An astronaut becomes stranded during a space walk after her jet pack malfunctions. Fortunately, there are two objects close to her that she can push to propel herself back to the International Space Station (ISS). Object A has the same mass as the astronaut, and Object \(\mathrm{B}\) is 10 times more massive. To achieve a given momentum toward the ISS by pushing one of the objects away from the ISS, which object should she push? That is, which one requires less work to produce the same impulse? Initially, the astronaut and the two objects are at rest with respect to the ISS.

The astronaut and the two objects are initially at rest. The mass of Object A is equal to the mass of the astronaut, and the mass of Object B is 10 times greater than the mass of Object A (and thus the astronaut). Let the mass of the astronaut be \(m\).

The impulse experienced by the astronaut will equal the change in her momentum. The impulse-momentum theorem states that the impulse (\(J\)) equals the product of the average force (\(F_{avg}\)) and the time interval (\(\Delta t\)) during which the force was applied: $$ J = F_{avg} * \Delta t $$

In both cases, the astronaut applies the same force on both objects. The force applied on Object A (\(F_A\)), its final velocity (\(v_{A'}\)), and the final velocity of the astronaut (\(v_{A}\)) to give the same impulse are: $$ F_A * \Delta t_A = m * (v_{A'} - v_{A}) $$ The force applied on Object B (\(F_B\)), its final velocity (\(v_{B'}\)), and the final velocity of the astronaut (\(v_{B}\)) to give the same impulse are: $$ F_B * \Delta t_B = m * (v_{B'} - v_{B}) $$

The work-energy theorem states that the work done (\(W\)) on an object is equal to the change in its kinetic energy: $$ W = \Delta KE = \frac{1}{2} m * (v^2 - u^2) $$ Here, \(v\) is the final velocity, and \(u\) is the initial velocity of the astronaut.

The work done on the astronaut when pushing Object A is: $$ W_A = \frac{1}{2} m * (v_A^2 - 0^2) $$ The work done on the astronaut when pushing Object B is: $$ W_B = \frac{1}{2} m * (v_B^2 - 0^2) $$

Since the astronaut and Object A have the same mass, and the force applied is the same, their velocities after pushing will be equal but in opposite directions (from Newton's third law of motion). Thus, the work done (\(W_A\)) in this case will be: $$ W_A = \frac{1}{2} m * (v_{A'}^2 - 0^2) $$ In the second case, Object B is 10 times more massive than the astronaut. When she pushes Object B with the same force, the velocity change will be divided by 10 for Object B, while the astronaut's velocity will be 10 times greater due to the difference in masses. Thus, the work done (\(W_B\)) in this case will be: $$ W_B = \frac{1}{2} m * (10 * v_{B'}^2 - 0^2) $$ Comparing the two cases, we see that: $$ W_B = 10 * W_A $$ Since the work done on the astronaut in the first case (\(W_A\)) is 10 times smaller than the work done in the second case (\(W_B\)), she should choose to push Object A to require less work to produce the same impulse.