Today, we saw in class the types of collision and how we can identify them. What I want to do here specifically is to show the cases of collisions and the conditions

• Explosions (one to many)
  A single object can explode into multiple objects (explosions).
  $$\Delta\overrightarrow{\text{P}}=0\text{ \& }\Delta\text{K}>0$$
  The increase in the kinetic energy can be attributed to the potential energy the system has. For instance, in the cases of explosives, the chemical potential energy is converted into the kinetic energy of the shrapnel. The energy is not created out of thin air.

• Inelastic Collisions
  In this case, objects colliding have their momentum conserved but the kinetic energy decreases after impact.
  $$\Delta\overrightarrow{\text{P}}=0\text{ \& }\Delta\text{K}<0$$
  A special case of inelastic collisions exists called perfectly inelastic collisions such that the kinetic energy after impact is the lowest possible in that situation and the colliding objects stick together(many to one).

• Elastic Collisions(many to many)
  In this case, different objects collide and the total kinetic energy of the system stays constant.
  $$\Delta\overrightarrow{\text{P}}=0\text{ \& }\Delta\text{K}=0$$
  Perhaps, this is the one we need to discuss in a greater detail because the parameters are related in more than 1 manner. Since both the kinetic energy and linear momentum are conserved here, we have two starting equations to work with(assuming two colliding bodies):
  $$\Delta\overrightarrow{\text{P}}=0\implies\overrightarrow{\text{P}_i}=\overrightarrow{\text{P}_f}$$
  
  m₁v_1, i + m₂v_2, i = m₁v_1, f + m₂v_2, f
  
  ΔK = 0 ⟹ K_i = K_f
  
  $$\dfrac{1}{2}m_1v_{1,i}^2+\dfrac{1}{2}m_2v_{2,i}^2=\dfrac{1}{2}m_1v_{1,f}^2+\dfrac{1}{2}m_2v_{2,f}^2 \label{2}$$
  Notice from equation [1] that we can rearrange the terms along with those on [2] to come up with the following equations
  m₁v_1, i − m₁v_1, f = m₂v_2, f − m₂v_2, i
  
  $$\dfrac{1}{2}m_1v_{1,i}^2-\dfrac{1}{2}m_1v_{1,f}^2=\dfrac{1}{2}m_2v_{2,f}^2-\dfrac{1}{2}m_2v_{2,i}^2 \label{4}$$
  After taking common variables into consideration and multiplying both sides of [4] with 2, we get the following
  m₁(v_1, i − v_1, f) = m₂(v_2, f − v_2, i)
  
  m₁(v_1, i² − v_1, f²) = m₂(v_2, f² − v_2, i²)
  Now that we have simplified the equations to where we can do simple algebraic computations, we can divide equation [6] by equation [5] to get the following result
  $$\dfrac{v_{1,i}^2-v_{1,f}^2}{v_{1,i}-v_{1,f}}=\dfrac{v_{2,f}^2-v_{2,i}^2 }{v_{2,f}-v_{2,i}} \label{7}$$
  Which is a very simple term on both sides
  v_1, i + v_1, f = v_2, f + v_2, i
  Now that we have equation [8], we can go back to equation [1] to substitute the values of the variables and find the final conditions after impact. Even better, the modified form of equation [8] can be modified to be
  v_2, f = v_1, i + v_1, f − v_2, i
  Now, let’s put the equivalent of v_2, f into equation [1], we get
  m₁v_1, i + m₂v_2, i = m₁v_1, f + m₂(v_1, i + v_1, f − v_2, i)
  Simplifying equation [10] as follows
  m₁v_1, i + m₂v_2, i = m₁v_1, f + m₂v_1, i + m₂v_1, f − m₂v_2, i
  Taking the common variables into account, we get
  (m₁ − m₂)v_1, i + 2m₂v_2, i = (m₁ + m₂)v_1, f
  Solving for v_1, f, we get the following
  $$v_{1,f}=\dfrac{(m_1-m_2)v_{1,i}+2m_2v_{2,i}}{m_1+m_2} \label{11}$$
  Following the same steps above for v_2, f, we get the following equation
  $$v_{1,f}=\dfrac{(m_2-m_1)v_{2,i}+2m_1v_{1,i}}{m_1+m_2} \label{11}$$