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Conservation of momentum and Impulse

by leftyguitarjo
Tags: conservation, impulse, momentum
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leftyguitarjo
#1
Apr11-08, 12:54 PM
P: 54
1. The problem statement, all variables and given/known data
A large plate is dropped and breaks into three large pieces. The pieces fly apart parallel to the floor. As the plate falls, its momentum has only vertical components, none parallel to the floor. After the collision, the component of momentum parallel to the floor must remain zero since the external force acting on the plate has no parallel component. As viewed from above, piece one has a component velocity of 3m/s at an angle of 115 degrees to the horizontal. Piece two has a velocity of 1.79m/s at an angle of 45 degrees. The third has a velocity of 3.07m/s at -90 degrees and a mass of 1.3 Kg.

What is the mass of the other two pieces?



2. Relevant equations
P=mv
J=F[tex]\Delta[/tex]t=[tex]\Delta[/tex]p
conservation of momentum


3. The attempt at a solution

First, I drew a picture.



then I solved for the velocity in the X and Y directions using sin and cos

I am stuck on the meat of the problem

(x direction)0=mv1+mv2+mv3
=m1.26+m1.26+1.3(0)

y direction 0=m2.71+m1.26+1.3(3.07)

I am lost at this point.
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Dick
#2
Apr11-08, 02:16 PM
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The velocities in your momentum equation should have signs. Two velocities should have opposite signs when they are pointed in opposite directions. Furthermore the masses are not necessarily all the same, are they?
leftyguitarjo
#3
Apr11-08, 03:01 PM
P: 54
Quote Quote by Dick View Post
The velocities in your momentum equation should have signs. Two velocities should have opposite signs when they are pointed in opposite directions. Furthermore the masses are not necessarily all the same, are they?
This was just the way my teacher had it set up. I may have copied it down wrong.

If someone would be as kind as to tell me the correct setup, that would be great.

Dick
#4
Apr11-08, 03:06 PM
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Conservation of momentum and Impulse

Call the two unknown masses m1 and m2 (instead of both m). Now for the velocity components call the velocity component positive if it is up or to the right and negative if it is down or to the left. With these changes your equations are correct. Now just solve them for m1 and m2.
leftyguitarjo
#5
Apr11-08, 03:15 PM
P: 54
Quote Quote by Dick View Post
Call the two unknown masses m1 and m2 (instead of both m). Now for the velocity components call the velocity component positive if it is up or to the right and negative if it is down or to the left. With these changes your equations are correct. Now just solve them for m1 and m2.
some of velocities should have negative values then!!

I am familiar with adding vectors, so this concept is not new to me. Its the whole conservation of momentum part I've yet to fully grasp.

BUT, do I solve for both X any Y directions?
Dick
#6
Apr11-08, 03:25 PM
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You have two unknowns, m1 and m2. So you need two equations to solve for them. The x and y momentum components are those two equations.
leftyguitarjo
#7
Apr11-08, 03:37 PM
P: 54
Quote Quote by Dick View Post
You have two unknowns, m1 and m2. So you need two equations to solve for them. The x and y momentum components are those two equations.
0=m(1)2.71+m(2)1.26+1.3(-3.07)
0=m(1)-1.26+m(2)1.26+1.3(0)

does this look like an good start then?
Dick
#8
Apr11-08, 04:13 PM
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Looks great. Write m(1)*(-1.26) instead of m(1)-1.26, ok? Otherwise the '-' looks like a subtraction instead of a sign on the 1.26.


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