Classical Mechanics collission particles

[matt222]

Homework Statement

particle of mass m1 collides with particle m2 at rest. The out come of the reaction were m3 and m4 which leave the collision at angles 3 and angle 4 withthe original path. find the energy of the reaction Q in terms of the masses and angles and p1

Homework Equations

x-direction p1=p3cos(theata3)+p4cos(theata4)

y-direction 0=p3sin(theata3)-p4sin(theata4)

T1+Q=T3+T4

T1=p1^2/2m1

T3=p3^2/2m3

The Attempt at a Solution

BY eliminating theata 4 by sequaring and adding x and y equations the out come where

P4^2=P1^2+P3^2-2P1P3COS(theata3)

so Q=p3^2/2m3+p4^2/2m4-p1^2/2m1

but I couldn't find the Q with respect to angles and only p1 can anyone help be

 

[fzero]

You can use the equation

y-direction 0=p3sin(theata3)-p4sin(theata4)

to solve for p4 in terms of p3 and the angles, then substitute the result into

x-direction p1=p3cos(theata3)+p4cos(theata4)

to determine p3 in terms of p1 and the angles.

 

[matt222]

I have got the answer which is
Q= P^2/2m3+2p1^2-2p1^2cos(theata3)/2m4 -p1^2/2m1

do you agree witth my solution so far, but in the book answer is totally different, it is Q=P1^2/2m1{((m1/m3)sin^2(theata4)+(m1/m4)sin^2(theata3))/sin^2(theata3+theata4) -1}

 

[fzero]

I find the same answer as your book. You might want to explain what you did when you said "BY eliminating theata 4 by sequaring." I can see no way to completely eliminate \theta_4. Therefore your equation

P4^2=P1^2+P3^2-2P1P3COS(theata3)

is probably wrong. All you have to do is solve the equations for the y and x directions for p3 and p4, then substitute those results into the energy conservation equation. There's one trig identity that you need to use to compare with the book.

 

[matt222]

i mean by eliminating theata4 is just x and y direction equatation move the first term to the left then sequaring and adding up the two equation the results came like this, which trig you had used because i have tried a lot but i couldnt, i did your way i think need your help

 

[matt222]

my problem now how to eliminate P3 ,

 

[fzero]

i mean by eliminating theata4 is just x and y direction equatation move the first term to the left then sequaring and adding up the two equation the results came like this,

OK, I see now. If you solve for p3 you'll reintroduce the \theta_4 dependence.

which trig you had used because i have tried a lot but i couldnt, i did your way i think need your help

The only trig identity to compare to the book answer is the one for \sin(\theta_3+\theta_4). Try to come up with the expressions for p3 and p4 in terms of p1 and the angles. Once you get those, it's pretty easy to use the conservation of energy equation to get Q.

 

[matt222]

perfect i got it finally,thank you very much for your help