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Collisions and Conservation of Momentum

There are two types of collisions:

• Elastic

• Inelastic

Inelastic Collisions occur when only momentum is conserved.  Some kinetic energy is transformed into heat or other forms of energy. The maximum amount of energy lost in a collision occurs when two bodies collide and subsequently stick together. This is an example of an inelastic collision.

Elastic collisions occur when both Kinetic Energy and Momentum are conserved.
Consider an object m₁ with velocity v₁ colliding with a second object m₂ with velocity v₂ traveling in the opposite direction of m₁. After the collision the two objects will rebound and have velocities v₁' and v₂' .

This type of collision is called an elastic collision.

Before the collision the first object will have momentum p₁ = m₁v₁ and the second object will have momentum p₂ = m₂v_2.

After the collision the first object will have momentum p₁' = m₁v₁' and the second object will have momentum p₂' = m₂v₂'.

The Law of Conservation of Momentum tells us that the momentum of a closed system before a collision is equal to the momentum after the collision.

i.e.:     p₁ +  p₂  =  p₁'  +  p₂'
m₁v₁ +  m₂v₂ =  m₁v₁' + m₂v₂'

This simple equation is a a re-statement of  Newton's Third Law of Motion:  ("For every Action there exists an equal but opposite reaction") and can be rewritten in terms of forces:

In the following illustration the balloon moves in the opposite direction of the air flow.

Conservation of Momentum example:

A proton (mass = 1.67 X 10^-27 Kg ) with velocity of 1 X 10⁷ m/s collides with a motionless He nucleus.  The proton bounces back at 6 X 10⁶ m/s.  The He nucleus move forward with a velocity of 4 X 10⁶ m/s .  The collision lasts 1 s.
Calculate:

a) the mass of the helium nucleus
b) The force that acts during the collision

Solution:

v_He = 0 m/s    v_He^' = 4 X 10⁶ m/s
m_p  = 1.67 X 10^-27 Kg
 v_p = 1 X 10⁷ m/s (before collision)     v_p^' = -6 X 10⁶ m/s (AFTER collision)  

a)  Find:  m_He = ?

m_pv_{p +} m_Hev_{He =} m_pv_p^' ₊ m_Hev_He^'       _BUT     m_Hev_{He = 0}

(m_pv_p - m_pv_p^'  ) / v_He^'   =  m_He =  7 x 10^-27 Kg

b) F∆t = m∆v 
... F = (m∆v) / ∆t       
... Where ∆t = 1 s

Now, F = (m_Hev_He^' - m_Hev_He) / ∆t = 2.8 X 10^-20 N

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