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Chapter 7: Problem 37

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.

Short Answer

Expert verified
The speed of the rocket case is 689.77 m/s, and the speed of the payload is 1599.77 m/s after they separate.

Step by step solution

01

Identify the given values and apply the conservation of momentum

Given initial speed of the combined rocket and payload: \[v_0 = 7600 \; \mathrm{m/s}\] Mass of the rocket case: \[m_1 = 290.0 \; \mathrm{kg}\] Mass of the payload capsule: \[m_2 = 150.0 \; \mathrm{kg}\] Relative speed between the rocket case and the payload after separation: \[v_{rel} = 910.0 \; \mathrm{m/s}\] Let's denote the final speed of the rocket case as \(v_1\) and the final speed of the payload as \(v_2\). Applying conservation of momentum: The total momentum before separation equals the total momentum after separation: \[ (m_1 + m_2) v_0 = m_1 v_1 + m_2 v_2 \]
02

Express the relative speed equation

The relative speed equation can be written as: \[ v_2 - v_1 = v_{rel} \] This can be rearranged to: \[ \quad v_2 = v_1 + v_{rel} \]
03

Substitute the expression for \(v_2\) back into the momentum conservation equation

Substitute \[v_2 = v_1 + 910.0 \; \mathrm{m/s} \] into the momentum conservation equation: \[ (290.0 \; \mathrm{kg} + 150.0 \; \mathrm{kg}) 7600 \; \mathrm{m/s} = 290.0 \; \mathrm{kg} \; v_1 + 150.0 \; \mathrm{kg} (v_1 + 910.0 \; \mathrm{m/s}) \]
04

Solve for \(v_1\)

Expand and simplify the equation to solve for \(v_1\): \[ 440,000 \; \mathrm{kg \cdot m/s} = 290.0 \; \mathrm{kg} \; v_1 + 150.0 \; \mathrm{kg} \; v_1 + 150.0 \; \mathrm{kg} \cdot 910.0 \; \mathrm{m/s} \] \[ 440,000 \; \mathrm{kg \cdot m/s} = 440.0 \; \mathrm{kg} \; v_1 + 136,500 \; \mathrm{kg \cdot m/s} \] Subtract \( 136,500 \; \mathrm{kg \cdot m/s} \) from both sides: \[ 440,000 \; \mathrm{kg \cdot m/s} - 136,500 \; \mathrm{kg \cdot m/s} = 440.0 \; \mathrm{kg} \; v_1 \] \[ 303,500 \; \mathrm{kg \cdot m/s} = 440.0 \; \mathrm{kg} \; v_1 \] Divide both sides by 440.0 kg: \[ v_1 = 689.77 \; \mathrm{m/s} \]
05

Calculate \(v_2\)

Use the relative speed equation to find \(v_2\): \[ v_2 = v_1 + 910.0 \; \mathrm{m/s} \] \[ v_2 = 689.77 \; \mathrm{m/s} + 910.0 \; \mathrm{m/s} \] \[ v_2 = 1599.77 \; \mathrm{m/s} \]

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Key Concepts

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

Relative Velocity
In physics, relative velocity refers to the velocity of an object as observed from a particular reference frame. It is crucial for understanding the motion of objects in relation to each other. For example, in the context of the rocket separation problem:
  • The relative speed between the rocket case and the payload capsule is given as 910.0 m/s.
  • This value indicates how fast one part is moving away from the other post-separation.
The concept helps us bridge the gap between the initial combined state and the separated state of the rocket parts. The formula for relative velocity in one-dimensional motion can be written as:
\[ \text{Relative velocity} \ (V_\text{rel}) = V_2 - V_1 \]
This equation ensures that we can relate the velocities of two bodies that were initially moving together.
Rocket Separation
Rocket separation involves the division of a rocket's final stage into its component parts, often due to mechanisms like a compressed spring or explosive bolts. This separation is essential for releasing payloads, reducing weight, and improving efficiency. In our example:
  • The rocket and payload are initially moving together at 7600 m/s.
  • After separation, they move apart at a relative speed of 910.0 m/s.
Momentum conservation is key to analyzing post-separation speeds. According to the law of conservation of momentum:
\[ (M_1 + M_2) V_0 = M_1 V_1 + M_2 V_2 \]
This equation confirms that the total momentum before and after separation must be constant. The separation occurs along the original line of motion, simplifying calculations. By applying these principles, we estimate the new speeds of the separated parts.
Physics Problem Solving
Solving physics problems often requires a step-by-step approach, leveraging fundamental principles. Here are essential steps:
  • Identify given values: Speeds, masses, and relative velocities.
  • Apply conservation laws: Momentum conservation is particularly useful here.
For the rocket separation problem:
  • Start by noting initial conditions: combined speed and masses.
  • Set up momentum conservation equations to relate initial and final states.
  • Use relative velocity to find the relationship between final speeds.
  • Substitute values and solve for unknowns.
Finally, we calculate individual post-separation speeds. This method ensures clarity and accuracy in solving complex physics problems. Following these steps facilitates a deeper understanding and application of physical principles to real-world situations.

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Most popular questions from this chapter

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