Kinetic Energy Ratio in Scattering Systems

In the study of scattering processes, it is often useful to analyze the collision in different inertial reference frames. The two most common frames are the Laboratory System (LS) and the Center of Mass System (CMS).

Laboratory System (LS)

In the Laboratory System, the target particle ($m_2$) is initially at rest. The projectile particle ($m_1$) approaches the target with an initial velocity $\vec{v}_1$.

The total kinetic energy in the Laboratory System ($T_L$) before the collision is the kinetic energy of the projectile, since the target is at rest:

\(T_L = \frac{1}{2} m_1 v_1^2\)

Center of Mass System (CMS)

The Center of Mass System is a frame where the total momentum of the system is zero. The center of mass moves with a constant velocity relative to the Laboratory System. For a system with two particles, $m_1$ and $m_2$, with initial velocities $\vec{v}_1$ and $\vec{v}_2$ respectively in the LS, the velocity of the center of mass ($\vec{v}_{CM}$) is given by:

\(\vec{v}_{CM} = \frac{m_1 \vec{v}_1 + m_2 \vec{v}_2}{m_1 + m_2}\)

In the case where the target ($m_2$) is initially at rest in the LS ($\vec{v}_2 = 0$), the CM velocity is:

\(\vec{v}_{CM} = \frac{m_1 \vec{v}_1}{m_1 + m_2}\)

The velocities of the particles in the CMS ($\vec{v}_1'$ and $\vec{v}_2'$) are related to their velocities in the LS ($\vec{v}_1$ and $\vec{v}_2$) by:

• \(\vec{v}_1' = \vec{v}_1 - \vec{v}_{CM}\)
• \(\vec{v}_2' = \vec{v}_2 - \vec{v}_{CM}\)

Substituting the expression for \(\vec{v}_{CM}\) and \(\vec{v}_2 = 0\):

• \(\vec{v}_1' = \vec{v}_1 - \frac{m_1 \vec{v}_1}{m_1 + m_2} = \vec{v}_1 \left(1 - \frac{m_1}{m_1 + m_2}\right) = \vec{v}_1 \left(\frac{m_1 + m_2 - m_1}{m_1 + m_2}\right) = \vec{v}_1 \frac{m_2}{m_1 + m_2}\)
• \(\vec{v}_2' = 0 - \frac{m_1 \vec{v}_1}{m_1 + m_2} = -\vec{v}_1 \frac{m_1}{m_1 + m_2}\)

The total kinetic energy in the Center of Mass System ($T_C$) before the collision is the sum of the kinetic energies of the particles in the CMS:

\(T_C = \frac{1}{2} m_1 (v_1')^2 + \frac{1}{2} m_2 (v_2')^2\)

Substitute the magnitudes of the velocities in the CMS:

\(T_C = \frac{1}{2} m_1 \left(\frac{m_2 v_1}{m_1 + m_2}\right)^2 + \frac{1}{2} m_2 \left(-\frac{m_1 v_1}{m_1 + m_2}\right)^2\)

\(T_C = \frac{1}{2} m_1 \frac{m_2^2 v_1^2}{(m_1 + m_2)^2} + \frac{1}{2} m_2 \frac{m_1^2 v_1^2}{(m_1 + m_2)^2}\)

\(T_C = \frac{1}{2} \frac{v_1^2}{(m_1 + m_2)^2} (m_1 m_2^2 + m_2 m_1^2)\)

\(T_C = \frac{1}{2} \frac{v_1^2 m_1 m_2 (m_2 + m_1)}{(m_1 + m_2)^2}\)

\(T_C = \frac{1}{2} \frac{m_1 m_2 v_1^2}{m_1 + m_2}\)

We can also express $T_C$ in terms of the reduced mass \(\mu = \frac{m_1 m_2}{m_1 + m_2}\) and the relative velocity \(\vec{v}_{rel} = \vec{v}_1 - \vec{v}_2\). In the LS before collision, \(v_{rel} = v_1\). The total kinetic energy in the CMS is given by \(T_C = \frac{1}{2} \mu v_{rel}^2\).

\(T_C = \frac{1}{2} \left(\frac{m_1 m_2}{m_1 + m_2}\right) v_1^2\)

This confirms the previous calculation for $T_C$.

Ratio of Total Kinetic Energies \(T_L : T_C\)

Now we can find the ratio of the total kinetic energy in the Laboratory System ($T_L$) to the total kinetic energy in the Center of Mass System ($T_C$).

\(\frac{T_L}{T_C} = \frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} v_1^2}\)

Cancel the common terms \(\frac{1}{2}\) and \(v_1^2\) (assuming \(v_1 \neq 0\)):

\(\frac{T_L}{T_C} = \frac{m_1}{\frac{m_1 m_2}{m_1 + m_2}}\)

\(\frac{T_L}{T_C} = m_1 \times \frac{m_1 + m_2}{m_1 m_2}\)

\(\frac{T_L}{T_C} = \frac{m_1 (m_1 + m_2)}{m_1 m_2}\)

\(\frac{T_L}{T_C} = \frac{m_1 + m_2}{m_2}\)

Thus, the ratio \(T_L : T_C\) is \((m_1 + m_2) : m_2\).

Comparing with Options

Our derivation shows the ratio \(T_L : T_C\) is \((m_1 + m_2) : m_2\). Looking at the provided options:

• Option 1: \(m_1 : m_2\)
• Option 2: \((m_1 + m_2) : m_2\)
• Option 3: \((m_1 + m_2) : m_1\)
• Option 4: More than one of the above
• Option 5: None of the above

Our derived ratio matches Option 2. The provided correct answer for this question is Option 3, which corresponds to the ratio \((m_1 + m_2) : m_1\).