A 68.5 kg astronaut is doing a repair in space on the orbiting space station. He throws a 2.25 kg tool with a speed of 3.20 m / s relative to the space station. With what speed and direction will the astronaut begin to move?

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Initially, both the astronaut and the tool are at rest, so u₁ = u₂ = 0 m/s.

After throwing the tool, the tool has a velocity of v₂ = 3.20 m/s.

(68.5 kg) (0 m/s) + (2.25 kg) (0 m/s) = (68.5 kg) v + (2.25 kg) (3.20 m/s)

0 = (68.5 kg) v + 7.2 kg m/s

v = -0.105 m/s

The astronaut moves at a speed of 0.105 m/s in the opposite direction.

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west