# Reaction Kinematics - Lab and Center of Mass Systems

Nuclear reactions abide to conservation laws (energy and momentum along with mass and charge). To analyze the kinematics of reactions, two frames of reference can be used.

The Lab System analyzes particles from the perspective of their nominal location and velocity as in within a laboratory.

The Center of Mass System analyzes particles from the calculated center of mass perspective.

We will analyze the scenarios when a particle collides with another at rest.

## Lab System

In the lab system, when a particle collides with another at rest, their position and velocity are solved using conservation of momentum and energy:

$$m_1 \vec{v_1} = m_1 \vec{v_{1f}} + m_2 \vec{v_{2f}}$$

$$\frac{1}{2} m_1\vec{v_1^2} = \frac{1}{2} m_1\vec{v_{1f}^2} + \frac{1}{2} m_2\vec{v_{2f}^2}$$

 Particle Initial Position Initial Velocity Final Velocity 1 $$\vec{r_1}$$ $$\vec{v_1}$$ $$\vec{v_{1f}}=\cfrac{m_1-m_2}{m_1+m_2}\vec{v_1}$$ 2 $$\vec{r_2}$$ $$\vec{v_2}=0$$ $$\vec{v_{2f}}=\cfrac{2m_1}{m_1+m_2}\vec{v_1}$$

## Center of Mass System

In the center of mass, when a particle collides with another at rest, their position and velocity are also solved using conservation of momentum and energy but after the coordinate system:

$$\vec{r_{CM}} = \cfrac{m_1 \vec{r_1} + m_2 \vec{r_2}}{m_1 + m_2}$$

$$\vec{v_{CM}} = \cfrac{m_1 \vec{v_1}}{m_1 + m_2}$$

In the center of mass system, the total momentum is zero:

$$m_1\vec{v_{1,CM}} + m_2\vec{v_{2,CM}} = m_1\cfrac{m_2\vec{v_1}}{m_1+m_2} - m_2\cfrac{m_1\vec{v_1}}{m_1+m_2}=0$$

 Particle Initial Position Initial Velocity Final Velocity 1 $$\vec{r_{1,L}} - \vec{r_{CM}}$$ $$\cfrac{m_2\vec{v_1}}{m_1+m_2}$$ $$\vec{v_{1f}}=$$ 2 $$\vec{r_{2,L}} - \vec{r_{CM}}$$ $$-\cfrac{m_1\vec{v_1}}{m_1+m_2}$$ $$\vec{v_{2f}}=$$

The center of mass reference makes it convenient to solve for the velocities when an atom decays into two. In the center of mass system the momentum is zero, thus $$\vec{v_{CM}} = \vec{v_0}$$ and $$\vec{v_{0,CM}} = 0$$ (i.e., the atom is not moving in the center of mass system.

The final momentum with particles 1 and 2 is:

$$m_1 \vec{v_1} = -m_2\vec{v_2}$$

The final kinectic energy of the daughters will equal the binding energy released from the reaction:

$$\frac{1}{2} m_1\vec{v_1^2} + \frac{1}{2} m_2\vec{v_2^2} = Q$$

 Particle Initial Velocity Final Velocity 0 $$0$$ N/A 1 N/A $$\vec{v_{1f}}= \sqrt{\cfrac{2Q}{m_1(1+\frac{m_1}{m_2})}}$$ 2 N/A $$\vec{v_{2f}}= -\sqrt{\cfrac{2m_1Q}{m_2(m_1+m_2)}}$$

The particles will move in opposite directions to conserve momentum (initally zero).