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Billiard ball A of mass 1.80 kg is given a velocity 6.10 m/s in +x direction...

Billiard ball A of mass 1.80 kg is given a velocity 6.10 m/s in +x direction as shown in the figure. Ball A strikes ball B of the same mass, which is at rest, such that after the impact they move at angles ΘA = 60.0 ° and ΘB respectively. The velocity of ball A after impact is 3.30 m/s in the direction indicated in the figure.
Figure showing a layout of the problem as described in text
What is the total momentum after impact in x-direction?

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What is the velocity of ball B after impact in x-direction?

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What is the velocity of ball B after impact in y-direction?

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What is the momentum of ball B after impact in the direction shown?

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Answer #1

Given, mA 1.80 g M4 6.lo s mg 1.80kg er= 60° Ya= 3.30 rys. 8A Total mamentum be-fone Collsion: P fix-malas(1.80)(6-10)= lo. 9Alright Dude, If that worked for you... dont forget to give THUMBS UP.(that will work for me!)
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If I missed something feel free to leave a comment.
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