Astronaut question Help Physics

Question

An astronaut with a mass astr = 65 kg and wearing a tool belt full of tools that have a combined mass of 15 kg (so initial mass is 80 kg) is floating beside a space station 22 meters away. The safety line has been cut by someone closing a door and catching the line in the door.

 

How fast would the astronaut have to throw a hammer of mass 1.2 kg to obtain the same speed as when the wrench was thrown in part c?

 

here is part C: 

Assume the astronaut and tools are initially stationary. To get back to the spaceship, the astronaut throws a small wrench of masswrench = .40 kg away from the space station with a velocity of 28 m/s.

c) What will the final velocity of the astronaut be after the throw?

= 0.141 m/s .

Answer 1

Initial state:

 

m = 80 kg

v_i = 0

p_i = 0

 

Final state:

 

Δm = mass of hammer = 1.2 kg

m' = final astronaut total mass = m - Δm = (80 - 1.2) kg

v_f = final speed of astronaut = 0.141 m/s  (given)

u = speed of hammer in direction away from space station

p_f = (m - Δm)v_f - (Δm)u

 

Conservation of linear momentum:

 

p_i = p_f

 

0 = (m - Δm)v_f - (Δm)u

 

u = (m - Δm)v_f / Δm = [(80 - 1.2) kg] (0.141 m/s) / (1.2 kg) = 9.259 m/s