Question

Two particles of masses m$_1$, m$_2$ move with initial velocities u$_1$ and u$_2$. On collision, one of the particles get excited to higher level, after absorbing energy $\varepsilon$. If final velocities of particles be v$_1$ and v$_2$ then we must have

• A. $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2 -\varepsilon =\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$
• B. $\frac{1}{2}m_1^2u_1^2+\frac{1}{2}m_2^2u_2^2 + \varepsilon =\frac{1}{2}m_1^2v_1^2+\frac{1}{2}m_2^2v_2^2$
• C. $m_!^2u_1+m_2^2u_2-\varepsilon =m_1^2 v_1 +m_2^2v_2$
• D. $\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2 =\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2 -\varepsilon $

Answer 1

The Correct Option is A

We need to determine the correct expression relating the energies before and after a collision, where one of the particles gets excited by absorbing energy $\varepsilon$.

The problem involves a change of kinetic energy during collision, taking into account the energy absorbed by a particle. The principles of conservation of momentum and energy can help us derive the correct relationship.

Firstly, consider the principle of conservation of energy. Initially, the total kinetic energy of the system is given by the sum of the kinetic energies of the two particles:

$\text{Initial Kinetic Energy} = \frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2$

After the collision, the energy system changes as one particle absorbs energy $\varepsilon$, and the particles move with velocities $v_1$ and $v_2$. The final kinetic energy plus the absorbed energy must equal the initial kinetic energy:

$\text{Final Kinetic Energy} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

This leads us to the energy conservation equation:

$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \varepsilon = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

The equation correctly accounts for the change in energy because it compensates for the energy $\varepsilon$ absorbed in the system.

The option that matches this expression is the first one provided in the problem statement. Therefore, the correct answer is:

$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \varepsilon = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$