = mass *velocity In elastic collisions, momentum is conserved, therefore "initial momentum = final momentum" . Also kinetic energy is conserved. initial x momentum = 2.2*m initial y momentum = 0 final x momentum = m(vf1 cos 45) + m*vf2 cos(-45) final y momentum = m (vf1 sin 45)+ m (vf2 sin(-45)) 2.2 m = m*vf1 cos 45 + m*vf2 cos(-45) 0 = m (vf1 sin 45)+ m (vf2 sin(-45)) 1/2m*2.2^2 = 1/2 m* vf1^2 + 1/2* m*vf2^2 canceling out mass m 2.2 = vf1 cos 45 +vf2 cos(-45) 0 = (vf1 sin 45)+ (vf2 sin(-45)) 1/2*2.2^2 = 1/2 vf1^2 + 1/2*vf2^2 from the second equation vf1 sin45 = vf2 sin(45) vf1 = vf2

  16. perl
    • one year ago
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    now plug that into the first equation

  17. anonymous
    • one year ago
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    ohh okay ...ermm i got 0.7071 :/ that doesn't look right though L:/

  18. perl
    • one year ago
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    2.2 = vf1 cos 45 +vf2 cos(-45) vf1 = vf2 2.2 = vf1 cos 45 +vf1 cos(45) because cos(-45) = cos(45) 2.2 = 2vf1 * cos(45) 2.2 /( 2 cos45) = vf1

  19. anonymous
    • one year ago
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    ohhh okay 1.555 so our solution is 1.6 m/s ?

  20. perl
    • one year ago
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    yes, approximately

  21. perl
    • one year ago
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    to the tenth place , or 2 sig figs

  22. anonymous
    • one year ago
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    yay! thanks so much!

  23. perl
    • one year ago
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    so it turns out that the speed is the same for both balls after the collision. this is because they are both the same mass and diverge at symmetric angles

  24. anonymous
    • one year ago
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    :)

  25. perl
    • one year ago
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    |dw:1433019716902:dw|

  26. anonymous
    • one year ago
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    yes! thank you!

  27. perl
    • one year ago
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    nice, so it turns out we didn't have to use kinetic energy in this problem. but if they are different masses, then you have to use kinetic energy

  28. perl
    • one year ago
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    your welcome :)

  29. anonymous
    • one year ago
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    :)

  30. perl
    • one year ago
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    but it doesnt hurt to pull out all the equations , just in case

  31. perl
    • one year ago
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    any questions on the trig part?

  32. anonymous
    • one year ago
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    no, it all makes sense now :) thank you!!

  33. perl
    • one year ago
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    :)

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