Answer:
The velocity of the astronaut after throwing the wrench is 8.571 × 10⁻⁴ m/s
Step-by-step explanation:
The given parameters are;
The mass of the wrench the astronaut releases, m₁ = 4 grams = 0.004 kg
The speed with which the wrench is travelling, v₁ = -15 m/s
The mass of the astronaut, m₂ = 70 kg
The velocity of the astronaut after throwing the wrench = v₂
Whereby the astronaut and the wrench were initially at rest, we have the total initial momentum = 0 kg·m/s
The total final momentum = m₁×v₁ + m₂×v₂ = 0.004 × (-15) + 70 × v₂
By the conservation of momentum principle, we have;
The total initial momentum = The total final momentum
Which gives;
0 = 0.004 × (-15) + 70 × v₂
0.004 × (-15) = -70 × v₂
∴ v₂ = 0.004 × (-15)/-70 = 0.0008571
The velocity of the astronaut after throwing the wrench = v₂ = 0.0008571 m/s = 8.571 × 10⁻⁴ m/s