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Chapter 9: Problem 97

Two astronauts, one of mass \(65 \mathrm{~kg}\) and the other \(85 \mathrm{~kg}\), are initially at rest in outer space. They then push each other apart. How far apart are they when the lighter astronaut has moved \(12 \mathrm{~m} ?\)

Short Answer

Expert verified
The astronauts are 21.18 m apart when the lighter one has moved 12 m.

Step by step solution

01

Understand the problem and given data

We are given the masses of two astronauts and the distance the lighter astronaut moves after they push each other in outer space. The question asks for the distance between the two astronauts at this point. Let the mass of the first astronaut be \( m_1 = 65 \text{ kg} \) and the mass of the second astronaut \( m_2 = 85 \text{ kg} \). The distance the first astronaut covers is \( 12 \text{ m} \).
02

Apply the conservation of momentum

In an isolated system like astronauts in space, momentum is conserved. Initially, both astronauts are at rest, so the initial momentum is zero. After they push each other, the momentum is still zero. If \( v_1 \) and \( v_2 \) are the velocities of astronauts 1 and 2 after the push, then:\[ m_1 \cdot v_1 + m_2 \cdot v_2 = 0 \]
03

Determine the relationship between velocities

From the conservation of momentum:\[ m_1 \cdot v_1 = -m_2 \cdot v_2 \]We can express \( v_2 \) in terms of \( v_1 \):\[ v_2 = -\frac{m_1}{m_2} \cdot v_1 \]
04

Apply the relationship to distances

The distance traveled by each astronaut is proportional to their velocities because time is the same for both. Let \( d_2 \) be the distance the second astronaut travels.From the relation between velocities, we have:\[ \frac{d_2}{d_1} = \frac{|v_2|}{|v_1|} = \frac{m_1}{m_2} \]Substituting \( d_1 = 12 \text{ m} \):\[ d_2 = \frac{65}{85} \times 12 \text{ m} \]
05

Calculate the distance \( d_2 \) and total distance

Calculate \( d_2 \):\[ d_2 = \frac{65}{85} \times 12 = \frac{780}{85} = 9.18 \text{ m} \]The total distance between the astronauts is \( d_1 + d_2 = 12 + 9.18 = 21.18 \text{ m} \).

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Key Concepts

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

Isolated System
An isolated system in physics is a concept where no external forces influence the objects within the system. This means that the total momentum within the system remains constant before and after any event like a collision or a push. In the context of our problem, the two astronauts and their interaction in space form an isolated system. There are no other forces acting on them: no air resistance, no friction, and no gravitational pulls from nearby objects. This isolation allows us to apply the conservation of momentum, a powerful tool that helps us predict the outcomes of such interactions.
Velocity
Velocity is a vector quantity that describes the speed of an object in a specific direction. In the context of our problem, after the astronauts push each other apart, each begins to move at a certain velocity. These velocities depend on their masses and how much force they exert on each other. For instance, if astronaut 1 with mass \( m_1 \) moves at velocity \( v_1 \), then astronaut 2 with mass \( m_2 \) will move at velocity \( v_2 \) in the opposite direction. Since momentum is conserved, the product of mass and velocity for each astronaut must balance each other out to keep the initial momentum of zero:
  • Formally, this is expressed as \( m_1 \cdot v_1 + m_2 \cdot v_2 = 0 \).
  • The negative sign indicates that the velocities are in opposite directions.
  • This relationship allows us to solve for unknowns like distances and velocities themselves.
Mass
Mass is a fundamental property of matter and is often mistaken simply for 'weight'. However, in physics, mass refers to the amount of matter in an object and does not change based on location. The mass of the astronauts directly influences their velocities after the push due to the conservation of momentum. Here:
  • Astronaut 1 has a mass \( m_1 = 65 \text{ kg} \).
  • Astronaut 2 has a mass \( m_2 = 85 \text{ kg} \).
  • Heavier objects (more mass) accelerate less, for the same amount of force applied.
This means that for the conservation of momentum, the heavier astronaut will move slower than the lighter one. This relationship crucially determines the distances each travels after the push.
Distance Calculation
Distance calculation stems from the understanding of velocity and the mass connected through momentum conservation. Once the velocities are related by their masses, you can determine the distances they travel. If time is the same for both astronauts since their motion starts and ends simultaneously from the push, then:
  • The distance \( d_1 \) covered by astronaut 1 is given as \( 12 \text{ m} \).
  • For astronaut 2, the distance \( d_2 \) is calculated using the ratio \( \frac{m_1}{m_2} \).
  • This gives \( d_2 = \left( \frac{65}{85} \right) \times 12 = 9.18 \text{ m} \).
  • The sum of the distances \( d_1 + d_2 = 21.18 \text{ m} \) gives the total separation.
Understanding the role of each variable in these calculations is key to predicting the outcome of isolated interactions.

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Most popular questions from this chapter

A meteor whose mass was about \(2.0 \times 10^{8} \mathrm{~kg}\) struck the Earth \(\left(m_{\mathrm{E}}=6.0 \times 10^{24} \mathrm{~kg}\right)\) with a speed of about \(25 \mathrm{~km} / \mathrm{s}\) and came to rest in the Earth. (a) What was the Earth's recoil speed (relative to Earth at rest before the collision)? (b) What fraction of the meteor's kinetic energy was transformed to kinetic energy of the Earth? (c) By how much did the Earth's kinetic energy change as a result of this collision?

(II) A mass \(m_{\mathrm{A}}=2.0 \mathrm{~kg},\) moving with velocity \(\overrightarrow{\mathbf{v}}_{\mathrm{A}}=\) \((4.0 \hat{\mathbf{i}}+5.0 \hat{\mathbf{j}}-2.0 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s},\) collides with mass \(m_{\mathrm{B}}=3.0 \mathrm{~kg}\) which is initially at rest. Immediately after the collision, mass \(m_{\mathrm{A}}\) is observed traveling at velocity \(\overrightarrow{\mathbf{v}}_{\mathrm{A}}^{\prime}=(-2.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s}\) Find the velocity of mass \(m_{\mathrm{B}}\) after the collision. Assume no outside force acts on the two masses during the collision.

(II) You drop a 12 -g ball from a height of \(1.5 \mathrm{~m}\) and it only bounces back to a height of \(0.75 \mathrm{~m}\). What was the total impulse on the ball when it hit the floor? (Ignore air resistance).

(II) A bullet of mass \(m=0.0010 \mathrm{~kg}\) embeds itself in a wooden block with mass \(M=0.999 \mathrm{~kg},\) which then compresses a spring \((k=120 \mathrm{~N} / \mathrm{m})\) by a distance \(x=0.050 \mathrm{~m}\) before coming to rest. The coefficient of kinetic friction between the block and table is \(\mu=0.50 .\) (a) What is the initial speed of the bullet? (b) What fraction of the bullet's initial kinetic energy is dissipated (in damage to the wooden block, rising temperature, etc.) in the collision between the bullet and the block? \(?\)

(I) An atomic nucleus at rest decays radioactively into an alpha particle and a smaller nucleus. What will be the speed of this recoiling nucleus if the speed of the alpha particle is \(2.8 \times 10^{5} \mathrm{~m} / \mathrm{s}\) ? Assume the recoiling nucleus has a mass 57 times greater than that of the alpha particle.
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