(II) A 0.450-kg hockey puck, moving east with a speed of 5.80 m/s, has a head-on collision with a 0.900-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each puck after the collision?

The second puck moves with a speed of 3.867 m/s due east and the first puck moves with a speed of 1.933 m/s, but in the Western direction after collision.

Explanation:

Let the mass of the first hockey puck = m₁ = 0.45 kg

Mass of the resting puck = m₂ = 0.90 kg

Velocity of the first hockey puck = v₁ = 5.80î m/s; magnitude = 5.80 m/s

Velocity of resting puck = v₂ = 0 m/s

Mass of the two pucks are the same after collision.

Velocity of first puck after collision = v₃ = ?

Velocity of second puck after collision = v₄ = ?

In an elastic collision, both momentum and kinetic energy are conserved.

Momentum before collision = momentum after collision

Momentum before collision

m₁v₁ + m₂v₂ = (0.45 × 5.8) + (0.9 × 0) = 2.61 kgm/s

Momentum after collision

m₁v₃ + m₂v₄ = 0.45v₃ + 0.9v₄

So, according to the law of conservation of Momentum,

2.61 = 0.45v₃ + 0.9v₄ (eqn 1)

Kinetic energy before collision = kinetic energy after collision

Kinetic energy before collision = (m₁v₁²/2) + (m₂v₂²/2) = (0.45 × 5.8²/2) + (0.9 × 0²/2) = 7.569 J

Kinetic energy after collision = (m₁v₃²/2) + (m₂v₄²/2) = (0.45v₃²/2) + (0.9v₄²/2) = 0.225v₃² + 0.45v₄²

Since kinetic energy is conserved,

7.569 = 0.225v₃² + 0.45v₄² (eqn 2)

2.61 = 0.45v₃ + 0.9v₄ (eqn 1)

7.569 = 0.225v₃² + 0.45v₄² (eqn 2)

Make v₃ the subject of formula from eqn 1,

v₃ = (2.61 - 0.9v₄)/0.45 = (5.8 - 2v₄)

Substitute the value of v₃ into eqn 2

7.569 = 0.225(5.8 - 2v₄)² + 0.45v₄²

7.569 = 0.225(33.64 - 23.2v₄ + 4v₄²) + 0.45v₄²

7.569 = 7.569 - 5.22v₄ + 0.9v₄² + 0.45v₄²

1.35v₄² - 5.22v₄ = 0

v₄² - 3.867v₄ = 0

v₄ (v₄ - 3.867) = 0

v₄ = 0 m/s or v₄ = 3.867 m/s

If v₄ = 0 m/s,

v₃ = (5.8 - 2v₄) = [5.8 - 2(0)] = 5.8 m/s

If v₄ = 3.867 m/s

v₃ = (5.8 - 2v₄) = [5.8 - 2(3.867)] = - 1.933 m/s

The (v₃,v₄) answers are (5.8,0) and (-1.933,3.867). Since the collision was stated to be elastic, the realistic answer is the (-1.933, 3.867) answer.

So, the answer is interpreted as,

The second puck moves with a speed of 3.867 m/s due east and the first puck moves with a speed of 1.933 m/s, but in the Western direction after collision.

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