A body of mass m moving with velocity v collides head-on elastically with another body of mass 2m which is initially at rest. The ratio of K. E. of mass m before and after the collision will be

Answer

9

Explanation

K.E. of the colliding body before the collision `= (1)/(2) mv ^(2)`
After collision its velocity becomes
`v = (( m _(1) - m _(2)))/( ( m _(1) + m _(2))) v = ( m )/( 3m ) v = (v)/(3)`
` therefore ` K.E. after collision `(1)/(2) mv ^(.2) = (1)/(2) ( mv ^(2))/(9)`
The ratio of kinetic energy
`= ( K. E _("before"))/( K.E _("after"))= ((1)/(2) mv ^(2))/( (1)/(2) ( mv ^(2))/( 9)) = 9:1`