`➢` Consider first a completely inelastic collision in one dimension. Then, in Fig. 6.10,
`color {blue}{θ_ 1 = θ _2 = 0}`
`m_1v_(1i) = (m_1+m_2)v_f` (momentum conservation)
`color {blue}{v_f =(m_1)/(m_1+m_2) v_(1i)}`................(6.23)
`➢` The loss in kinetic energy on collision is
`color{orange} {Δ K = 1/2 m_1v_(1i)^2 -1/2 (m_1 +m_2)v_f^2}`
`=1/2 m_1v_(1i)^2 -1/2 (m_1^2)/(m_1+m_2) v_(1i)^2` [using Eq. (6.23)]
`=1/2 m_1v_(1i)^2 [ 1- (m_1)/(m_1+m_2)]`
`=1/2 (m_1m_2)/(m_1+m_2) v_(1i)^2`
which is a positive quantity as expected.
`➢` Consider next an elastic collision. Using the above nomenclature with `θ_1 = θ_2 = 0`, the momentum and kinetic energy conservation
equations are
`color {blue}{m_1v_(1i)= m_1v_(1f)+m_2v_(2f)}`...........(6.24)
`color {blue}{m_1v_(1i)^2 =m_1v_(1f)^2 +m_2v_(2f)^2}`.............(6.25)
From Eqs. (6.24) and (6.25) it follows that,
`m_1v_(1i)(v_(2f)-v_(1i))=(m_1v_(1f) (v_(2f) -v_(1f))`
or `v_(2f)(v_(1i) -v_(1f))(v_(1i)+v_(1f))`
Hence,
`color {blue}{:. v_(2f) = v_(1f) + v_(1f)}`
...............(6.26)
Substituting this in Eq. (6.24), we obtain
`color {blue}{v_(1f)=((m_1-m_2))/(m_1+m_2) v_(1i)}`.............(6.27)
and `color {blue}{v_(2f)=(2m_1v_(1i))/(m_1+m_2)}`.............(6.28)
`➢` Thus, the unknowns `{v_(1f), v_(2f)}` are obtained in terms of the knowns `{m_1, m_2, v_(1i)}`.
`color{red} ►` Special casesof our analysis are interesting.
`color {blue}{{"Case I"}` : If the two masses are equal
`v_(1f)=0`
`color {blue}{v_(2f)=v_(1i)}`
`➢` The first mass comes to rest and pushes off the second mass with its initial speed on collision.
`color {blue}{{"Case II"}` : If one mass dominates, e.g. `m_2 > > m_1`
`color {blue}{v_(1f) ≃ v_(1i) quad v_(2f) ≃ 0}`
`➢` The heavier mass is undisturbed while the lighter mass reverses its velocity
`
`➢` Consider first a completely inelastic collision in one dimension. Then, in Fig. 6.10,
`color {blue}{θ_ 1 = θ _2 = 0}`
`m_1v_(1i) = (m_1+m_2)v_f` (momentum conservation)
`color {blue}{v_f =(m_1)/(m_1+m_2) v_(1i)}`................(6.23)
`➢` The loss in kinetic energy on collision is
`color{orange} {Δ K = 1/2 m_1v_(1i)^2 -1/2 (m_1 +m_2)v_f^2}`
`=1/2 m_1v_(1i)^2 -1/2 (m_1^2)/(m_1+m_2) v_(1i)^2` [using Eq. (6.23)]
`=1/2 m_1v_(1i)^2 [ 1- (m_1)/(m_1+m_2)]`
`=1/2 (m_1m_2)/(m_1+m_2) v_(1i)^2`
which is a positive quantity as expected.
`➢` Consider next an elastic collision. Using the above nomenclature with `θ_1 = θ_2 = 0`, the momentum and kinetic energy conservation
equations are
`color {blue}{m_1v_(1i)= m_1v_(1f)+m_2v_(2f)}`...........(6.24)
`color {blue}{m_1v_(1i)^2 =m_1v_(1f)^2 +m_2v_(2f)^2}`.............(6.25)
From Eqs. (6.24) and (6.25) it follows that,
`m_1v_(1i)(v_(2f)-v_(1i))=(m_1v_(1f) (v_(2f) -v_(1f))`
or `v_(2f)(v_(1i) -v_(1f))(v_(1i)+v_(1f))`
Hence,
`color {blue}{:. v_(2f) = v_(1f) + v_(1f)}`
...............(6.26)
Substituting this in Eq. (6.24), we obtain
`color {blue}{v_(1f)=((m_1-m_2))/(m_1+m_2) v_(1i)}`.............(6.27)
and `color {blue}{v_(2f)=(2m_1v_(1i))/(m_1+m_2)}`.............(6.28)
`➢` Thus, the unknowns `{v_(1f), v_(2f)}` are obtained in terms of the knowns `{m_1, m_2, v_(1i)}`.
`color{red} ►` Special casesof our analysis are interesting.
`color {blue}{{"Case I"}` : If the two masses are equal
`v_(1f)=0`
`color {blue}{v_(2f)=v_(1i)}`
`➢` The first mass comes to rest and pushes off the second mass with its initial speed on collision.
`color {blue}{{"Case II"}` : If one mass dominates, e.g. `m_2 > > m_1`
`color {blue}{v_(1f) ≃ v_(1i) quad v_(2f) ≃ 0}`
`➢` The heavier mass is undisturbed while the lighter mass reverses its velocity
`