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anonymous
 one year ago
A 1,600 kg car traveling north at 10.0 m/s crashes into a 1,400 kg car traveling east at 15 m/s at an unexpectedly icy intersection. The cars lock together as they skid on the ice. What is their speed after the crash?
8.8 m/s , 12 m/s , 18 m/s , or 26 m/s ?
anonymous
 one year ago
A 1,600 kg car traveling north at 10.0 m/s crashes into a 1,400 kg car traveling east at 15 m/s at an unexpectedly icy intersection. The cars lock together as they skid on the ice. What is their speed after the crash? 8.8 m/s , 12 m/s , 18 m/s , or 26 m/s ?

This Question is Closed

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here we have to apply the total momentum conservation law

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1433014089573:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to find the momentum p_1 and the momentum p_2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we have: \[\begin{gathered} {p_1} = 1600 \times 10 = ... \hfill \\ \hfill \\ {p_2} = 1400 \times 15 = ... \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so we get 16000 and 21000?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yay!! what happens now?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now, after collision, the system car #1 + car#2 has two components of its momentum, namely P_1 and P_2 such that the subsequent condition holds: \[\Large \begin{gathered} {P_1} = \left( {{m_1} + {m_2}} \right){V_1} = {p_1} \hfill \\ \hfill \\ P2 = \left( {{m_1} + {m_2}} \right){V_2} = {p_2} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1where m_1 and m_2 are the masses of the two cars respectively, and V_1 and V_2 are the components of the velocity of the system car #1 + car#2, namely: dw:1433014582296:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes :) how do we find the speed from that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to divide the formulas above, by m_1+m_2, like this: \[\Large \begin{gathered} {V_1} = \frac{{{p_1}}}{{{m_1} + {m_2}}} \hfill \\ \hfill \\ {V_2} = \frac{{{p_2}}}{{{m_1} + {m_2}}} \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok! what do we plug in? :/

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1it is simple: p_1=10,000, p_2=21,000, m_1=1,600 and m_2=1,400

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and the othe velocity?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is 3.333 (oops sorry forgot to write this one earlier haha)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ohh yes sorry!! so from there what do we do?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! so the requested velocity, has the subsequent magnitude: \[V = \sqrt {V_1^2 + V_2^2} = \sqrt {{{5.33}^2} + {7^2}} = ...\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.08.79? so our solution is 8.8 m/s?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! That's right!
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