# CM momentum equations

my notes tell me the following, speaking about a two body collision with one particle initially stationary. However i have no clue where lines 7.23 and 7.24 come from. (big "V" is CM velocity, v* is velocity of particle relative to the CM)

Taking a Galilean transformation to the lab frame, before the collision we get:
p1= m1 v1 = m1(v∗1 + V) = p∗ + m1 V
p2= m2 v2 = m2(v∗2 + V) = −p∗ + m2 V
while after the collision we get:
q1= q∗ + m1 V (7.21)
q2= −q∗ + m2 V. (7.22)
Exploiting the fact that p2= 0 we find

p1= p∗(1 + m1/m2) (7.23)
q1= q∗ + p∗m1/m2 (7.24)
q2= p∗ − q∗. (7.25)

## Answers and Replies

Andrew Mason
Science Advisor
Homework Helper
my notes tell me the following, speaking about a two body collision with one particle initially stationary. However i have no clue where lines 7.23 and 7.24 come from. (big "V" is CM velocity, v* is velocity of particle relative to the CM)

Taking a Galilean transformation to the lab frame, before the collision we get:
p1= m1 v1 = m1(v∗1 + V) = p∗ + m1 V
p2= m2 v2 = m2(v∗2 + V) = −p∗ + m2 V
while after the collision we get:
q1= q∗ + m1 V (7.21)
q2= −q∗ + m2 V. (7.22)
Exploiting the fact that p2= 0 we find

p1= p∗(1 + m1/m2) (7.23)
q1= q∗ + p∗m1/m2 (7.24)
q2= p∗ − q∗. (7.25)
If p2 = 0, p* = m2V. Substitute that value for p* in the first equation and you get

p1 = m2V + m1V = m2V(1 + m1/m2) = p*(1+m1/m2)

Do the same substitution in 7.22 to get 7.25. To get 7.24, let m1V =(m2V)(m1/m2) = p*(m1/m2)

AM

thanks for your help, that makes perfect sense