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QAIPM-2015-Ph-1325: NEET 2015 Physics - Class 11 AIPM-2015-Ph-1325 Solution: Two particles of masses \(m_1 m_2\) move with initial…

Two particles of masses m1m2m_1 m_2 move with initial velocities u1u_1 and u2u_2. On collision one of the particles get excited to higher level after absorbing energy ϵ\epsilon. If final velocities of particles be v1v_1 and v2v_2 then we must have: 12m1u12+12m2u22−ϵ=12m1v12+12m2v22\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 12m12u12+12m22u22+ϵ=12m12v12+12m22v22\frac{1}{2}m_1^2u_1^2 + \frac{1}{2}m_2^2u_2^2 + \epsilon = \frac{1}{2}m_1^2v_1^2 + \frac{1}{2}m_2^2v_2^2 m12u1+m22u2−ϵ=m12v1+m22v2m_1^2u_1 + m_2^2u_2 - \epsilon = m_1^2v_1 + m_2^2v_2 12m1u12+12m2u22=12m1v12+12m2v22−ϵ\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

Multiple Choice Options

  • A 12{1 \over 2} 2 1 _ m 1 u 12_1^2 1 2 _ + 12{1 \over 2} 2 1 _ m 2 u 22_2^2 2 2 _ −- _ ε\varepsilon _ = 12{1 \over 2} 2 1 _ m 1 v 12_1^2 1 2 _ + 12{1 \over 2} 2 1 _ m 2 v 22_2^2 2 2 _
  • B 12{1 \over 2} 2 1 _ m 12_1^2 1 2 _ u 12_1^2 1 2 _ + 12{1 \over 2} 2 1 _ m 22_2^2 2 2 _ u 22_2^2 2 2 _ + ε\varepsilon _ = 12{1 \over 2} 2 1 _ m 12_1^2 1 2 _ v 12_1^2 1 2 _ + 12{1 \over 2} 2 1 _ m 22_2^2 2 2 _ v 22_2^2 2 2 _
  • C m 12_1^2 1 2 _ u 1 + m 22_2^2 2 2 _ u 2 −- _ ε\varepsilon _ = m 12_1^2 1 2 _ v 1 + m 22_2^2 2 2 _ v 2
  • D 12{1 \over 2} 2 1 _ m 1 u 12_1^2 1 2 _ + 12{1 \over 2} 2 1 _ m 2 u 22_2^2 2 2 _ = 12{1 \over 2} 2 1 _ m 1 v 12_1^2 1 2 _ + 12{1 \over 2} 2 1 _ m 2 v 22_2^2 2 2 _ −- _ ε\varepsilon _

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