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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?

Sagot :

Supernova

Answer:

4 kg → +4 m/s

5 kg → -5 m/s

Explanation:

The law of conservation of momentum states that:

  • m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
  • left side → velocities before collision
  • right side → velocities 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. [✓]

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