Question

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?

Answer 1

The true statement is;

Neither momentum or kinetic energy is conserved

Step-by-step 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;

`P_(initial)` = m₁·v₁ + m₂·v₂ = 100 kg × 5 m/s + 100 kg × 0 m/s = 500 kg·m/s

`P_(initial)` = 500 kg·m/s

The sum of the final momentum of the astronauts is given as follows;

`P_(final)` = m₁·v₃ + m₂·v₃ = (m₁ + m₂) × v₃ = (100 kg + 100 kg) × 3 m/s = 600 kg·m/s

`P_(final)` = 600 kg·m/s

∴
`P_(initial)` = 500 kg·m/s ≠
`P_(final)` = 600 kg·m/s

`P_(initial)` <
`P_(final)`, 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;

`K.E._(initial)` = 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

`K.E._(initial)` = 1250 Joules

The sum of the final kinetic energy of the astronaut is given as follows;

`K.E._(final)` = 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

`K.E._(final)` = 900 joules

`K.E._(initial)` >
`K.E._(final)`, therefore, the kinetic energy is not conserved

From which we get that neither momentum or kinetic energy is conserved.