Answer:
860 cm/s
Explanation:
Let P = momentum of wrench-man system before throwing the wrench and P' = momentum of wrench-man system before throwing the wrench.
From the law of conservation of momentum,
P = P'
P = 0 (since both Astronaut and wrench were initially stationary) and P' = m₁v₁ + m₂v₂ where m₁ = mass of wrench = 4g = 0.004 kg v₁ = velocity of wrench after releasing the wrench = -15 m/s m₂ = mass of Astronaut = 70 kg and v₂ = velocity of Astronaut after releasing the wrench.
So, P = P'
0 = m₁v₁ + m₂v₂
-m₁v₁ = m₂v₂
dividing both sides by m₂, we have
v₂ = -m₁v₁/m₂
substituting in the values of the variables, we have
v₂ = -0.004 kg × (-15 m/s)/70 kg
= 0.06 kgm/s ÷ 70 kg
= 8.6 × 10⁻⁴ m/s
= 860 × 10⁻² m/s
= 860 cm/s
