Physics 121
Elastic Collisions: Zero Total Momentum
Section 10.6 of Knight describes a special case of
perfectly elastic collisions between two objects in which
one object is initially at rest. I want to look at another
special case. This is the case of perfectly elastic head-on
collisions between two objects in which the total
momentum is zero.
r
r
r
p i1 + p i 2 = 0
(H.1)
Since the collision is perfectly elastic, the momentum is conserved and the kinetic energy
after the collision is equal to the kinetic energy before the collision.
momentum conservation:
energy conservation:
r
r
r
r
p i1 + p i 2 = p f 1 + p f 2
r
r
r
r
m1 v i1 + m2 v i 2 = m1 v f 1 + m2 v f 2
(H.2)
Ki1 + Ki2 = Kf1 + Kf2
½m1vi12 + ½m2vi22 = ½m1vf12 + ½m2vf22
(H.3)
Combining equations H.1 and H.2, we see that
r
r
m1 v i1 + m2 v i 2 =
r
r
m1 v f 1 + m2 v f 2 =
r
Solving equation H.4 for v i1 we find
r
m r
v i1 = − 2 v i 2
m1
r
Solving equation H.5 for v f 1 we find
r
m r
vf1 = − 2 vf 2
m1
r
0
r
0
(H.4)
(H.5)
(H.6)
(H.7)
r
Remembering that vi1 = | v i1 |, substituting equations H.6 and H.7 into equation H.3 gives
2
 m r 
 m r
½m1  − 2 v i 2  + ½m2vi22 = ½m1  − 2 v f 2
 m1

 m1



2
2
2

 + ½m2vf22


m

m

½m1  2 v i 2  + ½m2vi22 = ½m1  2 v f 2  + ½m2vf22
 m1

 m1

m
Multiplying both sides by 2 1 gives
m2
m2vi22 + m1vi22 = m2vf22 + m1vf22
(m2 + m1)vi22 = (m2 + m1)vf22
vi22 = vf22
Since vi2 and vf2 are speeds, and thus are always positive
vi2 = vf2
(H.8)
We can perform the same analysis and solve for vi1 and vf1 to show that
vi1 = vf2
(H.9)
Equations H.8 and H.9 tell us that the speed of each particle after the collision is the same as the
speed of that particle before the collision.
What about the directions of the particles? Equation H.7 tells us that the final direction of
object 1 is exactly opposite the final direction of object 2. This must be so if the two objects have
r
r
r
r
a total momentum of zero. If the collision is head-on, v f 1 = − v i1 and v f 2 = − v i 2
To summarize, in the case of a perfectly elastic head-on collision between two objects in
which the total momentum is zero, the velocity of an object after the collision is the same as
the velocity of that object before the collision, except in exactly the opposite direction.
Using Reference Frames
This is all very interesting, but very few collisions occur in such a way that the total
momentum is zero. How can this help us analyze collisions in which the total momentum is not
zero? We can use the Galilean transformation to transform the initial velocities of the objects
from the “lab frame” S to a reference frame S' in which the total momentum is zero. In the S'
reference frame, the speeds after the collision are the same as they were before the collision.
r
How do we calculate the velocity V of the “zero momentum” reference frame? This is the
velocity the objects would have after a perfectly inelastic collision.
r m1 vr i1 + m 2 vr i 2
V=
m1 + m 2
(H.10)
r
This V is called the “center of mass velocity”. We will study center of mass when we study
rotation.
TACTICS: Analyzing head-on elastic collisions between two objects.
r
1. Use equation H.10 to calculate the center of mass velocity V.
2. Use the Galilean transformation to transform the initial velocities of objects 1
and 2 from the “lab frame” S to the center of mass reference frame S', which is
r
moving with a velocity V.
r
r
r
r
3. In the center of mass reference frame S', v′f 1 = − v′i1 and v′f 2 = − v′i 2 .
4. Transform the final velocities back to the “lab frame” S.
Let’s look at an example. A 180 g puck slides
across a frictionless surface at 5 m/s. Directly
ahead of this puck slides a 420 g puck which is moving at 2 m/s in the same direction as the first
puck. After they collide elastically, what are the velocities of the pucks?
r
First we find the center of mass velocity V. Let’s define the positive x direction as the
direction both pucks are initially moving (to the right in the diagram). Since there are no
components in the y direction, we can restrict ourselves to the x components.
m1 = 180 g
m2 = 420 g
(vix)1 = 5 m/s
(vix)2 = 2 m/s
m1 (v ix )1 + m 2 (v ix )2
m1 + m 2
180 g (5 m/s) + 420 g (2 m/s)
Vx =
180 g + 420 g
Vx = 2.9 m/s
Vx =
r r r
r r r
Second, make the Galilean transformation to S'. Remember that v = v′ + V or v′ = v − V .
(v ix )1′ = (v ix )1 − Vx
(v ix )1′ = 5 m/s – 2.9 m/s = 2.1 m/s
(v ix )′2 = (v ix ) 2 − Vx
(v ix )′2 = 2 m/s – 2.9 m/s = -0.9 m/s
After the collision, the velocities have just changed directions. So
(v fx )1′ = -2.1 m/s
(v fx )′2 = 0.9 m/s
r r r
Finally, make the Galilean transformation back to S using v = v′ + V
(v fx )1 = (v fx )1′ + Vx
(v fx )1 = -2.1 m/s + 2.5 m/s = 0.4 m/s
(v fx ) 2 = (v fx )′2 + Vx
(v fx ) 2 = 0.9 m/s + 2.5 m/s = 3.4 m/s
We see that after the collision, the 180 g puck is moving to the right at 0.4 m/s, while the 420 g
puck is moving to the right at 3.4 m/s. These results make sense. After the collision the 420 g
puck must be moving to the right faster than it was before the collision. After the collision the
180 g puck must be either be moving to the left, be at rest, or be moving to the right slower than
it was before the collision.