Two particles of masses `m_(1)` and `m_(2)` and velocities `u_(1)` and `alphau_(1)(alpha!=0)` make an elastic head on collision. If the initial kinetic energies of the two particles are equal and `m_(1)` comes to rest after collision, then

Options

A. `u_(1)/u_(2)=sqrt(2)+1`
B. `u_(1)/u_(2)=sqrt(2)-1`
C. `m_(2)/m_(1)=3+2sqrt(2)`
D. `m_2/m_1=3-2sqrt2`

Answer

A, C, D

Explanation

Since both masses have equal kinetic energy, so `1/2m_(1)u_(1)^(2)=1/2m_(2)(alphau_(1))^(2)`
`impliesm_(2)=(m_(1))/(alpha^(2))`
By law of conservation of momentum
`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)u_(2)`
`m_(1)u_(1)+(m_(1))/(alpha^(2))(alphau_(1))=0+((m_(1))/(alpha^(2)))v_(2)`
`v_(2)=alphau_(1)(1+alpha)`
Further by law of conservation of energy, we have
`1/2m_(1)u_(1)^(2)+1/2((m_(1))/(alpha^(2)))(alphau_(1))^(2)=0+1/2((m_(1))/(alpha^(2)))v_(2)^(2)`
`m_(1)u_(1)^(2)=1/2((m_(1))/(alpha^(2))) v_(2)^(2)`
`v_(2)=sqrt(2)(alphau_(1))`
From Eqs ii and iii we get
`(v_(2))/(alphau_(1))=sqrt(2)=1+alphaimpliessqrt(2)-1`
`(u_(2))/(u_(1))=sqrt(2)-1`
`(u_(1))/(u_(2))=sqrt(2)+1`[on rotationaliting]
`(m_(2))/(m_(1))=1/(alpha^(2))=(1/(sqrt(2)-1))^(2)=(sqrt(2)+1)^(2)`
`(m_(2))/(m_(1))+3+2sqrt(2)`