5.3 Elastic and Inelastic Collisions
Physics Tool box

Elastic Collision –a collision in which the total kinetic energy after the
collision equals the total kinetic energy before the collision.
v1 f 

 m  m2 
2m2
v2i   1
 v1i
mT
 mT 
 2m 
m  m1
v2 f   1  v1i  2
v2i
m
m
 T 
T



v10  v20   v1 f  v2 f  v2 f  v1 f

Inelastic Collision – a collision in which the total kinetic energy after the
collision is different from the total kinetic energy before the collision.
Completely Inelastic Collision – a collision in which there is a maximum
decrease in kinetic energy after the collision since the objects stick together
and move at the same velocity
In all elastic, inelastic, and completely inelastic collisions involving an
isolated system, the momentum is conserved.
Elastic collisions can be analyzed by applying both the conservation of kinetic
energy and the conservation of momentum simultaneously.
For all collisions involving two objects on which the net force is zero,




momentum is conserved:

mv1  mv2  mv1'  mv2'
For elastic collision, the total kinetic energy before the collision equals the
total kinetic after the collision:
1 2 1 2 1 '2 1 '2
mv1  mv2  mv1  mv2
2
2
2
2
In applying the laws of conservation of momentum and of energy, keeping track of
signs is extremely important. It is best to follow the standard sign system, where right,
east , up , north are designated as positive.
The are four possible outcomes in a collision between two objects




The objects collide and rebound
The objects collide and move in the same direction
The objects collide, stick together, and continue moving in the same direction
One of the objects stops after collision.
Elastic Collision: two billiard balls colliding
EKi  EK p
 
pi  p p
The forces between the bodies are also conservative, so that no mechanical energy is
lost or gained in the collision.
Inelastic Collision: two tennis balls colliding
EKi  EK p
 
pi  p p
Completely Inelastic Collision: two silly putty balls colliding
EKi  EK p
 
pi  p p
Note: It is a common misconception that the only inelastic collisions are those in which
the colliding bodies stick together. If two cars collide and bouce off each other in a
“fender-bender”, the work done to deform the fenders cannot be recovered as kinetic
energy of the cars, so the collision is inelastic.
Problem Solving Strategy
The first step is to distinguish between elastic; inelastic, and completely inelastic
collisions.
In any collision in which external forces can be neglected (net force is zero),
momentum is conserved and the total momentum before equals the total momentum
after; in elastic collisions only, the total kinetic energy before equals the total kinetic
energy after.
Example
A 5.0 kg mass is moving to the right at 3.0 m/s. It collides with a 9.0 kg mass that was
stationary. Given that the collision was perfectly elastic, determine the final velocities of
the masses.
Solution:
Method 1: The long way
Let’s label the moving mass as
Since the collision is elastic,
m1 , and the stationary mass as m2 .
EKi  EK p
Therefore
1
1
1
1
m1v120  m2v220  m1v12f  m2v22f
2
2
2
2
2
2
 
 
1
m
1
m
1
1
 5.0kg   3.0    9.0kg   0    5.0kg  v12f   9.0kg  v22f
2
s 2
2

 s 2
45.0kg
 
 
m2
  5.0kg  v12f   9.0kg  v22f
s2
 
pi  p p
And
m1v10  m2v20  m1v1 f  m2 f
 5.0kg   3.0

Now
v1 f 
 
 
 
 
m
 m
   9.0kg   0    5.0kg  v1 f   9.0kg  v2 f
s
 s
kg  m
15
  5.0kg  v1 f   9.0kg  v2 f
s
15
 
kg  m
  9.0kg  v2 f
s
5.0kg
 
 kg  m
 15 s   9.0kg  v2 f
m2
45.0kg 2   5.0kg  
s
5.0kg


 
2


2
   9.0kg  v2 f


 
 kg  m

15
  9.0kg  v2 f 
2

m
s
  9.0kg v 2
45.0kg 2  

 2f
s
5.0kg
225kg 2
270
2
 
 
 
2
m2
kg 2m
2 m

225
kg

270
v2 f  81kg 2 v2 f
s2
s2
s
 
 
 
kg 2 m
v2 f  126kg 2 v2 f
s
m
270  126 v2 f
s
m
v2 f  2.1429
s
2
 
 45kg 2 v22f
2
 
kg  m
m

  9.0kg   2.1429 
s
s

Now v1 f 
5.0kg
m
 0.85722
s
15
Therefore final velocity of the moving ball was -0.86 m/s to the left and the final
velocity of the resting mass was 2.1 m/s to the right.
Method 2: The equations
 m  m2 
v1 f   1
 v1o
 mT 
 5.0kg  9.0kg  
m

 3.0 
14kg
s


m
 0.857
s
 2m 
v2 f   1  v1o
 mT 
 2  5.0kg   
m

  3.0 
s
 14kg  
m
 2.1429
s
Therefore final velocity of the moving ball was -0.86 m/s to the left and the final
velocity of the resting mass was 2.1 m/s to the right.
Method 3:
Conservation of momentum:
 5.0kg   3.0

 
 
 
 
m
 m
   9.0kg   0    5.0kg  v1 f   9.0kg  v2 f
s
 s
kg  m
15
  5.0kg  v1 f   9.0kg  v2 f
s
Elastic Collision:
v10  v20  v2 f  v1 f
3.0
m
 v2 f  v1 f
s
 
kg  m
m


  5.0kg  v1 f   9.0kg   3.0  v1 f 
s
s


kg  m
kg  m
15
  5.0kg  v1 f  27
 9.0kg v1 f
s
s
kg  m
12
 14  kg  v1 f
s
m
v1 f  0.8571
s
15
 
 
m 
m
  0.8571 
s 
s
m
 2.1429
s
v2 f  3.0
 