anonymous
  • anonymous
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!
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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perl
  • perl
can we make a diagram of the situation
anonymous
  • anonymous
ok, i am not quite sure how!
perl
  • perl
|dw:1433018245578:dw|

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perl
  • perl
|dw:1433018360172:dw|
anonymous
  • anonymous
yes:) what do we do from here?
perl
  • perl
note that the collision is not 'head on' , thats why it goes off an angle
anonymous
  • anonymous
yes:) so in calculating, would it be moving at the speed of 3.1 m/s?
anonymous
  • anonymous
ok! what do i plug in for m? :/
perl
  • perl
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
perl
  • perl
we get a system of three equations, (we can cancel m out)
perl
  • perl
|dw:1433019149286:dw|
anonymous
  • anonymous
okay! and so we get left with vf1 and vf2? we are looking for vf2 becuase of the second billiard\?
perl
  • perl
right
anonymous
  • anonymous
how can i solve that value?
perl
  • perl
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
perl
  • perl
now plug that into the first equation
anonymous
  • anonymous
ohh okay ...ermm i got 0.7071 :/ that doesn't look right though L:/
perl
  • perl
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
anonymous
  • anonymous
ohhh okay 1.555 so our solution is 1.6 m/s ?
perl
  • perl
yes, approximately
perl
  • perl
to the tenth place , or 2 sig figs
anonymous
  • anonymous
yay! thanks so much!
perl
  • perl
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
anonymous
  • anonymous
:)
perl
  • perl
|dw:1433019716902:dw|
anonymous
  • anonymous
yes! thank you!
perl
  • perl
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
perl
  • perl
your welcome :)
anonymous
  • anonymous
:)
perl
  • perl
but it doesnt hurt to pull out all the equations , just in case
perl
  • perl
any questions on the trig part?
anonymous
  • anonymous
no, it all makes sense now :) thank you!!
perl
  • perl
:)

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