(b) Head-On Inelastic Collision of Two Particles Which Stick Together
Consider two particles of masses \( m_1 \) and \( m_2 \). The first particle is moving with velocity \( u_1 \), while the second particle is initially at rest. They collide head-on and, after collision, stick together and move with a common velocity \( v \).
(i) In the Laboratory Frame of Reference (L-Frame)
Let the initial velocity of particle \( m_1 \) be \( u_1 \), and that of particle \( m_2 \) be zero.
According to the law of conservation of linear momentum,
\[ m_1 u_1 + m_2(0) = (m_1 + m_2)v \]Therefore,
\[ \boxed{v = \dfrac{m_1 u_1}{m_1 + m_2}} \tag{1} \]Loss of Kinetic Energy
The kinetic energy before collision is
\[ K_i = \dfrac{1}{2} m_1 u_1^2 \]and the kinetic energy after collision is
\[ K_f = \dfrac{1}{2}(m_1 + m_2)v^2. \]Using Eq. (1),
\[ \begin{align} K_f &= \dfrac{1}{2}(m_1 + m_2) \left( \dfrac{m_1 u_1}{m_1 + m_2} \right)^2 \\ &= \boxed{\dfrac{m_1^2 u_1^2}{2(m_1 + m_2)}} \tag{2} \end{align} \]Hence,
\[ \begin{align} \dfrac{K_f}{K_i} &= \dfrac{ \dfrac{m_1^2 u_1^2}{2(m_1 + m_2)} }{ \dfrac{1}{2} m_1 u_1^2 } \\ &= \boxed{\dfrac{m_1}{m_1 + m_2} < 1} \tag{3} \end{align} \]Thus,
\[ \boxed{K_f < K_i} \]It is clear from Eq. (3) that the kinetic energy is not conserved in an inelastic collision. The loss of kinetic energy appears in other forms of energy, such as heat, sound, deformation, etc.
(ii) In the Centre-of-Mass Frame of Reference (C-Frame)
The velocity of the centre of mass is
\[ V_C = \dfrac{m_1 u_1 + m_2(0)}{m_1 + m_2} = \boxed{\dfrac{m_1 u_1}{m_1 + m_2}} \tag{4} \]The initial velocity of particle \( m_1 \) in the C-frame is
\[ \begin{align} u_1' &= u_1 - V_C \\ &= u_1 - \dfrac{m_1 u_1}{m_1 + m_2} \\ &= \boxed{\dfrac{m_2 u_1}{m_1 + m_2}} \end{align} \]The initial velocity of particle \( m_2 \) in the C-frame is
\[ \begin{align} u_2' &= 0 - V_C \\ &= \boxed{-\dfrac{m_1 u_1}{m_1 + m_2}} \end{align} \]After collision, the two particles stick together and move with a common velocity \( v' \) in the C-frame.
Using the law of conservation of linear momentum,
\[ (m_1 + m_2)v' = m_1 u_1' + m_2 u_2' \]Substituting the values of \( u_1' \) and \( u_2' \),
\[ \begin{align} (m_1 + m_2)v' &= m_1\left( \dfrac{m_2 u_1}{m_1 + m_2} \right) + m_2\left( -\dfrac{m_1 u_1}{m_1 + m_2} \right) \\ &= \dfrac{m_1 m_2 u_1 - m_1 m_2 u_1}{m_1 + m_2} \\ &= 0 \end{align} \]Therefore,
\[ \boxed{v' = 0} \tag{5} \]Hence, in the centre-of-mass frame, after a perfectly inelastic collision, the two particles remain stuck together and are at rest with respect to the centre of mass.