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In introductory mechanics, the momentum of a particle is its mass times its velocity. In electrodynamics, the momentum of a field is proportional to the cross-product of the electric field with the magnetic field. In special relativity, momentum is generalized to four-momentum.

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+ m_2v_2^2 $ $I_0w_0^2 = I_1w_1^2 + I_2w_2^2 $ where $v_0$, $v_1$, and $v_2$ are the magnitudes of the velocities $v = \sqrt{v_x^2 + v_y^2}$ You also need to conserve linear and angular momentum: $m_0 …
answered Apr 3 '15 by Tetradic