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Chapter 8: Problem 30

An astronaut in space cannot use a conventional means, such as a scale or balance, to determine the mass of an object. But she does have devices to measure distance and time accurately. She knows her own mass is 78.4 kg, but she is unsure of the mass of a large gas canister in the airless rocket. When this canister is approaching her at 3.50 m/s, she pushes against it, which slows it down to 1.20 m/s (but does not reverse it) and gives her a speed of 2.40 m/s. What is the mass of this canister?

Short Answer

Expert verified
The mass of the canister is approximately 81.8 kg.

Step by step solution

01

Understand the Law of Conservation of Momentum

The law of conservation of momentum states that if no external forces act on a system, the total momentum before an interaction is equal to the total momentum after. In mathematical terms, this can be written as: \ \[m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\] \ where \(m_1\) and \(m_2\) are the masses of the astronaut and the canister, respectively, and \(v_{1i}\), \(v_{2i}\), \(v_{1f}\), and \(v_{2f}\) are their initial and final velocities.
02

Identify the Initial and Final Velocities

Given: - The initial velocity of the canister \(v_{2i} = 3.50\, \text{m/s}\) - The initial velocity of the astronaut \(v_{1i} = 0\, \text{m/s}\) (since she is initially at rest) - The final velocity of the canister \(v_{2f} = 1.20\, \text{m/s}\) - The final velocity of the astronaut \(v_{1f} = 2.40\, \text{m/s}\).
03

Write the Momentum Equation

Substitute the given values into the conservation of momentum equation: \[78.4 \times 0 + m_2 \times 3.50 = 78.4 \times 2.40 + m_2 \times 1.20\] \ This simplifies to: \[3.50m_2 = 188.16 + 1.20m_2\].
04

Solve for the Unknown Mass

Rearrange the equation for \(m_2\): \[3.50m_2 - 1.20m_2 = 188.16\] Simplify to get: \[2.30m_2 = 188.16\] Thus, \[m_2 = \frac{188.16}{2.30}\] Calculate \(m_2\): \[m_2 \approx 81.8 \, \text{kg}\].

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Key Concepts

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

Astronaut Physics
In the fascinating realm of astronaut physics, motion and interaction in space present unique challenges and scenarios. Unlike on Earth, astronauts operate in a microgravity environment, where conventional methods of measuring and manipulating mass do not apply. The absence of gravity means no scales or balances can be used.

To comprehend how astronauts solve such issues, consider the example of an astronaut interacting with a gas canister in the void of space. Here, the principles of momentum come into play significantly. An astronaut can use their understanding of motion to indirectly gauge the mass of objects. By measuring velocity changes and utilizing the conservation of momentum, astronauts can determine the mass of an object without relying on traditional weighing devices. These principles form a crucial part of their training and operations in space.
Velocity Calculations
Velocity calculations are fundamental to solving problems associated with the movement of objects in space. Velocity, by definition, is the speed of something in a given direction. When dealing with two objects such as an astronaut and a gas canister, understanding their respective velocities before and after an interaction helps us analyze their motion.

In the given exercise, initially, the gas canister was moving towards the astronaut at 3.50 m/s, while the astronaut was at rest with a velocity of 0 m/s. After the astronaut pushed the canister, its velocity decreased to 1.20 m/s, while the astronaut gained a velocity of 2.40 m/s in the opposite direction. These changes are pivotal in solving for unknown variables, as they reflect the action-reaction nature of the momentum exchange between the two bodies. Correctly identifying and substituting these velocities into the momentum equation allows us to deduce other quantities such as mass.
Momentum Equation
The momentum equation plays a central role in understanding and solving the problem at hand. The law of conservation of momentum tells us that the total momentum before an interaction equals the total momentum after, provided no external forces act on the objects involved.

For this specific exercise, we start with the equation:
  • Initial total momentum: \[m_1v_{1i} + m_2v_{2i}\], where each term refers to the mass and initial velocity of the astronaut and canister, respectively.
  • Final total momentum: \[m_1v_{1f} + m_2v_{2f}\], reflecting their masses and final velocities after interaction.
By substituting known values into this formula, you can rearrange and solve for the unknown mass of the canister. The solution requires some simple algebra, isolating the variable \(m_2\), and then performing the division to find the result. Utilizing this equation showcases the beauty of physics in which theoretical principles can provide practical solutions in the space environment.

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Most popular questions from this chapter

A 10.0-g marble slides to the left at a speed of 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head- on, elastic collision with a larger 30.0-g marble sliding to the right at a speed of 0.200 m/s (\(\textbf{Fig. E8.48}\)). (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all motion is along a line.) (b) Calculate the \(change\) \(in\) \(momentum\) (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the \(change\) \(in\) \(kinetic\) \(energy\) (the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

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.

Starting at \(t\) = 0, a horizontal net force \(\vec{F} = (0.280 N/s)t\hat{\imath} + (-0.450 N/s^2)t^2\hat{\jmath}\) is applied to a box that has an initial momentum \(\vec{\rho} = (-3.00 kg \cdot m/s)\hat{\imath}+ (4.00 kg \cdot m/s)\hat{\jmath}\). What is the momentum of the box at \(t\) = 2.00 s?

A 0.150-kg glider is moving to the right with a speed of 0.80 m/s on a frictionless, horizontal air track. The glider has a head-on collision with a 0.300-kg glider that is moving to the left with a speed of 2.20 m/s. Find the final velocity (magnitude and direction) of each glider if the collision is elastic.

Three odd-shaped blocks of chocolate have the following masses and center-of- mass coordinates: (1) 0.300 kg, (0.200 m, 0.300 m); (2) 0.400 kg, (0.100 m, \(-\)0.400 m); (3) 0.200 kg, (\(-\)0.300 m, 0.600 m). Find the coordinates of the center of mass of the system of three chocolate blocks.
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