Elastic Collision of Two Bodies
in One Dimension:
The Generalized Case
Paul Robinson
Initial Conditions
Block 1, of mass m1, moves across a
frictionless surface with speed v1i. It collides
elastically with block 2, of mass m2, which is
at rest. After the collision, block 1 moves
with speed v1f, while block 2 moves with
speed v2f. What are v1f and v2f?
Part 1: Isolate v2f
Using the
conservation of
momentum, we
isolate v2f in terms
of the other
variables.
m1v1i  m1v1 f  m2v2 f
m2 v2 f  m1v1 f  m1v1i
v2 f
m1

v1 f  v1i 

m2
Part 2: Solve for v1f
Take our v2f expression, and plug it
into the conservation of kinetic
energy.
m1v1i  m1v1 f  m2v2 f
2
m1v1i  m1v1 f
2
2
2
2
m
 m2 
m
2
1
2
2
v
1f
 v1i 
2



Part 2 cont…
m1v1i  m1v1 f
2
m1v1i  m1v1 f
2
2
Now, take the
updated KE equation
and solve for v1f in
terms of the given
constants. First we
cancel out, then
group the v12 on one
side, and then factor
and cancel out again.
2

m1 v1i  v1 f
2
 v1i  v1 f
2
m12
 2  v1 f  v1i  m2
m2

2
m1 2

 v1 f  v1i 
m2
  v1i  v1 f  
v1i  v1 f 
v1 f  v1i
2
2
m12
 2  v1 f  v1i  m2
m2
m1
v1 f  v1i 

m2
m1
v1 f  v1i 

m2
m1  m2
m1  m2
2
Solve for v2f
Take the v1f
expression and
plug into the
conservation of
momentum
equation. Then,
simply solve for v2f
m1v1i  m1v1 f  m2 v2 f
 m  m2 
m1v1i  m1  1
 v1i  m2 v2 f
 m1  m2 
 m  m2 
m2 v2 f  m1v1i  m1  1
 v1i
 m1  m2 
m2 v2 f
v2 f

 m1  m2  
 v1i  m1  m1 
 

 m1  m2  

 2m1 

 v1i
 m1  m2 