# Two particles of masses m1, m2 move with initial velocities u1 and u2. On collision, one of the particles get excited to higher level, after absorbing energy  (E). If final velocities of particles be v1 and v2 then we must have: Option 1) $m_1^{2}u_1 +m_2^{2}u_2 - \epsilon = m_1^{2}v_1 +m_2^{2}v_2$ Option 2) $\frac{1}{2}m_1u_1^{2}+ \frac{1}{2}m_2u_2^{2}= \frac{1}{2}m_1v_1^{2}+ \frac{1}{2}m_2v_2^{2}-\epsilon$ Option 3) $\frac{1}{2}m_1u_1^{2}+ \frac{1}{2}m_2u_2^{2}-\epsilon =\frac{1}{2}m_1v_1^{2}+ \frac{1}{2}m_2v_2^{2}$ Option 4) $\frac{1}{2}m_1u_1^{2}+ \frac{1}{2}m_2u_2^{2}+\epsilon =\frac{1}{2}m_1v_1^{2}+ \frac{1}{2}m_2v_2^{2}$

P Plabita

As we learnt in

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}$

$u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}$

Initial energy of two particles = $\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

final energy of two particles = $\frac{1}{2}m_{2}v_{2}^{2}+\frac{1}{2}m_{1}v_{1}^{2}+\varepsilon$

from conservation of energy.

$\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}=\frac{1}{2}m_{2}v_{2}^{2}+\frac{1}{2}m_{1}v_{1}^{2}+\varepsilon$

$\therefore \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}-\varepsilon =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

Option 1)

$m_1^{2}u_1 +m_2^{2}u_2 - \epsilon = m_1^{2}v_1 +m_2^{2}v_2$

Incorrect

Option 2)

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

Incorrect

Option 3)

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

Correct

Option 4)

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

Incorrect

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