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Chapter 9: Problem 87

An \(80-\mathrm{kg}\) astronaut has become detached from the safety line connecting her to the International Space Station. She's \(200 \mathrm{m}\) from the station, at rest relative to it, and has 4 min of air remaining. To get herself back, she tosses a 10 -kg tool kit away from the station at \(8.0 \mathrm{m} / \mathrm{s} .\) Will she make it back in time?

Short Answer

Expert verified
Yes, the astronaut will make it back in time because it takes her 3.33 minutes to return and she has 4 minutes of air remaining.

Step by step solution

01

Calculate the Astronaut's Velocity

According to the law of conservation of momentum, the total momentum before and after tossing the toolkit remains constant. This can be represented as \(80 \, \text{kg} \times 0 \, \text{m/s}+ 10 \, \text{kg} \times 0 \, \text{m/s} = 80 \, \text{kg} \times v_a + 10 \, \text{kg} \times 8.0 \, \text{m/s}\), where \(v_a\) is the velocity of the astronaut. On solving, we get \(v_a=1 \, \text{m/s}\).
02

Calculate Time to Return

We need to find out the time taken for the astronaut to cover a distance of 200 m traveling at 1 m/s. This can be calculated using the formula: time = distance / speed. Hence, time = 200 / 1 = 200 seconds or 3.33 minutes.

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Most popular questions from this chapter

A block of mass \(m_{1}\) undergoes a one-dimensional elastic collision with an initially stationary block of mass \(m_{2} .\) Find an expression for the fraction of the initial kinetic energy transferred to the second block, and plot your result for mass ratios \(m_{1} / m_{2}\) from 0 to 20.

Wildlife biologists fire \(20-\mathrm{g}\) rubber bullets to stop a rhinoceros charging at \(0.81 \mathrm{m} / \mathrm{s}\). The bullets strike the rhino and drop vertically to the ground. The biologists' gun fires 15 bullets each second, at \(73 \mathrm{m} / \mathrm{s},\) and it takes \(34 \mathrm{s}\) to stop the rhino. (a) What impulse does each bullet deliver? (b) What's the rhino's mass? Neglect forces between rhino and ground.

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