Neet Solved Paper 2015 Question 9
Question: Two particles of masses $ m _1,m _2 $ move with initial velocities $ u _1 $ and $ u _2 $ . On collision, one of the particles get excited to higher level, after absorbing energy $ \varepsilon $ . If final velocities of particles be $ v _1 $ and $ v _2, $ then we must have
Options:
A) $ m_1^{2}u _1+m_2^{2}u _2-\varepsilon =m_1^{2}v _1+m_2^{2}v _2 $
B) $ \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}-\varepsilon $
C) $ \frac{1}{2}m _1u_1^{2}+\frac{1}{2}m _2u_2^{2}-\varepsilon =\frac{1}{2}m _1v_1^{2}+\frac{1}{2}m _2v_2^{2} $
D) $ \frac{1}{2}m_1^{2}u_1^{2}+\frac{1}{2}m_2^{2}u_2^{2}+\varepsilon =\frac{1}{2}m_1^{2}v_1^{2}+\frac{1}{2}m_2^{2}v_2^{2} $
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Answer:
Correct Answer: C
Solution:
- Total initial energy = $ \frac{1}{2}m _1u_1^{2}+\frac{1}{2}m _2u_2^{2} $ Since, after collision one particle absorb energy $ \varepsilon $ .
$ \therefore $ Total final energy = $ \frac{1}{2}m _1v_1^{2}+\frac{1}{2}m _2v_1^{2}+\varepsilon $ From conservation of energy, $ \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}+\varepsilon $
$ \Rightarrow \frac{1}{2}m _1u_1^{2}+\frac{1}{2}m _2u_2^{2}-\varepsilon $ $ =\frac{1}{2}m _1v_1^{2}+\frac{1}{2}m _2v_2^{2} $
