Magnitude of momentum between each pair of adjacent locations

[Axking]

Homework Statement

You will calculate the change in the magnitude of momentum between each pair of adjacent locations. Begin by calculating ∆ |~pBC|,the change between locations B and C

~pB = 2.03, 2.83, 0 kg · m/s
~pC =1.55, 0.97, 0 kg · m/s
~pD = 2.24, −0.57, 0 kg · m/s
~pE = 7.97, −1.93, 0 kg · m/s

Homework Equations

magnitude = sqaretoot(x^2 + y^2+z^2)

The Attempt at a Solution

The momentum is given above, so I am confused If i need to see them as vectors, calculate magnitude of the B and C and just subtract?

Thank you for your input

 

[NascentOxygen]

Homework Statement

You will calculate the change in the magnitude of momentum between each pair of adjacent locations. Begin by calculating ∆ |~pBC|,the change between locations B and C

~pB = 2.03, 2.83, 0 kg · m/s
~pC =1.55, 0.97, 0 kg · m/s
~pD = 2.24, −0.57, 0 kg · m/s
~pE = 7.97, −1.93, 0 kg · m/s

Homework Equations

magnitude = sqaretoot(x^2 + y^2+z^2)

The Attempt at a Solution

The momentum is given above, so I am confused If i need to see them as vectors, calculate magnitude of the B and C and just subtract?

Thank you for your input

Hi Axking. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

I would interpret ~pB = 2.03, 2.83, 0 [/size]

to be a vector, of x value = +2.03, y value = +2.83, and z value = 0

Since all your z values are = 0, the 3D task simplifies to a 2 dimensional task in x and y.

The vector difference in going from (2.03, 2.83) to (1.55, 0.97) can be determined by subtracting components. Once you have determined these differences as an ordered pair, then apply your square-root formula.