Question

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

a rocket-propelled F-4 Phantom jet aircraft was crashed head-on into a
concrete barrier at high speed in Sandia, New Mexico, on April 19, 1988. The
F-4 phantom had a mass of 19,100 kg, while the retainment wall's mass was
469,000 kg. The wall sat on a cushion of air that allowed it to move during
impact. If the wall and F-4 moved together at 8.41 m/s during the collision,
what was the speed of the F-4 Phantom upon impact?

Answer 1

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

Step-by-step 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.