anonymous
  • anonymous
A 6,600 kg train car moving at +2.0 m/s bumps into and locks together with one of mass 5,400 kg moving at -3.0 m/s. What is their final velocity? A. -0.25 m/s B. -1.0 m/s C. +0.50 m/s D. +2.5 m/s
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Michele_Laino
  • Michele_Laino
since the system composed by the two trains, is isolated along the direction of the motion, namely no external forces, along the direction of motion are acting on our trains, then we can write: \[\Large {M_1}{V_1} + {M_2}{V_2} = \left( {{M_1} + {M_2}} \right)V\]
Michele_Laino
  • Michele_Laino
V is the final velocity
Michele_Laino
  • Michele_Laino
namely: \[\Large V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}}\]

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anonymous
  • anonymous
okay! what do i plug in?
Michele_Laino
  • Michele_Laino
here is the next step: \[\Large V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}} = \frac{{\left( {6600 \times 2} \right) + \left\{ {5400 \times \left( { - 3} \right)} \right\}}}{{6600 + 5400}}\]
anonymous
  • anonymous
i cannot see the last part you wrote!! i just see (6600*2) + [5400 ------------------- 6600+5400 what was the rest of it?
Michele_Laino
  • Michele_Laino
sorry, here is the formula: \[V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}} = \frac{{\left( {6600 \times 2} \right) + \left\{ {5400 \times \left( { - 3} \right)} \right\}}}{{6600 + 5400}}\]
anonymous
  • anonymous
ohh okay!! i see it now:) so we get -0.495 ?
Michele_Laino
  • Michele_Laino
I got a different result
anonymous
  • anonymous
oh oops sorry i got -0.25 !! i entered something wrong earlier!
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
yay! so our solution is choice A?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
yay!! thanks!
Michele_Laino
  • Michele_Laino
thanks!! :)

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