Answer:
v₃ = 9.62[m/s]
Explanation:
To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.
We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.
Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.
In this way, we can construct the following equation.

where:
m₁ = mass of the astronaut = 157 [kg]
m₂ = mass of the toolkit = 5 [kg]
v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]
v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]
v₃ = velocity at which the toolkit should be thrown [m/s]
Now replacing:
![-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]](https://tex.z-dn.net/?f=-%28157%2B5%29%2A0.2%3D%28157%2A0.1%29-%285%2Av_%7B3%7D%29%5C%5C%285%2Av_%7B3%7D%29%3D%2015.7%2B32.4%5C%5Cv_%7B3%7D%3D9.62%5Bm%2Fs%5D)