p45_011 - − 2 K 2 E 1 E 1 − E 2 2 = E 2 3 This is now...

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11. (a) Conservation of energy gives Q = K 2 + K 3 = E 1 E 2 E 3 where E refers here to the rest energies ( mc 2 ) instead of the total energies of the particles. Writing this as K 2 + E 2 E 1 = ( K 3 + E 3 ) and squaring both sides yields K 2 2 +2 K 2 E 2 2 K 2 E 1 +( E 1 E 2 ) 2 = K 2 3 +2 K 3 E 3 + E 2 3 . Next, conservation of linear momentum (in a reference frame where particle 1 was at rest) gives | p 2 | = | p 3 | (which implies ( p 2 c ) 2 =( p 3 c ) 2 ). Therefore, Eq. 38-51 leads to K 2 2 +2 K 2 E 2 = K 2 3 +2 K 3 E 3 which we subtract from the above expression to obtain
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Unformatted text preview: − 2 K 2 E 1 + ( E 1 − E 2 ) 2 = E 2 3 . This is now straightforward to solve for K 2 and yields the result stated in the problem. (b) Setting E 3 = 0 in K 2 = 1 2 E 1 h ( E 1 − E 2 ) 2 − E 2 3 i and using the rest energy values given in Table 45-1 readily gives the same result for K µ as computed in Sample Problem 45-1....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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