Magnitude of momentum between each pair of adjacent locations

  • Thread starter Axking
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  • #1
Axking
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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
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
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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 [Broken]

I would interpret ~pB = 2.03, 2.83, 0

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.
 
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