Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

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

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

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

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

Step-by-Step Solution

This problem is governed by the Law of Conservation of Energy, which states that the total energy of an isolated system remains constant .

Initial Energy: Before the collision, the total energy of the system is the sum of the kinetic energies of the two particles: $KE_{initial} = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2$ .
Final Energy: After the collision, the total energy consists of the final kinetic energies of the particles plus the internal excitation energy $E$ absorbed by one of the particles: $E_{final} = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 + E$ .
Energy Balance: Equating the initial and final states: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 + E$
Rearranging: To match the provided options, we subtract $E$ from both sides: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - E = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

This describes an inelastic collision where kinetic energy is not conserved because a portion of it is transformed into internal potential/excitation energy .

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