v1'= 972 m/s [E5.1*N]
v2'=944 m/s [E5.9*S]
Determine their original speed when they were linked together.
Pti = Ptf
m1v1 + m2v2 = m1v1' + m2v2'
Like I said, im not very good at trig so my attempt for a diagram got me even more confused (i dont think im doing it right). This is what I came up with:
p1' = m1v1'
p2' = m2v2'
p1' is at an angle of 5.1, p2' is at an angle of 5.9 (which is in the triangle because of the alternate pattern) and 180-5.1-5.9=169.
I know im wrong, but I hope that by showing my work, you can help me get things right.
You don't need a cosine law to add vectors. You just add them. But I may have been misjudging the level of your course. Have you done vectors? If you know the initial direction of both craft is east then you are doing the right thing. You are equating the eastward components of momentum before and after the collision. But I get an answer of more like 954m/s. You might not be keeping enough decimal places. You have the velocities to three decimal places. Try keeping three decimal places. But to two decimal places, your first answer is correct.
Did they tell you that both craft were heading exactly east before the separation?
But you aren't using vectors in your solution.
Oh. I guess when you said "Add the two final momentum vectors and equate that to the initial momentum vector and solve for the vector v" that was before you knew they were traveling east?
If that's the case, how would I approach a graphical trig solution?
You would work out the two components of the vectors you labelled p1' and p2', add them and figure out the two components of the vector you labelled p'. Then find it's length.