Particle 1 moving with velocity `v=10m//s` experienced a head-on collision with a stationary particle `2` of the same mass. As a result of the collision, the kinetic energy of the system decreased by `eta=1.0%`. Find the magnitude and direction of the velocity of particle 1 after the collision.

Explanation

Since, the collision is head on, the particle 1 will continue moving along the same line as before the collision, but there will be a change in the magnitude of it's velocity vector.
Let it starts moving with velocity `v_1` and particle 2 with `v_2` after collision, then from the conservation of momentum
`m u=mv_1+mv_2` or, `u=v_1+v_2` (1)
And from the condition, given,
`eta=(1/2m u^2-(1/2mv_1^2+1/2mv_2^2))/(1/2 m u^2)=1-(v_1^2+v_2^2)/(u^2)`
or, `v_1^2+v_2^2=(1-eta)u^2` (2)
From (1) and (2),
`v_1^2+(u-v_1)^2=(1-eta)u^2`
or, `v_1^2+u^2-2uv_1+v_1^2=(1-eta)u^2`
or, `2v_1^2-2v_1u+etau^2=0`
So, `v_1=2u+-(sqrt(4u^2-8etau^2))/(4)`
`=1/2[u+-sqrt(u^2-2etau^2)]=1/2u(1+-sqrt(1-2eta))`
Positive sign gives the velocity of the 2nd particle which lies ahead. The negative sign is correct for `v_1`.
So, `v_1=1/2u(1-sqrt(1-2eta))=5m//s` will continue moving in the same direction.
Note that `v_1=0` if `eta=0` as it must.