Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

two 100 kg astronauts are floating in space. the first astronaut is moving at 5 m/s while the second is at rest. the two astronauts collide. both astronauts are moving at 3 m/s. based on the information provided, which of the following statements is true?

Two 100 Kg Astronauts Are Floating In Space The First Astronaut Is Moving At 5 Ms While The Second Is At Rest The Two Astronauts Collide Both Astronauts Are Mov class=

Sagot :

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.

We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.