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Chapter 6: Problem 60

A 95 -tonne \((1 \mathrm{t}=1000 \mathrm{~kg})\) spacecraft moving in the \(+x=\) direction at \(0.34 \mathrm{~m} / \mathrm{s}\) docks with a 75 -tonne craft moving in the \(-x\) -direction at \(0.58 \mathrm{~m} / \mathrm{s}\). Find the velocity of the joined spacecraft.

Short Answer

Expert verified
The velocity of the joined spacecraft is approximately -0.0659 m/s.

Step by step solution

01

Set Up the Conservation of Momentum Equation

Since no external forces are acting, the total momentum before and after the docking event must be conserved. The equation is:\[m_1v_1 + m_2v_2 = (m_1 + m_2)v_f\]where \(m_1 = 95000\, \text{kg}\) (the first spacecraft's mass), \(v_1 = 0.34\, \text{m/s}\) (the first spacecraft's velocity), \(m_2 = 75000\, \text{kg}\) (the second spacecraft's mass), \(v_2 = -0.58\, \text{m/s}\) (the second spacecraft's velocity), and \(v_f\) is the final velocity of the joined spacecraft.
02

Substitute the Known Values

Substitute the given values into the momentum equation:\[95000 \times 0.34 + 75000 \times (-0.58) = (95000 + 75000) \times v_f\]Calculating the left side:\[32300 - 43500 = (170000) \times v_f\]
03

Simplify the Equation

Simplify the left-hand side of the equation:\[32300 - 43500 = -11200\]Now the equation is:\[-11200 = 170000 \times v_f\]
04

Solve for Final Velocity \(v_f\)

Solve for \(v_f\) by dividing both sides of the equation by the total mass:\[v_f = \frac{-11200}{170000}\]Calculating gives:\[v_f \approx -0.0659\, \text{m/s}\]

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Key Concepts

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

Spacecraft Docking
Spacecraft docking is a fascinating process where two spacecraft join together in space. This procedure is vital for various space missions, including resupplying stations or joint missions. Imagine docking being similar to two puzzle pieces fitting together perfectly. This operation needs precision, as both spacecraft must align their velocities and positions. Key concepts in docking include:
  • Relative velocity: The difference in speed and direction between the two spacecraft.
  • Alignment: Ensuring both spacecraft's docking ports are correctly positioned.
  • Synchronization: Coordinating movements to achieve a seamless connection.
Vision systems and guided systems often assist, ensuring everything fits perfectly. Understanding docking helps in evaluating how spacecraft work together during missions.
Calculating Velocity
Calculating velocity is an integral part of solving physics problems involving moving objects, like spacecraft. Velocity is a vector, meaning it has both magnitude and direction. It's crucial to distinguish this from speed, which only has magnitude. When you calculate the velocity of two joining spacecraft, consider the direction of each craft's movement. In the problem, the first spacecraft moves in the positive direction, while the second moves in the negative direction. This impacts the final combined velocity as:
  • Positive velocity means moving forward.
  • Negative velocity implies moving backward.
  • The net effect of both velocities determines the direction and speed of the joined craft.
Understanding these concepts ensures accurate results when dealing with such physical situations.
Momentum Equation
The momentum equation is foundational in physics for analyzing motion, especially when no external forces act on a system. Conservation of momentum tells us:\[m_1v_1 + m_2v_2 = (m_1 + m_2)v_f\]In this formula:
  • \(m_1\) and \(m_2\) are the masses of the two objects, and \(v_1\) and \(v_2\) are their velocities.
  • The left side represents the total momentum before docking.
  • The right side shows the combined momentum post-docking.
This equation effectively calculates the final velocity of the combined spacecraft system. It shows how two opposing momentums balance to determine the resultant motion after docking. Grasping this equation is key to solving various physics problems.
Physics Problem Solving
Problem-solving in physics, like this spacecraft docking exercise, involves a series of steps to reach a solution. Here's a simplified approach:
  • Identify given data, such as masses and velocities.
  • Apply appropriate physics principles, like conservation of momentum here.
  • Set up equations using known values.
  • Solve equations step-by-step, checking accuracy at each stage.
  • Interpret results in context, understanding physical implications.
This systematic method ensures students tackle each component methodically, making even complex problems manageable. Confidence in following these steps helps in approaching other challenging physics scenarios.

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Most popular questions from this chapter

In Compton scattering, a photon scatters elastically off an electron initially at rest. Suppose the incoming photon has momentum of \(1.0 \times 10^{-21} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \hat{\imath}\). The scattered photon's momentum is \(1.8 \times 10^{-22} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \hat{\imath}-3.1 \times 10^{-28} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} \hat{\jmath}\). Find the mo- mentum of the recoiling electron.

A bomb explodes into three identical pieces. One flies in the \(+y\) -direction at \(13.4 \mathrm{~m} / \mathrm{s},\) a second in the \(-x\) -direction at \(16.1 \mathrm{~m} / \mathrm{s} .\) What's the velocity of the third piece?

High-speed photos of a \(220-\mu\) g flea jumping vertically indicate that the jump lasts \(1.2 \mathrm{~ms}\) and involves an average vertical acceleration of \(100 g .\) What average (a) force and (b) impulse does the ground exert on the flea during its jump?

An eagle's dive. When diving for prey, bald eagles can reach speeds of \(200 \mathrm{mph}(322 \mathrm{~km} / \mathrm{h})\). Suppose a \(6.0-\mathrm{kg}\) eagle flying horizontally at \(322 \mathrm{~km} / \mathrm{h}\) suddenly grasps and holds a stationary \(5.0-\mathrm{kg}\) jackrabbit in her talons. (a) What's the eagle's speed just after grasping the rabbit? (b) What percent of her kinetic energy is lost during the process? (c) What would be the eagle's final speed if the rabbit were initially running toward her at \(2.0 \mathrm{~m} / \mathrm{s} ?\)

A net force \(0.340 \mathrm{~N} \hat{\imath}+0.240 \mathrm{~N} \hat{\jmath}\) is applied to a \(170-\mathrm{g}\) hockey puck for \(4.50 \mathrm{~s}\). (a) If the puck was initially at rest, find its final momentum. (b) If the puck was initially moving with velocity \(2.90 \mathrm{~m} / \mathrm{s} \hat{\imath}+1.35 \mathrm{~m} / \mathrm{s} \hat{\jmath},\) find its final momentum.
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