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Example: Idealization of an elastic collision
Mass m2 has a spring attached to it
Spring gets compressed during the collision (an interchange of
linear momentum and mechanical energy between the
individual particles occur.)
Spring released again (after the collision). Particles
m1 and m2 having acquired new linear momentum
and mechanical energy values.
We would like to know how much the spring compresses (based on the
initial values of linear momentum that the particles have before the collision,
what are the final values of the linear momentum, etc.
To make the problem more concrete, let’s assign some values to the masses
and initial velocities (see figure below.)
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BEFORE
the collision
At the time C (during the collision)
when the spring experiences
max compression
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5 m/s + (- 5 m/s) = 0
5 m/s + ( 2 m/s) = 7 m/s
Velocities measured
with respect to the floor
Thus,
NOTE: Notice in the example above that when the
collision is observed in the CM reference,
the blocks just reverse their velocities after
the collision
Indeed, using the results obtained above, we notice
 Before the collision
10 m/s
3 m/s
Observations with respect
to the floor
5 m/s
2 m/s
Observations with respect
to the CM
 After the collision
5 m/s
2 m/s
Observation with respect
to the CM
3-D case of elastic collision between two particles observed in
the CM reference
The reversal of the linear momentum of two particles colliding
elastically (i.e. no change in their internal states) is a quite general
result valid not only in the one-dimension case (as shown above,)
but also in 3-D.
p2f
p1i
p2i
Elastic collision
observed in the
CM reference
p1f
In the CM reference, the total linear momentum is zero. Hence
p1i =p2i , or equivalently m1v1i = m2 v2i
After the collision, we must also have p1f =p2f , or equivalently
m1v1f = m2 v2f .
The conservation of energy implies that p1i =p1f and p2i =p2f
General relationship between relative velocities
u
CM
vCM
v
v
velocity of the object with
respect to the Lab-reference
u
velocity of the object with
respect to the CM-reference
vCM velocity of the object with
Lab-reference
respect to the Lab-reference
v = vCM + u
u = v
- vCM
We have then a convenient shortcut way to solve a
problem involving collisions:
a) Given a problem described in the Lab reference:
LAB REFERENCE
3 m/s
10 m/s
2 Kg
5 Kg
Before the collision
VCM = 5 m/s
b) Describe it in the CM reference
CM REFERENCE
10 m/s - VCM
= 5 m/s
2 Kg
3 m/s - VCM
= -2 m/s
5 Kg
Before the collision
c) In the CM, the bocks suffer just a reverse in their orientation of
the velocity
CM REFERENCE
- 5 m/s
2 m/s
2 Kg
After the collision
5 Kg
d) The velocity of the CM, measured with respect to the LAB
reference, does not change (VCM = 5 m/s). Accordingly,
LAB REFERENCE
0 m/s
2 Kg
7 m/s
5 Kg
VCM = 5 m/s
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NOTE: Kinetic energy available for compressing
the spring
We wish to calculate the maximum compression the spring will
experience during the collision.
We may wonder:
Which value of kinetic energy value should we used in
order to evaluate the compression of the spring? The kinetic
energy calculated by observer “A”, or the kinetic energy
calculated by the observer B?
Naively we could initially speculate that observer “A” would
measure a greater compression of the spring than observer B,
simply because KA is greater than KB. We know this is wrong.
It is wrong because not all the kinetic energy is available for
compression, unless that kinetic energy of the system is evaluated
from the CM reference. (Below we address this issue in more
detail.)
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This term involves velocities
measured from reference A
This term involves
velocities measured
from the CM reference
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Notice, had we chosen the reference of observer B we would have obtained
a similar expression
This term involves velocities
measured from reference B
This term involves
velocities measured
from the CM reference
Notice KA ≠ KB and VCM,A ≠ VCM,B ,
but the value of KCM that appears in expression 1 is the same as the one
appearing in expression 2.
Since this vale of KCM (appearing in expressions 1 and 2 above) is precisely
the value that should be made equal to the variation of the spring’s potential
energy (1/2)k(x)2, (since, with respect to the CM reference, at the point of
max compression the kinetic energy becomes zero) then both observers A
and B will predict the same compression.
Particular case:
m2 at rest
Detailed solution of these equations is given in the
next page. Here we just give the final expressions:
Analysis of particular cases
Case A m1 = m2
Case B
m1 << m2
Case C
m1 >> m2
Case A
Case B
Case C
2 v1i
m2
m1
m2
For the case m1 = m2
show that θ = 900
Solution