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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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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Is 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.
@ivancsc1996 is right
This is my understanding, that A is the one deflected north and B is 90° right/clockwise of the deflection. |dw:1437222224101:dw|

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The 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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