/*! 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 5 An 80 -kg astronaut becomes sepa... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An 80 -kg astronaut becomes separated from his spaceship. He is \(15.0 \mathrm{~m}\) away from it and at rest relative to it. In an effort to get back, he throws a \(500-\mathrm{g}\) object with a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) in a direction away from the ship. How long does it take him to get back to the ship? a) \(1 \mathrm{~s}\) b) \(10 \mathrm{~s}\) c) \(20 \mathrm{~s}\) d) \(200 \mathrm{~s}\) e) \(300 \mathrm{~s}\)

Short Answer

Expert verified
Answer: e) 300 seconds

Step by step solution

01

Identify the given values

In this problem, we have the following given values: - Mass of astronaut (M): 80 kg - Mass of the object (m): 500 g = 0.5 kg (converted to kg) - Initial distance from the ship (d): 15.0 m - Velocity of the thrown object (v): 8.0 m/s
02

Apply the conservation of momentum

According to the conservation of momentum, the momentum before throwing the object is equal to the momentum after throwing it away. Let's represent the astronaut's initial velocity as V_i and final velocity as V_f, and the final velocity of the object as v_f. Since both astronaut and object are initially at rest, their initial velocities are 0. Initially, Total momentum = M * V_i + m * V_i = 0 (because V_i = 0) After throwing, Total momentum = M * V_f + m * v_f By conservation of momentum, these two momenta are equal: 0 = M * V_f + m * v_f Now, we can solve this equation to find V_f, the final velocity of the astronaut.
03

Find the astronaut's final velocity

We have the equation 0 = M * V_f + m * v_f. Now we can substitute the given values and solve for V_f. 0 = (80 kg) * V_f + (0.5 kg) * (8.0 m/s) Rearranging the terms to isolate V_f: V_f = - (0.5 kg * 8.0 m/s) / 80 kg Now, we can calculate V_f: V_f = - (4.0 kg * m/s) / 80 kg = -0.05 m/s The negative sign means the astronaut will move towards the spaceship, which is the desired direction.
04

Calculate the time taken to reach the spaceship

Now that we have the final velocity of the astronaut (V_f = -0.05 m/s), we can use it to find the time (t) it will take for him to cover the 15.0 m distance back to the ship. Using the linear motion equation with constant velocity: distance = velocity * time We can rearrange this equation to find the time: t = distance / velocity Now we substitute the known values: t = 15.0 m / -0.05 m/s The negative sign in the velocity will cancel out, because the distance is given in positive values and astronaut is moving towards the spaceship, as expected. Thus, we calculate the time as: t = 15.0 m / 0.05 m/s = 300 s Thus, it takes the astronaut 300 seconds to reach the spaceship. The correct answer is: e) \(300\,\text{s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the center of mass of a rectangular plate of length \(20.0 \mathrm{~cm}\) and width \(10.0 \mathrm{~cm} .\) The mass density varies linearly along the length. At one end, it is \(5.00 \mathrm{~g} / \mathrm{cm}^{2}\); at the other end, it is \(20.0 \mathrm{~g} / \mathrm{cm}^{2}\)

A catapult on a level field tosses a 3 -kg stone a horizontal distance of \(100 \mathrm{~m}\). A second 3 -kg stone tossed in an identical fashion breaks apart in the air into 2 pieces, one with a mass of \(1 \mathrm{~kg}\) and one with a mass of \(2 \mathrm{~kg} .\) Both of the pieces hit the ground at the same time. If the 1 -kg piece lands a distance of \(180 \mathrm{~m}\) away from the catapult, how far away from the catapult does the 2 -kg piece land? Ignore air resistance. a) \(20 \mathrm{~m}\) c) \(100 \mathrm{~m}\) e) \(180 \mathrm{~m}\) b) \(60 \mathrm{~m}\) d) \(120 \mathrm{~m}\)

A man with a mass of 55 kg stands up in a \(65-\mathrm{kg}\) canoe of length \(4.0 \mathrm{~m}\) floating on water. He walks from a point \(0.75 \mathrm{~m}\) from the back of the canoe to a point 0.75 m from the front of the canoe. Assume negligible friction between the canoe and the water. How far does the canoe move?

A spacecraft engine creates \(53.2 \mathrm{MN}\) of thrust with a propellant velocity of \(4.78 \mathrm{~km} / \mathrm{s}\). a) Find the rate \((d m / d t)\) at which the propellant is expelled. b) If the initial mass is \(2.12 \cdot 10^{6} \mathrm{~kg}\) and the final mass is \(7.04 \cdot 10^{4} \mathrm{~kg},\) find the final speed of the spacecraft (assume the initial speed is zero and any gravitational fields are small enough to be ignored). c) Find the average acceleration till burnout (the time at which the propellant is used up; assume the mass flow rate is constant until that time).

A 750 -kg cannon fires a 15 -kg projectile with a speed of \(250 \mathrm{~m} / \mathrm{s}\) with respect to the muzzle. The cannon is on wheels and can recoil with negligible friction. Just after the cannon fires the projectile, what is the speed of the projectile with respect to the ground?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.