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The answer is $\frac{1}{m_1}$ and $\frac{1}{m_2}$.

I used the conservation of momentum (taking into consideration initial momentum is zero) , coefficient of restitution , and conservation of kinetic energy (since it is elastic collision) and did not approach to a solution.

hardmath
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Powerful blaster
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  • just to clarify, the answers should be $\frac{I}{m_1}$ and $\frac{I}{m_2}$ – David Quinn Mar 31 '18 at 17:09
  • I made some minor edits, but much needs to be done to bring the Question up to the site standards. Please review [ask]. The body of your Question should give a reasonably complete statement of the problem you want help with, and putting it in your own words will help Readers to understand where you are having difficulty. – hardmath Apr 01 '18 at 15:01

1 Answers1

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With the usual notation, by conservation of momentum, $$0=m_1u_1-m_2u_2=-m_1v_1+m_2v_2$$ And by Newton's Law of Restitution, $$u_1+u_2=v_1+v_2$$

Eliminating $v_1$ gives $$v_2=\frac{m_{1} u_{1}+m_{1} u_{2}}{m_{1}+m_{2}}$$ $$=\frac{I+m_1\frac{I}{m_2}}{m_1+m_2}$$ $$=\frac{I}{m_2}$$

Similarly for $v_1$.

David Quinn
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