Question

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

Answer 1

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:
\[\Large {M_1}{V_1} + {M_2}{V_2} = \left( {{M_1} + {M_2}} \right)V\]

Answer 2

V is the final velocity

Answer 3

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

Answer 4

okay! what do i plug in?

Answer 5

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

Answer 6

i cannot see the last part you wrote!!

i just see (6600*2) + [5400
-------------------
6600+5400
what was the rest of it?

Answer 7

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

Answer 8

ohh okay!! i see it now:)

so we get -0.495 ?

Answer 9

I got a different result

Answer 10

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

Answer 11

that's right!

Answer 12

yay! so our solution is choice A?

Answer 13

yes!

Answer 14

yay!! thanks!

Answer 15

thanks!! :)