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Railway truck impulse physics homework

  • Thread starter nokia8650
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  • #1
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A railway truck S of mass 2000 kg is travelling due east along a straight horizontal track with constant speed 12 m s–1. The truck S collides with a truck T which is travelling 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
 

Answers and Replies

  • #2
rock.freak667
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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)?
 
  • #3
Doc Al
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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.
 
  • #4
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Thanks alot, I understand now.
 
  • #5
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so what is the answer guys???
 
  • #6
Doc Al
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so what is the answer guys???
That's for you to figure out.
 
  • #7
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plzzzzzz doc im so stuck plzzzzzzzzzzzzz
 
  • #8
Doc Al
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plzzzzzz doc im so stuck plzzzzzzzzzzzzz
Show what you did so far.
 
  • #9
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28800=2000 (12-v)
28800=24000-2000v
4800=2000v
2800=v

is the right????
 
  • #10
Doc Al
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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.)
 
  • #11
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ty dude u rock=P
 

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