Center of mass & collisions

1.Definitions

The velocity v of an object, or a particle i within this object, in the frame at rest, is equal to the velocity of the center of mass (CM) plus the volocity of this object or particle i in the frame of the CM, that is:
v_i = v'_i + v_cm
The momentum of a particle "i" in the center of mass frame is just its mass m_i times its velocity v'_i:
p'_i = m_iv'_i

2. Momentum in the CM frame

1. From the frame at rest

The momentum P of the object is:
Σ m_iv_i.

2. From the frame of the CM

The momentum P' of the object is: P' = Σm_iv'_i = Σm_i(v_i - v_cm) = Σm_iv_i - Σm_iv_cm = Σm_iv_i - Mv_cm) = P - P = 0.

We recall:
The sum of the momenta in the center of mass frame is zero

Let's consider the collision:
(m1,v1) + (m2,v2) → (m1,w1) + (m2,w2)
in the frame at rest, and
(m1,v'1) + (m2,v'2) → (m1,w'1) + (m2,w'2)
in the CM frame, we have:

P' = Σp_i = p'₁ + p'₂ = 0
Therefore:
p'₁ = - p'₂ before collision, and q'₁ = - q'₂ after collision

The conservation of the kinetic energy gives:
p'₁²/2m₁ + p'₂²/2m₂ = q'₁²/2m₁ + q'₂²/2m₂
p'₁²/2 [1/m₁ + 1/m₂] = q'₁²/2 [1/m₁ + 1/m₂]
(1/2) [1/m₁ + 1/m₂] [p'₁ ² - q'₁²] [1/m₁ + 1/m₂]

We have then two possibilities:
p'₁ = + q'₁ and
p'₂ = + q'₂
or
p'₁ = - q'₁ and
p'₂ = - q'₂


The first says that is no collision, and the second shows that the momenta of both objects reverse their directions. Since the masses do not change, we have for the related velocities:

v'₁ = - w'₁ and
v'₂ = - w'₂


We recall:
In the center of mass frame of two objects, the collision just reverses the direction of these two objects.

We have;
v_cm = [m₁v₁ + m₂ v₂ ]/M
Then:
v'₁ = v₁ - v_cm =
v₁ - [m₁v₁ + m₂ v₂ ]/M =
(1/M)[m₁v₁ + m₂v₁ - m₁v₁ - m₂ v₂] = (m₂/M)[v₁ - v₂]
hence:
w'₁ = -(m₂/M)[v₁ - v₂] Therefore:
w₁ = w'₁ + v_cm -v'₁ + v_cm = -v₁ + 2v_cm

w₁ = -v₁ + 2v_cm
w₂ = -v₂ + 2v_cm

That is:
w₁ = v₁(m₁ - m₂)/M + v₂ 2m₂/M
w₂ = v₂(m₂- m₁)/M + v₁ 2m₁/M

3. Energy in the CM frame

P' = p'₁ + p'₂ = 0 before explosion
Q' = q'₁ + q'₂ = 0 after explosion

Before the explosion, the total energy of the object in the center of mass frame is zero (because the original body was at rest in this frame) After the explosion, the total energy in this frame is:
ΔE = q'₁² /2m₁ + q'₂²/2m₂ = q'₁² [1/2m₁+ 1/2m₂]
Therefore:
q'₁ ²= 2 m₁m₂ΔE/(m₁ + m₂)
Then:
w'₂ = [q'₂²]^1/2/m² = [2 m₁ΔE/m₂M]^1/2
w'₂ = [2 m₁ΔE/m₂M]^1/2
Similarly,
m₁ w'₁ = - m₂ w'₂
Then:
w'₁ = - m₂ w'₂/m₁ = - [2 m₂ΔE/m₁M]^1/2

w'₂ = [2 m₁ΔE/m₂M]^1/2
w'₁ = - [2 m₂ΔE/m₁M]^1/2

We have:
v_cm = v
w₁ = w'₁ + v_cm
w₂ = w'₂ + v_cm

w₁ = w'₁ + v_cm
w₂ = w'₂ + v_cm

E_total = (1/2)m₁ w₁² + (1/2)m₂ w₁² =
m₁ w₁² = m₁[w'₁ + v_cm]²
and
m₂ w₂² = m₂[w'₂ + v_cm]²,
According to: w'₁ = - m₂ w'₂/m₁, we have: m₂ w₂ ² = m₂[(- m₁w'₁/m₂) + v_cm]² m₁[ 2 m₁ΔE/m₂M] + v_cm] Then:
m₁ w₁² + m₂ w₂² =
m₁[ w'₁² + v_cm² + 2w'₁v_cm] + m₂ [ (- m¹w'¹/m²)2 + v_cm ²-
2 m¹w'¹/m₂ v_cm]
We have:
m₁ 2w'₁v_cm = m₂ 2 m₁w'₁/m₂ v_cm
Letting:
m₁ w₁² + m₂ w₂² =
m₁[ w'₁² + v_cm² ] + m₂ [ (- m₁w'₁/m₂)² + v_cm²]
= m₁ w'₁ ² + m₂ w'₂²₂ + (m₁ + m₂)v_cm² =
2ΔE + Mv²_cm .

E_total = ΔE + (1/2) Mv_cm²

The total kinetic energy of a system is equal to the kinetic energy of the center of mass plus the sum of the kinetic energies of the objects of the sytem, as calculated in the center of mass frame.

Other way to express directly the total energy

v_i = v_cm + v'_i
E_total = Σ (1/2)m_iv_i² = Σ (1/2)m_i(v_cm + v'_i)² =
Σ (1/2)m_i[v_cm² + v'_i²+ 2 v'ⁱv^cm] = (1/2) Σ m_i[v_cm² + v'_i²+ 2 v'_iv_cm]
We have:
(1/2) Σ m_i2 v'_iv_cm
v_cmΣ m_i v'_i = v_cmΣ p'_i = 0 Letting:
E_total = (1/2) Σ m_iv²_cm + (1/2) Σ m_iv'²_i
= (1/2)M v²_cm + (1/2) Σ m_iv'²_i
= E_cm + E'

E_total = E_cm + E'

4. Example:

For the head-on collision in the rest frame:
(m,V) + (M,0) &arra; m w₁+ Mw₂, we have:
w = 2m/(m+M)
w'₂ = w₂ - v_cm
v_cm = mV/(m + M), so
w'₂ = w₂ - mV/(m + M) = mV/(m + M)
Therefore:
w'₂² = m²V²/(m + M)² = 2mΔE /M(m + M), so
ΔE = (1/2) m V² M/(m + M)

If K is the kinetic energy = (1/2) mV² of the projectile (m,V), we have:
ΔE = K M/(m + M)
Collision: (m,V) + (M,0)
The enrgy available to cause reaction is
ΔE = K M/(m + M)