If a particle elastically collides obliquely with a particle of same mass at rest then show that they move perpendicular to each other after collision in two dimension.

Explanation

After the collision, the rectangular components of the momentum of `m_(1)` are
(i) `m_(1)v_(1)costheta`, along positive x axis
(ii) `m_(1)v_(1)sintheta`, along positive y axis
After the collision, the rectangular components of the momentum of `m_(2)` are
(i) `m_(2)v_(2)costheta`, along positive x axis
`m_(2)v_(2)sintheta`, along positive y axis
Applying the law of conservation of momentum along x axis we get,
`m_(1)u_(1)=m_(1)v_(1)costheta_(1)+m_(2)v_(2)costheta_(2)" "...(1)`
The initial momentum of `m_(1)` or `m_(2)` along y axis is zero. Applying the principle of conservation of momentum along y axis we get,
`0=m_(1)thetasintheta_(1)+m_(2)v_(2)sintheta_(2)....(2)`
In this problem, a particle elastically collides obliquely with a particle of same mass at rest. So the two particles are identical.
`therefore m_(1)=m_(2)=m` (say)
By conservation of kinetic energy for elastic collision we get,
`(1)/(2)m u_(1)^(2)=(1)/(2)mv_(1)^(2)+(1)/(2)mv_(2)^(2)[becauseu_(2)=0]`
Dividing by `(1)/(2)m_(2)` we get
`therefore u_(1)=v_(1)^(2)+v_(2)^(2)" "...(3)`
By conservation of linear momentum we get
`mvecu_(1)=mvecv_(1)+mvecv_(2)`
Dividing by m we get
`vecu_(1)=vecv_(1)+vecv_(2)" "...(4)`
Then `vecu_(1).vecu_(1)=(vecv_(1)+vecv_(2))(vecv_(1)+vecv_(2))`
`=vecv_(1).vecv_(1)+vecv_(1).vecv_(2)+vecv_(2).vecv_(1)+vecv_(2).vecv_(2)`
or `u_(1)^(2)=v_(1)^(2)+v_(2)^(2)+2vecv_(1)vecv_(2)" "...(5)`
But from (3)
`v_(1)^(2)+v_(2)^(2)=u_(1)^(2)`
`therefore` Equation (5) becomes
`u_(1)^(2)=u_(1)^(2)+2vecv_(1)vecv_(2)`
`therefore vecv_(1)vecv_(2)=0`
This shows that the angle between `vecv_(1) and vecv_(2)` is `90^(@)`.
Hence two identical particles move perpendicular to each other after collision in two dimensions.