Discuss the completely inelastic collision in one dimension .

Explanation

Here particle of mass `m_(1)` is moving speed `v_(1f)` in x direction .
` :. theta_(1)=theta_(2) = 0 `
From the law of conservation of momentum , momentum before collision and momentum after collision is same .
` :. M_(1)v_(1f)=(m_(1)+m_(2))v_(1f) " "...(1)`
Energy is dissipated during collision , `DeltaK` = kinetic energy before collision - kinetic energy after collision
` 1/2 m_(1)v_(1)^(2)i - 1/2 (m_(1)+m_(2))v_(f)^(2) " " [ :. v_(2i)=0] `
Substituting the value of equation (1) ,
`DeltaK = 1/2m_(1)v_(1i)^(2)-1/2(m_(1)+m_(2))xx(m_(1)/(m_(1)+m_(2)))^(2)v_(1i)^(2)`
`=1/2m_(1)v_(1i)^(2)[1-(m_(1))/(m_(1)+m_(2))] `
` :. DeltaK = 1/2 m_(1)m_(2)v_(1i)^(2)` is positive .
Note If particles moves in the same direction and speed of particle of mass `m_(1) v_(1i)` and if `v_(1i) gt v_(2i)` them , after inelastic both having speed `v_(f)`
`v_(f) = (m_(1)v_(1i)+m_(2)v_(2i))/(m_(1)+m_(2))`
Kinetic energy before collision ,
` K_(i) = 1/2 m_(1)v_(1i)^(2) +1/2 m_(2)v_(2i)^(2)`
Kinetic energy after collision ,
`K_(f) =1/2 (m_(1)+m_(2))v_(f)^(2)`
` :. K_(f) = 1/2(m_(1)+m_(2))((m_(1)v_(1i)+m_(2)v_(2i))^(2))/((m_(1)+m_(2))^(2))`
`= 1/2((m_(1)v_(1i)+m_(2)v_(2i))^(2))/((m_(1)+m_(2)))`
` :. ` Energy dissipated after collision ,
`K_(i)-K_(f) =1/2 m_(1)v_(1i)^(2)+1/2m_(2)v_(2i)^(2)`
`-1/2(-m_(1)v_(1i)+m_(2)v_(2i))^(2)/(m_(1)+m_(2))`
`{m_(1)^(2)v_(1i)^(2)+m_(1)m_(2)v_(2i)^(2)+m_(1)m_(2)v_(1i)^(2)+m_(2)^(2)v_(2i)^(2)=1/2} (-(m_(1)^(2)v_(1i)^(2)+2m_(1)m_(2)v_(1i)v_(2i)+m_(2)^(2)v_(2i)^(2))/((m_(1)+m_(2))))`
`m_(1)m_(2)v_(2i)^(2)+m_(1)m_(2)v_(1i)^(2)=1/2((-2m_(1)m_(2)v_(1i)v_(2i)+v_(2i)^(2)))/(m_(1)+m_(2))`
`=(m_(1)m_(2))/(2(m_(1)+m_(2)))(v_(1i)^(2)-2v_(1i)^(2)-2v_(1i)v_(2i)+v_(2i)^(2))`
`=(m_(1)m_(2))/(2(m_(1)+m_(2)))(v_(1i)^(2)-2v_(1i)v_(2i)+v_(2i)^(2))`
`=(m_(1)m_(2))/(2(m_(1)+m_(2)))[v_(1i)-v_(2i)]^(2)`.
Hence ,kinetic energy is conserved in inelastic collision .