Totally Elastic Collisions
The Center of Mass of an object or collection of objects is the one point that behaves as if the total mass were located at that one point. In particular the total momentum of the system must be equal to the total mass times the velocity of the C.M.
We will soon be able to ‘prove’ the experimentally observed fact that for totally
elastic (bouncy) collisions, the speed at which the objects approach each other (or the
C.M.) is equal to the speed at which they separate.
We will use these two facts to calculate the final velocities of any two objects,
colliding with no loss of energy, moving in one dimension. The following two examples will illustrate the method.
M1 = 1 kg
M2 = 2 kg
1
2
Find total momentum
pcm = p1 + p2 = +12 kg m/s
Find Vcm = pcm/Mtotal = +4 m/s
V1i = + 6 m/s V2i = +3 m/s
How rapidly is each mass approaching the CM?
Vcm = +4 m/s
V1app = +2 m/s = –V1sep
V1f = 2 m/s
Therefore we can calculate the final velocities to be:
(Checking, the final momentum is +12 kg m/s.)
V2f = 5 m/s
M1 = 3 kg
M2 = 1 kg
3
1
Find total momentum
V2app = –1 m/s = –V2sep
pcm = p1 + p2 = +4 kg m/s
Find Vcm = pcm/Mtotal = +1 m/s
V1i = +2 m/s V2i = –2 m/s
How rapidly is each mass approaching the CM?
Vcm = +1 m/s
V1app = +1 m/s = –V1sep
V1f = 0 m/s
Therefore we can calculate the final velocities to be:
(Checking, the final momentum is +4 kg m/s.)
V2f = +4 m/s
V2app = –3 m/s = –V2sep
SUMMARY OF METHOD
1) Calculate the total momentum of the system (= pcm).
2) Calculate the Velocity of the C. M. = pcm/Mtotal.
3) Calculate the Approach Velocities of Mass1 and Mass 2 (Vapp = Vi – Vc.m).
4) Calculate the Final Velocities (Vf = Vsep + Vc.m.), knowing that the Separation
Velocities must be the negative of the Approach Velocities.
5) Check that the final total momentum still equals the initial total momentum.
PRACTICE PROBLEMS
M1 = 3 kg
M2 = 1 kg
3
1
Find total momentum pcm = ____ kg m/s
Find Vcm = pcm/Mtotal = ____ m/s
V1i = +8 m/s V2i = +4 m/s
How rapidly is each mass approaching the CM?
Vcm =___ m/s
V1app = ___ m/s = –V1sep
V2app = ___ m/s = –V2sep
Therefore we can calculate the final velocities to be:
V1f = ___ m/s V2f = ___ m/s (Checking, the final momentum is ____ kg m/s.)
M1 = 1 kg
M2 =2 kg
1
2
V1i = +3 m/s V2i = –3 m/s
V1f = ___ m/s
M1 = 2 kg
V2f = ___ m/s
M2 = 3 kg
2
3
V1i = +5 m/s V2i = –5 m/s
V1f = ___ m/s
M1 = 2 kg
V2f = ___ m/s
M2 = 3 kg
2
3
V1i = +5 m/s V2i = 0 m/s
V1f = ___ m/s
M1 = 1 kg
V2f = ___ m/s
M2 = 2 kg
1
2
V1i = +3 m/s V2i = 0 m/s
V1f = ___ m/s
V2f = ___ m/s
Instructor’s Notes.
1. This method replaces the two complex algebra equations with fairly straightforward physical reasoning.
2. It reinforces the Center of Mass work that we have been doing and introduces
(or reinforces) relative velocity calculations.
3. The approach and separation velocities are in the center of mass reference frame
and are found by
Vapp = Vi – Vc.m.
and
Vf = Vsep + Vc.m.
Since the approach and separation velocities are the negative of each other, a simple
substitution leads to
Vf = 2Vc.m. – Vi
This gives an incredibly simple way to calculate final velocities in totally elastic collisions, but it is much less conceptual than carrying out the operations outlined in the
instructions and shouldn’t be done. However, some of the better students should be
able to discover the formula. If they find it themselves, it should be used, because then
it will have meaning for them.
4. Any number of examples can be easily constructed and integer answers can always be produced as follows:
(1) Choose any two masses and velocities that you wish.
(2) Calculate the Vcm as explained above.
(3) If the Vcm is a fraction, as it most often will be, simply multiply EACH of
the initial velocities by the denominator of the fraction. This guarantees that
the answers will be an integer.
Chuck Britton
NCSSM, Durham, NC
February 1995
Name_________________________
Period_____________
M1 = 1 kg
M2 = 4 kg
1
4
V1i = +40 m/s V2i = +10 m/s
V1f = ___ m/s
M1 = 4 kg
V2f = ___ m/s
M2 = 2 kg
4
2
V1i = +24 m/s V2i = –6 m/s
V1f = ___ m/s
M1 = 3 kg
V2f = ___ m/s
M2 = 1 kg
3
1
V1i = +24 m/s V2i = –12 m/s
V1f = ___ m/s
M1 = 4 kg
V2f = ___ m/s
M2 = 3 kg
4
3
V1i = +35 m/s V2i = 21 m/s
V1f = ___ m/s
M1 = 1 kg
V2f = ___ m/s
M2 = 5 kg
1
25
V1i = –12 m/s V2i = –18 m/s
V1f = ___ m/s
V2f = ___ m/s