Question

4. An astronaut at rest in space with mass 84 kg fires a thruster that expels 35 g of hot gas at 875 m/s. What is the velocity of the astronaut after firing the thruster?

Answer 1

The impulse (change in momentum) experienced by the astronaut is given by the product of the force applied and the time over which it is applied. The force applied is given by the rate at which mass is expelled multiplied by the velocity at which it is expelled, which is (35 g * 875 m/s) = 30.625 N.

The impulse experienced by the astronaut is equal to the change in momentum, which is given by the initial mass multiplied by the final velocity, so we can set up the following equation:

m_initial * v_initial = m_final * v_final

where m_initial = 84 kg (mass of astronaut) and m_final = 84 kg (mass of astronaut + 35 g of hot gas).

Solving for v_final, we get:

v_final = m_initial * v_initial / m_final

= 84 kg * 0 m/s / 84 kg

= 0 m/s

So the astronaut's velocity does not change after the thruster is fired because the velocity of the expelled gas is in the opposite direction to the astronaut's motion, they are canceling out.