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A 68.5-kg astronaut is doing a repair in space on the orbiting space station. She throws a 2.25-kg tool away from her at 3.20 m/s relative to the space station. With what speed and in what direction will she begin to move?

Short Answer

Expert verified
The astronaut moves at 0.105 m/s in the opposite direction of the thrown tool.

Step by step solution

01

Understanding Conservation of Momentum in Space

In space, when an object is thrown away, there is no air resistance or other forces to consider, except for the conservation of momentum. The momentum before and after a tool is thrown must be equal because momentum is conserved. The momentum is given by the product of mass and velocity, expressed as \( p = m \times v \).
02

Calculating Initial Momentum

Initially, both the astronaut and the tool are stationary relative to the space station, so their combined momentum is \( 0 \). This means the initial momentum of both the tool and astronaut together is zero: \( p_{initial} = 0 \).
03

Calculating Final Momentum

After the tool is thrown, the momentum of the system must remain zero due to conservation: \( 0 = m_{astronaut} \cdot v_{astronaut} + m_{tool} \cdot v_{tool} \). Here, \( m_{astronaut} = 68.5 \) kg, \( m_{tool} = 2.25 \) kg, and \( v_{tool} = 3.20 \) m/s.
04

Solving for the Astronaut's Velocity

Rearrange the momentum conservation equation to find the speed of the astronaut: \( 0 = 68.5 \cdot v_{astronaut} + 2.25 \cdot 3.20 \). Solving for \( v_{astronaut} \), we get \( 68.5 \cdot v_{astronaut} = -2.25 \cdot 3.20 \), which simplifies to \( v_{astronaut} = -\frac{2.25 \times 3.20}{68.5} \). Calculating this gives \( v_{astronaut} \approx -0.105 \) m/s.
05

Determining the Direction

The negative sign in the calculation indicates that the astronaut moves in the opposite direction to the tool. Therefore, the astronaut will move backward relative to the direction in which the tool is thrown.

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Key Concepts

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

Momentum Calculation
Momentum is a fundamental concept in physics that describes the motion of objects. It is the product of an object's mass and its velocity, expressed with the formula: \[ p = m \times v \]where:
  • \( p \) is momentum,
  • \( m \) is mass, and
  • \( v \) is velocity.
In the provided exercise, the astronaut initially has no momentum because neither she nor the tool she is holding is moving. This is why we say the initial momentum is zero. After the tool is thrown, however, momentum is transferred from the astronaut to the tool. In such scenarios, conservation of momentum tells us that the total momentum before the event (throwing the tool) is equal to the total momentum after the event. This principle allows us to calculate unknown variables, such as the astronaut's velocity, using the conservation of momentum equation:\[ 0 = m_{astronaut} \times v_{astronaut} + m_{tool} \times v_{tool} \]By rearranging this formula, solving for the astronaut's final velocity becomes straightforward. Here, the computation shows that the astronaut's velocity is negative, indicating the direction opposite to which the tool is thrown.
Space Physics
Space physics deals with phenomena that occur in the expanse outside Earth's atmosphere, where traditional forces like air resistance are virtually non-existent. One pivotal aspect of space physics is the isolation from external forces that we typically contend with on Earth. This is most apparent in the conservation laws, such as the conservation of momentum. Within the vacuum of space, mechanical energy and momentum are particularly easy to observe because external interferences are absent. In this context, when the astronaut throws the tool, the absence of air resistance and friction means the only forces acting are those of the astronaut and the tool, defined by their interaction. Such an environment makes it an excellent setting to study pure conservation laws and observe them in action without additional factors complicating the scenario. Space also serves as a neutral playing field where experiments and practical tasks can reveal insights into how these laws work in isolation. For instance, when the astronaut throws the tool, the reaction propels her in the opposite direction, beautifully illustrating conservation concepts that sometimes seem abstract on Earth.
Newton's Laws
Newton’s laws of motion are the fundamental principles describing the relationship between a body and the forces acting on it, crucial to understanding physics and space dynamics. Each of the three laws plays a role in the scenario of the astronaut and the tool:
  • First Law (Inertia): An object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. Here, both the astronaut and the tool are initially at rest, and motion begins only once the tool is pushed.
  • Second Law (F=ma): The force exerted on an object is equal to the mass of the object times its acceleration. Although not calculated explicitly in this scenario, the forces here are internal but signify action causing motion, consistent with Newton's second law.
  • Third Law (Action-Reaction): For every action, there is an equal and opposite reaction. This principle directly answers why the astronaut moves when she throws the tool: the force used to throw the tool causes an equal and opposite reaction, pushing her in the opposite direction.
Each of these laws is demonstrated exquisitely when the astronaut throws her tool in space, allowing us to see Newtonian physics in its purest form, unaffected by friction or air resistance.

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Most popular questions from this chapter

In Section 8.5 we calculated the center of mass by considering objects composed of a \(finite\) number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$x_{cm} = {1\over M}\int x \space dm \space y_{cm} = {1 \over M}\int y \space dm$$ where \(x\) and \(y\) are the coordinates of the small piece of the object that has mass \(dm\). The integration is over the whole of the object. Consider a thin rod of length \(L\), mass \(M\), and cross-sectional area \(A\). Let the origin of the coordinates be at the left end of the rod and the positive \(x\)-axis lie along the rod. (a) If the density \(\rho = M/V\) of the object is uniform, perform the integration described above to show that the \(x\)-coordinate of the center of mass of the rod is at its geometrical center. (b) If the density of the object varies linearly with \(x-\)that is, \(\rho = ax\), where a is a positive constant\(-\)calculate the \(x\)-coordinate of the rod's center of mass.

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