Elastic collisions
One dimension
One mass initially at rest
The equations for conservation of momentum and energy are
m1v1i = m1v1 f + m2 v2 f and
1
1
1
m1v12i = m1v12f + m2 v22 f
2
2
2
We solve the first equation for v2 f :
v2 f =
m1v1i − m1v1 f
m2
and substitute into the second:
1
1
1  m v − m1v1 f 
m1v12i = m1v12f + m2  1 1i

m2
2
2
2 

2
Our task now is to simplify and solve for v1 f . To start,
2
m2 m12
m v = m v + 2 ( v1i − v1 f ) .
m2
2
1 1i
2
1 1f
Simplifying further
v12i = v12f +
2
m1
v1i − v1 f ) .
(
m2
Now, square the term in parentheses, and move everything to the right-hand side:
0 = − v12i + v12f +
2
m1
v1i − v1 f ) .
(
m2
Note that the first two terms can be written as
0 = ( v1 f − v1i )( v1 f + v1i ) +
2
m1
v1i − v1 f )
(
m2
or, rearranging signs
0 = − ( v1i − v1 f )( v1 f + v1i ) +
2
m1
v1i − v1 f ) .
(
m2
We now divide both sides by ( v1i − v1 f ) to obtain
0 = − ( v1 f + v1i ) +
0 = − v1 f − v1i +
m1
( v1i − v1 f ) or
m2
m1
m
v1i − 1 v1 f
m2
m2
Collecting terms,
 m 

m 
0 = − v1 f  1 + 1  + v1i  −1 + 1  .
m2 
 m2 

We move the first term to the left-hand side
 m 

m 
v1 f  1 + 1  = v1i  −1 + 1 
m2 
 m2 

and solve for v1 f
m1 
m1 
 −1 +
 m2
m2 
m2 
=
v1 f =
.
m1


m
1
1+
 1 +  m2
m2
 m2 
−1 +
Where in the last step we have multiplied top and bottom by m2 . Clearing the fractions, we
obtain
v1 f =
m1 − m2
v1i ,
m1 + m2
the result we are looking for! To find v2 f we substitute the above equation into our previous
result:
v2 f =
=
m1v1i − m1v1 f
m2
=
m1
( v1i − v1 f )
m2
m1 
m1 − m2 
v1i 
 v1i −
m2 
m1 + m2 
I leave it as an exercise to put the terms in the parentheses under a common denominator and
show
v2 f =

m1  2m2
2m1
v1i  =
v1i

m2  m1 + m2  m1 + m2
which is the result we want. We have now shown that for an elastic collision in 1D,
v1 f =
m1 − m2
v1i
m1 + m2
and v2 f =
2m1
v1i .
m1 + m2
Remember that these results apply only for elastic collisions—collision in which both
momentum and kinetic energy are conserved.