Railway truck impulse physics homework

[nokia8650]

A railway truck S of mass 2000 kg is traveling due east along a straight horizontal track with constant speed 12 m s–1. The truck S collides with a truck T which is traveling due west along the same track as S with constant speed 6 m s–1. The magnitude of the impulse of T on S is 28 800 Ns.


Calculate the speed of S immediately after the collision


My calculation goes:

28800 = 2000(v-12)

However, the markscheme has it as 28800 = 2000(12-v)

Both of these clearly yield different solutions. Is it due to the fact that the impulse is in fact -28800 (minus 28800)? How would one know that though?

Thanks

 

[rock.freak667]

Did you take into account that you take one direction as +ve and the other as -ve (i.e. take all vectors going due east as +ve and all due west as -ve)?

 

[Doc Al]

Both of these clearly yield different solutions. Is it due to the fact that the impulse is in fact -28800 (minus 28800)?

Yes.

How would one know that though?

By being careful with your signs. Let's call east to be positive. Thus the initial speed of S is + 12 m/s. The impulse is negative, since it must point west.

It doesn't matter whether you call east positive or negative. As long as you're consistent, you'll get the same answer for the speed.

 

[nokia8650]

Thanks a lot, I understand now.

 

[mighty jk]

so what is the answer guys?

 

[Doc Al]

so what is the answer guys?

That's for you to figure out.

 

[mighty jk]

pleasezzzzz doc I am so stuck pleasezzzzzzzzzzzz

 

[Doc Al]

pleasezzzzz doc I am so stuck pleasezzzzzzzzzzzz

Show what you did so far.

 

[mighty jk]

28800=2000 (12-v)
28800=24000-2000v
4800=2000v
2800=v

is the right?

 

[Doc Al]

28800=2000 (12-v)
28800=24000-2000v
4800=2000v
2800=v

is the right?

It was OK until the last line. (Divide, don't subtract.)

 

[mighty jk]

ty dude u rock=P