Collision Problem: Find w in Terms of u

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SUMMARY

The discussion focuses on solving the collision problem involving three particles (A, B, and C) of equal mass at the origin. Particle A moves with speed u in the direction of (1/sqrt(2))(-i-j), particle B moves with speed v in the direction of (sqrt(3)/2)i+(1/2)j, and particle C moves with speed w in the direction of -i. The conservation of linear momentum dictates that the total momentum before the collision must equal zero, leading to the equation for momentum: P = 0. The user seeks to express w in terms of u, utilizing the relationship between the momentum vectors of the particles.

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Homework Statement


Three particles, A,B,C all of equal mass m,collide at the origin. Prior to the collsion the particles are moving as follows:

A has speed u in direction (1/sqrt(2))(-i-j)
B has speed v in direction (sqrt(3)/2)i+(1/2)j
C has speed w in direction -i

After the collision all particles remain at the origin.

find w in terms of u.

Homework Equations


The Attempt at a Solution



I know that the momentum of each particle is the mass times the velocity. I know that the momentum of the system of particles at time t is P=0

I know that from the conservation of linear momentum that the sum of the individual momentums before the collision must also be 0.

I am probably being a bit thick in my mathematical thinking here because I don't see how to state w in terms of u only.

I just need a nudge in the right direction. No complete solutionsThanks
 
Last edited:
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wrote the linear momentum as
[tex]\frac{1}{m} \vec P = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{2}} \begin{pmatrix} -1 \\ -1 \\ 0 \end{pmatrix} \, u ~ + ~ \frac{1}{2} \begin{pmatrix} \sqrt{3} \\ 1 \\ 0 \end{pmatrix} \, v ~ + ~ \begin{pmatrix} -1 \\ 0 \\ 0 \end{pmatrix} \, w[/tex]​
so you see that the motion is confined in a plane! So any three vectors in a plane are linearly dependent!
for more information see : http://en.wikipedia.org/wiki/Linearly_independent
with best regards
 
Last edited:
Perfect! Thanks
 

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