I have a problem regarding a momentum related question.

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HazyMan
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Homework Statement
A car travels at a velocity of 20 meters per second and has a mass of 1200 kilograms. This car crashes with another car, which travels in the opposite direction. What's the momentum of the OTHER car BEFORE the crash?
Relevant Equations
p1+p2=P, Pinitial=Pfinal
I managed to solve the exercise, but I'm not sure if it's correct or not. I came up with this: [tex]p1+p2=P[/tex] then [tex]p1+p2=0[/tex] then [tex]p1=-p2[/tex] and therefore i solved for p1 so that i could find p2, which is the NEGATIVE value of p1, according to the previous equation. I'm just concerned about this because although a source i found shows the same solution, ANOTHER source says that [tex]p1=p2[/tex] and NOT [tex]p1=-p2[/tex]. Which is the correct equation for the problem? I know momentum is dealt with as a vector so p1=p2 makes no sense to me, but if I'm wrong please help and enlighten me. Thanks for reading!
 
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HazyMan said:
Problem Statement: A car travels at a velocity of 20 meters per second and has a mass of 1200 kilograms. This car crashes with another car, which travels in the opposite direction. What's the momentum of the OTHER car BEFORE the crash?
Relevant Equations: p1+p2=P, Pinitial=Pfinal

I managed to solve the exercise, but I'm not sure if it's correct or not. I came up with this: [tex]p1+p2=P[/tex] then [tex]p1+p2=0[/tex] then [tex]p1=-p2[/tex] and therefore i solved for p1 so that i could find p2, which is the NEGATIVE value of p1, according to the previous equation. I'm just concerned about this because although a source i found shows the same solution, ANOTHER source says that [tex]p1=p2[/tex] and NOT [tex]p1=-p2[/tex]. Which is the correct equation for the problem? I know momentum is dealt with as a vector so p1=p2 makes no sense to me, but if I'm wrong please help and enlighten me. Thanks for reading!

Momentum is a vector, so it has direction. For one-dimensional motion this amounts to having + for momentum one `direction and - for momentum in the other.

In this case ##p_1 = -p_2## in your problem.

However, you can also consider the magnitude of momentum. In this case you drop the ##\pm## and rely on a diagram to show the direction of each momentum. In this case you would have ##|p_1| = |p_2|## and as a shorthand can write ##p_1 = p_2##.

The important thing is that you know and understand the method you are using. I tend to use both, depending on the problem.
 
PeroK said:
Momentum is a vector, so it has direction. For one-dimensional motion this amounts to having + for momentum one `direction and - for momentum in the other.

In this case ##p_1 = -p_2## in your problem.

However, you can also consider the magnitude of momentum. In this case you drop the ##\pm## and rely on a diagram to show the direction of each momentum. In this case you would have ##|p_1| = |p_2|## and as a shorthand can write ##p_1 = p_2##.

The important thing is that you know and understand the method you are using. I tend to use both, depending on the problem.
I see. So, did i actually solve the problem correctly? Thank you for your response.
 
PeroK said:
I don't understand the problem. Do the cars end at rest immediately after the crash?
Oh, yes. 0.12 seconds after the crash they do rest.
 
HazyMan said:
Oh, yes. 0.12 seconds after the crash they do rest.

I guess that's trying to imply that the cars have equal and opposite momentum before the crash. In which case, you are correct.

Note that the phrase "equal and opposite" sums up what I was saying before: equal magnitude momentum but in opposite directions.
 
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PeroK said:
I guess that's trying to imply that the cars have equal and opposite momentum before the crash. In which case, you are correct.

Note that the phrase "equal and opposite" sums up what I was saying before: equal magnitude momentum but in opposite directions.
with magnitude are you referring to the absolute value?