Posted by CLUELESS on Sunday, May 6, 2012 at 8:44pm.
u1 = (m1-m2) •v/(m1+m2) – (m-16m) •250/(m+16m) = - 220.6 m/s,
Minus indicates that the proton is returned back due to the collision.
m1•v = m1•u1+m2•u2
u2 = (v+u1)m1/ m2 = 250 –(-220.6) •m/16•m = 29.4 m/s
in the direction of the initial direction of the proton.
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