Question

In a reference frame K, two particles travel along the x axis, one of mass $m_1$ with velocity $\vec{v_1}$, and the other of mass $m_2$ with velocity $\vec{v_2}$. Another reference frame K' moves with a velocity $\vec{v}$ relative to K. If $\mu$ is the reduced mass of the particles, then:

• A. The total kinetic energy wrt K' will be minimum if $\vec{v} = \vec{v_1}$ or $\vec{v_2}$
• B. The total kinetic energy wrt K' will be minimum if $\vec{v} = (\vec{v_1} + \vec{v_2})/2$
• C. The minimum value of kinetic energy wrt K' is $\frac{1}{2}\mu|\vec{v_1} - \vec{v_2}|^2$
• D. The minimum value of kinetic energy wrt K' is $\frac{1}{2}\mu|\vec{v_1} + \vec{v_2}|^2$

Answer 1

Answer: (C)

The minimum value of kinetic energy wrt K' is $\frac{1}{2}\mu|\vec{v_1} - \vec{v_2}|^2$

Answer 2

The kinetic energy of the two-particle system in frame K' is $T' = \frac{1}{2}m_1 (\vec{v_1} - \vec{v})^2 + \frac{1}{2}m_2 (\vec{v_2} - \vec{v})^2$. To find the minimum kinetic energy, we differentiate $T'$ with respect to $\vec{v}$ (or its component, since motion is 1D) and set the derivative to zero. This yields $\vec{v} = \frac{m_1 \vec{v_1} + m_2 \vec{v_2}}{m_1 + m_2}$, which is the velocity of the center of mass ($\vec{v}_{CM}$) of the system. Thus, the minimum kinetic energy occurs when K' is the center of mass frame. Substituting $\vec{v} = \vec{v}_{CM}$ back into the expression for $T'$ and simplifying, we get $T'_{min} = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}|\vec{v_1} - \vec{v_2}|^2$. Recognizing that $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass, the minimum kinetic energy is $\frac{1}{2}\mu|\vec{v_1} - \vec{v_2}|^2$.