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Two particles of masses m₁ m₂ move with initial velocities u₁ and u₂ .On collision, one of the particle s get excited to higher level, after absorbing energy E.If final velocities of particles be v₁ and v₂, then we must have
Options
(a) m₁²u₁+ m₂²u₂ -E = m₁²v₁+ m₂²v₂
(b) 1/2 m₁u₁²+m₂ u₂² = 1/2 m₁v₁²+ 1/2 m₂v₂² -E
(c) 1/2 m₁u₁²+m₂ u₂² -E = 1/2 m₁v₁²+ 1/2 m₂v₂²
(d) 1/2 m₁u₁²+m₂ u₂² +E = 1/2 m₁v₁²+ 1/2 m₂v₂²
Correct Answer:
1/2 m₁u₁²+m₂ u₂² -E = 1/2 m₁v₁²+ 1/2 m₂v₂²
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Topics: Work Energy and Power
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- The work done in which of the following processes is equal to the change in internal energy
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- The sum of the magnitudes of two forces acting at a point is 16 N
- If A=3i+4j and B=7i+24j the vector having the same magnitude as B and parallel to A is
- Which of the following is not a transverse wave?
Topics: Work Energy and Power (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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