The discussion centers on the conservation of momentum in a two-dimensional explosion scenario involving a shell traveling horizontally north at speed v0. After the explosion, one fragment moves vertically upward at speed v0, while the other fragment's velocity, V2, must be determined. The correct application of conservation of momentum leads to the conclusion that V2 equals v0 in the horizontal direction, confirming that momentum is conserved in both the x and y components. The participants emphasize the importance of a clearly defined coordinate system to avoid confusion in vector components.
PREREQUISITESStudents studying physics, particularly those focusing on mechanics and momentum, as well as educators looking to clarify concepts of vector motion and conservation laws.
Don't forget they have directions associated with them.heycoa said:p0=2m*v0
p1=m*v0
Yes ...heycoa said:so the direction of p0 is in the x-direction, and p1 is in the y-direction?
heycoa said:well p2=v2*m
and when i solve for v2 i get 2*v0(x hat) + v0(y hat)
Yes, for p2.heycoa said:Then it should be minus the second term (-m*v0*yhat), right?
Yes !heycoa said:for v2 i get 2*v0(x hat) + v0(y hat)
i calculated the magnitude to be v2=v0*sqrt(5)
does this appear to be correct?
You're welcome.heycoa said:Ok excellent!
So I apparently need to read these questions more carefully and define my coordinate systems.
I can't thank you enough for taking the time and having the patience to work with and follow up with me. Thank you very much