Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

Two particles of masses $m_1, m_2$ move with initial velocities $u_1$ and $u_2$ . On collision, one of the particles gets excited to a higher level, after absorbing energy $\varepsilon$ . If the final velocities of the particles are $v_1$ and $v_2$ , then we must have:

$m_1^2u_1 + m_2^2u_2 - \varepsilon = m_1^2v_1 + m_2^2v_2$

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

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

$\frac{1}{2}m_1^2u_1^2 + \frac{1}{2}m_2^2u_2^2 + \varepsilon = \frac{1}{2}m_1^2v_1^2 + \frac{1}{2}m_2^2v_2^2$

Step-by-Step Solution

Identify the System: The system consists of two particles undergoing a collision. This is an inelastic collision because kinetic energy is not conserved; a specific amount of energy ( $\varepsilon$ ) is absorbed into internal energy (excitation).
Conservation of Total Energy: The Principle of Conservation of Energy states that the total energy before collision must equal the total energy after collision. $E_{initial} = E_{final}$ $K_{initial} = K_{final} + E_{absorbed}$
Formulate Equations:

Initial Kinetic Energy ( $K_i$ ): $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2$
Final Kinetic Energy ( $K_f$ ): $\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$
Energy Absorbed: $\varepsilon$

Substituting these into the conservation equation: $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \left(\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\right) + \varepsilon$ 4. Rearrange: Subtract $\varepsilon$ from both sides to match the option format: $\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 - \varepsilon = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ (Refer to NCERT Class 11, Section 6.12.1 on Inelastic Collisions ).

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started