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  • Answer please! 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:

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}

 

Answers (1)

best_answer

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

Posted by

Plabita

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