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A particle of mass m _(1) makes a head on elastic collision with a stationary particle of mass m _(2). What fraction of kinetic energy is (a) retained by m _(1) (b) transferred onto m _(2) ?

Answer» <html><body><p></p>Solution :(a) ` v _(1) = (( m _(1) - m _(2)) u + <a href="https://interviewquestions.tuteehub.com/tag/2m-300757" style="font-weight:bold;" target="_blank" title="Click to know more about 2M">2M</a> _(2) (o))/( m _(1) + m _(2)) <a href="https://interviewquestions.tuteehub.com/tag/implies-1037962" style="font-weight:bold;" target="_blank" title="Click to know more about IMPLIES">IMPLIES</a> (( v _(1))/(u ))^(2) = ((m_(1) - m _(2))/(m _(1) + m _(2))) ^(2)` <br/> `implies ((1)/(2) m _(1) v _(1) ^(2))/( (1)/(2) m _(1) u ^(2)) = ((m _(1) - m _(2)) /( m _(1) +m _(2))) ^(2) (KE of 1 ^(st ) " particle finally")/(KE of 1 ^(st) "particle initially") = (( m _(1) - m _(2))/( m _(1) + m _(2))) ^(2)` <br/> `therefore` fraction of KE retained by `1 ^(st)` body `f _(1) = ((m _(1) - m _(2))/( m _(1) + m _(2)) ) ^(2)`<br/> (<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>) `v _(2) = (2 m _(1) u + ( m _(2) - m _(1)) 0o)/(m _(1) + m_(2)) implies(v _(2))/( u ) = (2m _(1))/( m _(1) + m _(2))` <br/> `f _(2) = ((1)/(2) m _(2) v _(2) ^(2))/( (1)/(2) m _(1) u ^(2))= (m_(2))/( m _(1)) [ ( 2m _(1))/( m _(1) + m _(2)) ] ^(2)` <br/> `therefore f _(2) = ( 4m _(1) m _(2))/( ( m _(1) + m _(2)) ^(2))` (fraction of KE given to `2 ^(nd)` body)</body></html>


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