A 4kg and 5kg bodies moving on a frictionless horizontal surface at a velocity of ( -6i )m/s and ( +3 )m/s respectively. Collide a head on elastic collision. What is the velocity ( magnitude and direction) of the each body after collision?

You'll notice that we have two missing variables: v₁' & v₂'. Assuming this is a perfectly elastic collision, we can use the conservation of kinetic energy to set the initial and final velocities of the individual bodies equal to each other.

v₁ + v₁' = v₂ + v₂'

Let's substitute all known variables into the first equation.

(4)(-6) + (5)(3) = (4)v₁' + (5)v₂'
-24 + 15 = 4v₁' + 5v₂'
-9 = 4v₁' + 5v₂'

Let's substitute the known variables into the second equation.

(-6) + v₁' = (3) + v₂'
-9 = -v₁' + v₂'
9 = v₁' - v₂'

Now we have a system of equations where we can solve for v₁ and v₂.

-9 = 4v₁' + 5v₂'
9 = v₁' - v₂'

Use the elimination method and multiply the bottom equation by -4.

-9 = 4v₁' + 5v₂'
-36 = -4v₁' + 4v₂'

Add the equations together.

-45 = 9v₂'
-5 = v₂'

The final velocity of the second body (5 kg) is -5 m/s. Substitute this value into one of the equations in the system to find v₁.

9 = v₁' - v₂'
9 = v₁' - (-5)
9 = v₁' + 5
4 = v₁'

The final velocity of the first body (4 kg) is 4 m/s.

We can verify our answer by making sure that the law of conservation of momentum is followed.

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
(4)(-6) + (5)(3) = (4)(4) + (5)(-5)
-24 + 15 = 16 - 25
-9 = -9

The combined momentum of the bodies before the collision is equal to the combined momentum of the bodies after the collision. [✓]