## anonymous one year ago 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

1. 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: $\Large {M_1}{V_1} + {M_2}{V_2} = \left( {{M_1} + {M_2}} \right)V$

2. Michele_Laino

V is the final velocity

3. Michele_Laino

namely: $\Large V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}}$

4. anonymous

okay! what do i plug in?

5. Michele_Laino

here is the next step: $\Large V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}} = \frac{{\left( {6600 \times 2} \right) + \left\{ {5400 \times \left( { - 3} \right)} \right\}}}{{6600 + 5400}}$

6. anonymous

i cannot see the last part you wrote!! i just see (6600*2) + [5400 ------------------- 6600+5400 what was the rest of it?

7. Michele_Laino

sorry, here is the formula: $V = \frac{{{M_1}{V_1} + {M_2}{V_2}}}{{{M_1} + {M_2}}} = \frac{{\left( {6600 \times 2} \right) + \left\{ {5400 \times \left( { - 3} \right)} \right\}}}{{6600 + 5400}}$

8. anonymous

ohh okay!! i see it now:) so we get -0.495 ?

9. Michele_Laino

I got a different result

10. anonymous

oh oops sorry i got -0.25 !! i entered something wrong earlier!

11. Michele_Laino

that's right!

12. anonymous

yay! so our solution is choice A?

13. Michele_Laino

yes!

14. anonymous

yay!! thanks!

15. Michele_Laino

thanks!! :)