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Summersnow8
 one year ago
Ball A of mass 0.55 kg has a velocity of 0.65 m/s east. It strikes a stationary ball, also of mass 0.55 kg. Ball A deflects off ball B at an angle of 37° north of A's original path. Ball B moves in a line 90° right of the final path of A. Find the momentum (in kg m/s) of Ball A after the collision.
Summersnow8
 one year ago
Ball A of mass 0.55 kg has a velocity of 0.65 m/s east. It strikes a stationary ball, also of mass 0.55 kg. Ball A deflects off ball B at an angle of 37° north of A's original path. Ball B moves in a line 90° right of the final path of A. Find the momentum (in kg m/s) of Ball A after the collision.

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ivancsc1996
 one year ago
Best ResponseYou've already chosen the best response.0Is there a picture you can supply? If ball B moves 37° north from the east, and 90° right of the final path of A, A must move 127° north from the east, that is in the north west direction. But the both A's and B's momenta would have a positive component in the north direction which would violate the law of conservation of momentum since at the beginning there was no component of momentum going north.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is my understanding, that A is the one deflected north and B is 90° right/clockwise of the deflection. dw:1437222224101:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The we can show momentum conservation in the horizontal and vertical directions. Horizontal: \(m_av_a=m_av'_a \cos(37°) + m_bv'_b \cos(53°)\) Vertical: \(0=m_av'_a \sin(37°) + m_bv'_b \sin(53°)\) The masses are the same, so they cancel and you're left with this system of equations \[0.65~ m/s = v'_a \cos 37° + v'_b \cos (53°)\] \[0 = v'_a \sin 37° + v'_b \sin (53°)\]
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