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A roller blader with a mass of 84 kg is traveling at 19 m/s [E]. A 21.2 kg child is riding a 7.8 kg bike at a speed of 6 m/s [W]. The roller blader and the cycling child collide in what can be assumed to be an elastic collision.

What are the final velocities of the roller blader and the cycling child after the collision?

Answer :

Final velocity of roller blader, v₁ = approximately 14.37 m/s [E], Final velocity of cycling child, v₂ = approximately 2.03 m/s [E]

To find the final velocities of the roller blader and the cycling child after an elastic collision, we can use the conservation of momentum and the conservation of kinetic energy.

Given Data:

  • Mass of the roller blader, m₁ = 84 kg
  • Initial velocity of the roller blader, u₁ = 19 m/s [E]
  • Mass of the child with bike, m₂ = 21.2 kg + 7.8 kg = 29 kg
  • Initial velocity of the cycling child, u₂ = -6 m/s [W] (negative since it is in the opposite direction)

Conservation of Momentum:

The total momentum before the collision must equal the total momentum after the collision:

[tex]m_{1} u_{1} +m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2}[/tex]

Where:

  • v₁ = final velocity of the roller blader
  • v₂ = final velocity of the cycling child

Conservation of Kinetic Energy:

Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision:

[tex]\frac{1}{2} m_{1} u_{1}^{2} +\frac{1}{2} m_{2} u_{2}^{2}= \frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}[/tex]

Step 1: Set up the equations

  1. Momentum Equation:
    84 × 19 +29 × (−6) =84v₁ + 29v₂ ​1596 − 174 =84v₁ + 29v₂​ 1422 =84v₁ +29v₂​ (Equation 1)

  2. Kinetic Energy Equation:
    1/2 × 84 × (19)²+ 1/2​ × 29 × (−6)² = 1/2 ​× 84v₁²​ + 1/2​ × 29v₂²

    1/2 ​× 84 × 361 + 1/2 ​× 29 × 36 = 21 ​× 84v₁²​ + 1/2 ​× 29v₂² ​15114 + 522 = 42v₁²​ + 14.5v₂² 15636 = 42v₁² ​+ 14.5v₂² (Equation 2)

Step 2: Solve the equations simultaneously

From Equation 1, express one variable in terms of the other:
29 v₂​=1422 − 84v₁

[tex]v_{2}=[/tex][tex]\frac{1422-84v_{1} }{29}[/tex]

Substituting v₂​ into Equation 2:
[tex]15636=42v_{1 }^{2} +14.5(\frac{1422-84v_{1} }{29})^{2}[/tex]

After solving this equation, you would find:

v₁ = 14.37 m/s [E], v₂ = 2.03 m/s [E]

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Rewritten by : Barada