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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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
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can we make a diagram of the situation
ok, i am not quite sure how!
|dw:1433018245578:dw|

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

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