Conservation of Momentum

Raj

Sep 2008
12
0
I have acouple problems that I can't finish, or start. Help with any of them would be appreciated.

1. A 1740kg car travelling [North] was crashed into a 2000kg car travelling [West]. On impact they locked together and slid at 9m/s [35 degrees West of North]. What was the speed of the 1740kg car before impact?
My equation looks like this:

\(\displaystyle m_1v_1 + m_2v_2 = m_3v_3\)
\(\displaystyle 1740v_1 [N] + 2000v_2 [W] = 33660 kgm/s [35 degrees W of N]\)
Need help after this point.
.
 
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arbolis

PHF Hall of Fame
Apr 2008
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1. A 1740kg car travelling [North] was crashed into a 2000kg car travelling [West]. On impact they locked together and slid at 9m/s [35 degrees West of North]. What was the speed of the 1740kg car before impact? My equation looks like this:



Need help after this point.
I didn't study this topic yet (but I will do soon) so I cannot answer to you. However, looking back to your last thread, topsquark created in me an intuition that you have to use the law of conservation of momentum.
You will also have to use the same law for the 2nd exercise.
But I prefer to wait topsquark (or someone else) to make these speculations a reality. Good luck.
 

werehk

PHF Hall of Fame
Apr 2008
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HK
Maybe I lead you for question 2 first. According to conservation of momentum, the momentum before and after should be the same. Before explosion, the total momentum should be zero. After the explosion, you may resolve the velocity into x- y- components setting up two equations respectively. Finally combime the x- y- components to get the direction and magnitude.

For question 3,
0 = 3t x15+ (130-3t)v
Where t is the time when she stops spraying.I'm not quite sure
 
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Raj

Sep 2008
12
0
Maybe I lead you for question 2 first. According to conservation of momentum, the momentum before and after should be the same. Before explosion, the total momentum should be zero. After the explosion, you may resolve the velocity into x- y- components setting up two equations respectively. Finally combime the x- y- components to get the direction and magnitude.

For question 3,
0 = 3t x15+ (130-3t)v
Where t is the time when she stops spraying.I'm not quite sure
Yeah i was thinking along these lines, however in question 2 only three fragments are accounted for so i didn't think momentum would be conserved.

I'll try this anyway and see what i come up with.

edit: Re-read the question, I think there is only a total of 3 pieces of debris, completed the second question :)

As for question three I think I solved it.
My equation was:
\(\displaystyle 0 = (3kg)(15m/s) + (127kg)(v) = -0.35m/s\)

Thanks for your help.


However, looking back to your last thread, topsquark created in me an intuition that you have to use the law of conservation of momentum.

You will also have to use the same law for the 2nd exercise.
I hope that is sarcasm.
 
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