Answer:
The true statement is;
Neither momentum or kinetic energy is conserved
Explanation:
The question relates to the verification of the conservation of linear momentum, and kinetic energy
The given parameters are;
The mass of each astronaut = 100 kg
From which, we have;
The mass of the moving astronaut, m₁ = 100 kg
The mass of the stationary astronaut, m₂ = 100 kg
The initial velocity of the moving astronaut, v₁ = 5 m/s
The initial velocity of the stationary astronaut, v₂ = 0 m/s
The final velocity of both astronauts, v₃ = 3 m/s
The sum of the initial momentum of both astronauts is given as follows;
[tex]P_{initial}[/tex] = m₁·v₁ + m₂·v₂ = 100 kg × 5 m/s + 100 kg × 0 m/s = 500 kg·m/s
[tex]P_{initial}[/tex] = 500 kg·m/s
The sum of the final momentum of the astronauts is given as follows;
[tex]P_{final}[/tex] = m₁·v₃ + m₂·v₃ = (m₁ + m₂) × v₃ = (100 kg + 100 kg) × 3 m/s = 600 kg·m/s
[tex]P_{final}[/tex] = 600 kg·m/s
∴ [tex]P_{initial}[/tex] = 500 kg·m/s ≠ [tex]P_{final}[/tex] = 600 kg·m/s
[tex]P_{initial}[/tex] < [tex]P_{final}[/tex], therefore, the sum of the linear momentum of both astronauts is not conserved
The sum of the initial kinetic energy of each astronaut is given as follows;
[tex]K.E._{initial}[/tex] = 1/2·m₁·v₁² + 1/2·m₂·v₂² = 1/2 × 100 kg × (5 m/s)² + 1/2 × 100 kg × (0 m/s)² = 1250 Joules
[tex]K.E._{initial}[/tex] = 1250 Joules
The sum of the final kinetic energy of the astronaut is given as follows;
[tex]K.E._{final}[/tex] = 1/2·m₁·v₃² + 1/2·m₂·v₃² = 1/2 × 100 kg × (3 m/s)² + 1/2 × 100 kg × (3 m/s)² = 900 joules
[tex]K.E._{final}[/tex] = 900 joules
[tex]K.E._{initial}[/tex] > [tex]K.E._{final}[/tex], therefore, the kinetic energy is not conserved
From which we get that neither momentum or kinetic energy is conserved.