A ball of mass `m` makes head-on elastic collision with a ball of mass urn which is initially at rest. Show that the fractional transfer of energy by the first ball is `4n/(1 + n)^(2)`. Deduce the value of `n` for which the transfer is maximum.

Answer

1

Explanation

Let `u` be the initial velocity of the ball of mass `m`. Then
`"mu"=mv_(1)+nmv_(2)impliesv_(1)+nv_(2)=u`……….i
For elastic collision, Newton's experimental formula is `(u_(2)=0)`
`v_(1)=v_(2)=-(u_(1)-u_(2))impliesv_(1)-v_(2)=-u`…ii
Solving Eqs. i and ii `v_(1)=(1-n)/(1+n)u`
Fractional loss is `KE` (`=` fractional transfer of `KE`)
`f=(K_(i)-K_(f))/(K_(i))-(1/2"mu"^(2)-1/2mv_(1)^(2))/(1/2"mu"^(2))=1-((v_(1))/u)^(2)`
`f=1-((1-n)/(1+n))^(2)=(4n)/((n+1)^(2))`
The transfer of energy is maximum when `f=1` or `100%`
`(4n)/((n+1)^(2))=(4n)/((n+1)^(2))`
That is the transfer of energy is maximum wen the mass ratio is unity.