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The last stage of a rocket, which is traveling at a speed of \(7600 \mathrm{~m} / \mathrm{s}\), consists of two parts that are clamped together: a rocket case with a mass of \(290.0 \mathrm{~kg}\) and a payload capsule with a mass of \(150.0 \mathrm{~kg}\). When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of \(910.0 \mathrm{~m} / \mathrm{s}\). What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line. Find the total kinetic energy of the two parts (c) before and (d) after they separate. (e) Account for the difference.

Short Answer

Expert verified
(a) Find rocket case velocity using conservation of momentum and relative velocity. (b) Use relative velocity for payload. (c) Calculate initial kinetic energy. (d) Calculate final kinetic energy. (e) Explain any energy difference.

Step by step solution

01

Understand the Scenario

The rocket case and the payload capsule are initially traveling together at a velocity of \(v_0 = 7600 \, \text{m/s}\). The relative speed between them after the clamp is released is \(\Delta v = 910 \, \text{m/s}\). The masses are \(m_1 = 290.0 \, \text{kg}\) for the rocket case and \(m_2 = 150.0 \, \text{kg}\) for the payload capsule.
02

Apply Conservation of Momentum

Since no external forces act in the direction of motion, the momentum of the system is conserved. The total initial momentum is \(P_i = (m_1 + m_2)v_0\). After separation, the momentum is split into two parts: \(P_f = m_1 v_1 + m_2 v_2\), where \(v_1\) and \(v_2\) are the final velocities of the rocket case and payload, respectively. Set \(P_i = P_f\): \[(m_1 + m_2) v_0 = m_1 v_1 + m_2 v_2.\]
03

Use Relative Velocity Equation

The relative velocity equation gives us another equation to work with: \(v_1 - v_2 = \Delta v\). This allows us to express one velocity in terms of the other. For example, \(v_2 = v_1 - 910\).
04

Solve for Rocket Case Velocity

Substitute the expression for \(v_2\) from the relative velocity equation into the momentum equation: \[(m_1 + m_2) v_0 = m_1 v_1 + m_2 (v_1 - 910)\]. Simplify and solve for \(v_1\).
05

Solve for Payload Capsule Velocity

After finding \(v_1\), use the relative velocity equation to solve for \(v_2\): \(v_2 = v_1 - 910\). This gives the final velocity of the payload.
06

Calculate Total Initial Kinetic Energy

The kinetic energy before separation is given by \(KE_i = \frac{1}{2}(m_1 + m_2)v_0^2\). Substitute the known values to find \(KE_i\).
07

Calculate Total Final Kinetic Energy

The total kinetic energy after separation is \(KE_f = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\). Use the velocities found in Steps 4 and 5 to calculate \(KE_f\).
08

Compare Energies and Account for the Difference

Determine the change in kinetic energy \(\Delta KE = KE_f - KE_i\). Comment on potential causes of this difference, such as the work done by the spring during separation.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fundamental concept in physics that describes the energy an object possesses due to its motion. It is given by the formula:\[ KE = \frac{1}{2}mv^2 \]where \( m \) is the mass of the object and \( v \) is its velocity. When dealing with systems, such as the rocket and payload in our scenario, the total kinetic energy can be calculated by summing the kinetic energies of each component.
In the original exercise, the initial kinetic energy of the combined rocket and payload is calculated using their total mass and initial velocity. After separation, each part has its own velocity; hence, the kinetic energy is recalculated individually for both the rocket case and the payload. The difference in kinetic energy before and after the event can often reveal insights into the process, like the work done by internal forces, such as a spring in this case. Understanding how kinetic energy transforms helps us in explaining real-world phenomena, like why a moving car stops faster on a dry road than on ice or why seat belts are crucial for safety.
Physics Problem Solving
Physics problem solving involves breaking down complex scenarios into simpler parts that can be analyzed and solved step by step. Let's consider solving the rocket separation scenario:
1. **Understand the problem**: Identify given data, such as masses, initial speed, and relative speed after separation. 2. **Apply appropriate principles**: Here, conservation of momentum and kinetic energy principles are crucial. 3. **Develop equations**: Use the conservation of momentum and relative velocity concepts to form the necessary equations. Using these steps, we determine separate velocities for the rocket and payload by applying the relative speed equation combined with the initial total momentum equation.
This methodical approach not only simplifies the problem but also builds a deeper understanding of the physical concepts and laws at play. Practicing such structured approaches strengthens problem-solving skills, making complex situations more manageable.
Relative Velocity
Relative velocity is a measure of how fast one object is moving in relation to another. It's crucial in scenarios where objects interact or move independently after a certain event. In the context of the rocket and payload, the relative velocity tells us how much faster or slower one part travels compared to the other after they separate.
The relative velocity between two objects, say object A with velocity \( v_A \) and object B with velocity \( v_B \), is given as:\[ v_{A/B} = v_A - v_B \]In our physics problem, this principle helps us establish a link between the final velocities of the rocket case and payload. Knowing that they differ by 910 m/s allows us to express one velocity in terms of another, simplifying the problem by reducing the number of unknowns.
Understanding relative velocity can also help explain real-world phenomena like the Doppler effect, where the apparent frequency of waves changes based on the moving observer or source. It's a versatile and essential concept in physics, aiding in the analysis of motion in everyday life and in advanced technological applications.

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