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3 years ago

To test the strength of a retainment wall designed to protect a nuclear reactor,

Physics

1 answer:

Hatshy [7]3 years ago

6 0

Answer:

The speed of the F-4 Phantom upon impact is approximately 214.92 m/s

Explanation:

The given parameter of the motion of the F-4 Phantom jet aircraft are;

The mass of the F-4 phantom rocket, m₁ = 19,100 kg

The mass of the retaining wall, m₂ = 469,000 kg

The velocity of the combined mass of the wall and the F-4 after collision, v₃ = 8.41 m/s

The retainment wall was initially at rest, therefore, v₂ = 0 m/s

Let 'v₁', represent the speed of the F-4 Phantom upon impact

By the principle of conservation of linear momentum, we have;

m₁·v₁ + m₁·v₂ = m₁·v₃ + m₂·v₃ = (m₁ + m₂)·v₃

Plugging in the values, we have;

19,100 × v₁ + 469,000 × 0 = (19,100 + 469,000) × 8.41

∴ v₁ = (19,100 + 469,000) × 8.41/(19,100) = 214.917329843

v₁ ≈ 214.92

The speed of the F-4 Phantom upon impact, v₁ ≈ 214.92 m/s.

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