In Space, an astronaut releases a wrench from his hand. The wrench has a mass of 4 grams and is traveling with a velocity of -15m/s. The Astronaut’s mass is 70kg. What is his Velocity?

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

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