Question

An astronaut of mass 210 kg including his suit and jet pack wants to acquire a velocity of\(2.0\;{\rm{m/s}}\)to move back toward his space shuttle. Assuming the jet pack can eject gas with a velocity of\(35\;{\rm{m/s}}\)what mass of gas will need to be ejected?

Short answer

The mass of the gas that should be ejected is 12 kg.

Step-by-step solution

Identification of the given data

The mass of the astronaut is\({m_1} = 210\;{\rm{kg}}\).

The velocity of the astronaut with respect to the space shuttle is\({v_1} = - 2\;{\rm{m/s}}\).

The eject velocity of the gas is\({v_2} = 35\;{\rm{m/s}}\).

Statement of the principle of conservation of linear momentum

The principle of conservation of linear momentum states that if two bodies collide with each other, the total linear momentum before and after the collision remains the same if no external force acts on the system.

\(\begin{aligned}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\{\left( {mv} \right)_{{\rm{before}}}} = {\left( {mv} \right)_{{\rm{after}}}}\end{aligned}\)

Determination of the mass of the gas to be ejected

Let the mass of the gas that needs to be ejected by the jet pack be\({m_2}\). The initial speed of both the astronaut and the gas is zero before coming out of the space shuttle.

Apply the law of conservation of linear momentum.

\(\begin{aligned}0 &= {m_1}{v_1} + {m_2}{v_2}\\0 &= \left( {210\;{\rm{kg}}} \right)\left( { - 2\;{\rm{m/s}}} \right) + {m_2}\left( {35\;{\rm{m/s}}} \right)\\{m_2} &= \frac{{420\;{\rm{kg}}\;{\rm{m/s}}}}{{35\;{\rm{m/s}}}}\\ &= 12\;{\rm{kg}}\end{aligned}\)