/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 An 80.0 -kg astronaut is working... [FREE SOLUTION] | 91Ó°ÊÓ

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An 80.0 -kg astronaut is working on the engines of his ship, which is drifting through space with a constant velocity. The astronaut, wishing to get a better view of the Universe, pushes against the ship and much later finds himself \(30.0 \mathrm{m}\) behind the ship. Without a thruster, the only way to return to the ship is to throw his \(0.500-\mathrm{kg}\) wrench directly away from the ship. If he throws the wrench with a speed of \(20.0 \mathrm{m} / \mathrm{s}\) relative to the ship, how long does it take the astronaut to reach the ship?

Short Answer

Expert verified
It takes the astronaut 240 seconds to reach the spaceship.

Step by step solution

01

Calculate Initial Momentum

According to the principle of conservation of linear momentum, the initial momentum of the astronaut-wrench system is \(0 \mathrm{kg} \cdot \mathrm{m/s}\) since the system's velocity is zero.
02

Calculate Final Momentum

After the astronaut throws the wrench, it gains momentum. This momentum is given as \(p_{\mathrm{wrench}} = m_{\mathrm{wrench}} \cdot v_{\mathrm{wrench}} = 0.500 \mathrm{kg} \cdot 20.0 \mathrm{m/s} = 10.0 \mathrm{kg\cdot m/s}\) in the direction away from the ship. According to conservation of linear momentum, the astronaut will gain momentum equal in magnitude, but opposite in direction. Let's denote the astronaut's velocity as \(v_{\mathrm{astro}}\). Therefore the momentum of the astronaut is \(p_{\mathrm{astro}} = m_{\mathrm{astro}} \cdot v_{\mathrm{astro}} = 80.0 \mathrm{kg} \cdot v_{\mathrm{astro}} = -10.0 \mathrm{kg\cdot m/s}\)
03

Find the velocity of the astronaut

Knowing the astronaut's momentum allows to find his velocity as \(v_{\mathrm{astro}} = \frac{p_{\mathrm{astro}}}{m_{\mathrm{astro}}} = \frac{-10.0 \mathrm{kg\cdot m/s}}{80.0 \mathrm{kg}} = -0.125 \mathrm{m/s}\). The negative sign indicates that the astronaut's direction is opposite to the direction of the tossed wrench.
04

Calculate Time

Using the formula \(t = \frac{d}{v}\) the time it takes for the astronaut to reach the spaceship can be calculated: \( t = \frac{30.0 \mathrm{m}}{0.125 \mathrm{m/s}} = 240 \mathrm{s}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Momentum
Linear momentum is a measure of how much 'motion' an object has and can be thought of as the object's tendency to keep moving at a constant speed and direction. It's a vector quantity, meaning it has both magnitude and direction, and is calculated by multiplying an object's mass by its velocity. Mathematically, linear momentum (\textbf{p}) is expressed as \[ \textbf{p} = m \times \textbf{v} \]
where \(m\) is mass and \(\textbf{v}\) is velocity. In the context of a drifting astronaut, as we see in the original exercise, the astronaut's linear momentum is directly related to how much effort (or impulse) they must exert to change their current state of motion, such as moving back towards the ship after being separated from it.
Momentum-Velocity Relationship
The momentum-velocity relationship connects the mass of an object with its velocity to define momentum. When an astronaut in space throws an object in one direction, they will move in the opposite direction. This intuitive concept aligns with Newton's third law: For every action, there's an equal and opposite reaction.
\[ \textbf{p} = m \times \textbf{v} \]
This equation tells us that for a given mass, an increase in velocity will result in a proportionally larger momentum and vice versa.

In our astronaut's scenario, by throwing the wrench with a certain velocity, the astronaut alters their momentum. Since their mass is constant, the change in momentum is due to the change in velocity. This relationship helps us to calculate the velocity of the astronaut after they've tossed the wrench.
Momentum Conservation in Space
Momentum conservation is one of the fundamental principles of physics, stating that if no external forces are acting on a system, the total momentum of that system remains constant. In the vacuum of space, where external forces like gravity or air resistance are negligible, conservation of momentum ensures that an astronaut's actions, like throwing a wrench, will have predictable results.

The astronaut's situation depicted in the exercise demonstrates this principle. When the astronaut throws the wrench, the total momentum before and after the throw must remain zero (assuming the astronaut and wrench system was at rest initially). With the known mass and velocity of the wrench, we can find the velocity at which the astronaut must move in order to conserve momentum. This velocity then allows us to calculate how long it would take the astronaut to return to the ship, showing a practical application of momentum conservation.

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Most popular questions from this chapter

A billiard ball moving at \(5.00 \mathrm{m} / \mathrm{s}\) strikes a stationary ball of the same mass. After the collision, the first ball moves, at \(4.33 \mathrm{m} / \mathrm{s},\) at an angle of \(30.0^{\circ}\) with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.

High-speed stroboscopic photographs show that the head of a golf club of mass \(200 \mathrm{g}\) is traveling at \(55.0 \mathrm{m} / \mathrm{s}\) just before it strikes a 46.0 -g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at \(40.0 \mathrm{m} / \mathrm{s} .\) Find the speed of the golf ball just after impact.

The mass of the Earth is \(5.98 \times 10^{24} \mathrm{kg},\) and the mass of the Moon is \(7.36 \times 10^{22} \mathrm{kg} .\) The distance of separation, measured between their centers, is \(3.84 \times 10^{8} \mathrm{m} .\) Locate the center of mass of the Earth-Moon system as measured from the center of the Earth.

An object of mass \(3.00 \mathrm{kg},\) moving with an initial velocity of \(5.00 \mathrm{i} \mathrm{m} / \mathrm{s},\) collides with and sticks to an object of mass \(2.00 \mathrm{kg}\) with an initial velocity of \(-3.00 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s} .\) Find the final velocity of the composite object.

A 10.0 -g bullet is fired into a stationary block of wood \((m=\) \(5.00 \mathrm{kg}) .\) The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is \(0.600 \mathrm{m} / \mathrm{s} .\) What was the original speed of the bullet?

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