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chris215
 one year ago

chris215
 one year ago


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chris215
 one year ago
Best ResponseYou've already chosen the best response.02 m/s, 3m/s 3 m/s, 2 m/s 2 m/s, 3m/s 2 m/s, 3 m/s

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here you have to apply the total momentum conservation law, furthermore, since the collision is elastic, you can apply the kinetic energy conservation law

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we can write the subsequent equations: \[\Large \left\{ \begin{gathered} m{\mathbf{u}} + m{\mathbf{v}} = m{\mathbf{u'}} + m{\mathbf{v'}} \hfill \\ \frac{1}{2}m{u^2} + \frac{1}{2}m{v^2} = \frac{1}{2}mu{'^2} + \frac{1}{2}mv{'^2} \hfill \\ \end{gathered} \right.\] where u, v are the initial velocities, namely before collision of the two balls, and u', v' are the corresponding velocities after collision. Furthermore m is the common mass of the two balls. Please note that the first equation is a vector equation. you can use this subsequent reference system: dw:1435934610110:dw
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