/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 193 The third and fourth stages of a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The third and fourth stages of a rocket are coasting in space with a velocity of \(18000 \mathrm{km} / \mathrm{h}\) when a small explosive charge between the stages separates them. Immediately after separation the fourth stage has increased its velocity to \(v_{4}=18060 \mathrm{km} / \mathrm{h} .\) What is the corresponding velocity \(v_{3}\) of the third stage? At separation the third and fourth stages have masses of 400 and \(200 \mathrm{kg},\) respectively.

Short Answer

Expert verified
The third stage's velocity is \(17970 \, \text{km/h}\).

Step by step solution

01

Identify the Conservation Law

In this problem, we use the principle of conservation of momentum, which states that in an isolated system, the total momentum before an event must equal the total momentum after the event. The equation for momentum is given by: \[ p = m \cdot v \] where \( m \) is the mass and \( v \) is the velocity. The system here is the combined third and fourth stages of the rocket before and after the separation.
02

Write Initial Momentum Equation

Before the explosive charge separates the stages, the two stages are moving together at a velocity of \( 18000 \, \text{km/h} \). The initial total momentum \( p_{i} \) of the system is: \[ p_{i} = (m_3 + m_4) \cdot v_i = (400 \, \text{kg} + 200 \, \text{kg}) \cdot 18000 \, \text{km/h} \] \[ p_{i} = 600 \, \text{kg} \cdot 18000 \, \text{km/h} \]
03

Calculate Initial Momentum

Calculate the initial total momentum using the values provided: \[ p_{i} = 600 \, \text{kg} \cdot 18000 \, \text{km/h} = 10800000 \, \text{kg·km/h} \]
04

Write Final Momentum Equation

After separation, the fourth stage increases its velocity to \( v_4 = 18060 \, \text{km/h} \) and the third stage has an unknown velocity \( v_3 \). The final total momentum \( p_{f} \) of the system is: \[ p_{f} = m_3 \cdot v_3 + m_4 \cdot v_4 \] where \( m_3 = 400 \, \text{kg} \) and \( m_4 = 200 \, \text{kg} \).
05

Apply Momentum Conservation

According to the conservation of momentum, \( p_{i} = p_{f} \). Substitute known values into this equation: \[ 10800000 \, \text{kg·km/h} = 400 \, \text{kg} \cdot v_3 + 200 \, \text{kg} \cdot 18060 \, \text{km/h} \]
06

Solve for Unknown Velocity \(v_3\)

Calculate the momentum contribution of the fourth stage: \[ 400 \cdot v_3 + 200 \cdot 18060 = 10800000 \] \[ 400v_3 + 3612000 = 10800000 \] \[ 400v_3 = 10800000 - 3612000 \] \[ 400v_3 = 7188000 \] \[ v_3 = \frac{7188000}{400} \] \[ v_3 = 17970 \, \text{km/h} \]
07

Conclusion

The velocity of the third stage immediately after separation is \( v_3 = 17970 \, \text{km/h} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Rocket Stages
When dealing with rockets, a single structure can actually consist of multiple stages, each with its own mass and function. These stages are often used in combination to carry out various tasks such as launching into orbit or traveling through space efficiently. The idea behind multiple stages is to discard parts of the rocket once they have fulfilled their purpose. This reduces the overall weight, conserving fuel, and allowing for greater speeds and distances.

In our exercise, we specifically look at the third and fourth stages of a rocket. They have been coasting together through space but are about to be separated by a small explosive charge. Each stage continues to move independently after separation, with their own unique velocities. Understanding how these stages interact and separate is crucial for calculating their post-separation velocities.
Velocity Calculation
To understand how we calculate velocities after the separation of stages in a rocket, we use the concept of velocity which is basically the speed of an object in a specified direction. When two stages of a rocket separate in space, the change in their velocities is a key point of analysis.

In the given problem, we start off with both stages moving at a combined velocity of 18000 km/h. After the separation, the velocity of the fourth stage increases to 18060 km/h. We were tasked with finding out the new velocity of the third stage. Remember, when two objects interact in space and one changes its velocity, the other usually does too, often in an opposite way, to conserve momentum.

This process involved doing some simple calculations by applying the initial and new known velocities in the momentum equation, which we will discuss next. The main takeaway here is the interconnectedness of the stages, and how changes in velocity affect the system as a whole.
Momentum Equation
The momentum of an object is its mass multiplied by its velocity, and the conservation of momentum is a vital principle in physics. It tells us that in an isolated system, the total momentum before any event is the same as the total momentum after the event.

For our rocket problem, the momentum before the stages separate was calculated using the momentum equation. Initially, both the third and fourth stages are moving together, carrying a combined mass of 600 kg at 18000 km/h. This gives us an initial momentum of 10800000 kg·km/h.

Once the fourth stage changes speed, we apply the momentum conservation principle by setting this initial momentum equal to the final momentum, which includes separate terms for each rocket stage with their respective velocities. The equation required rearranging and solving for the unknown velocity of the third stage. This demonstrates the power of the momentum equation in understanding motion in space and correlating the velocity changes of connected objects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The bungee jumper, an 80 -kg man, falls from the bridge at \(A\) with the bungee cord secured to his ankles. He falls \(20 \mathrm{m}\) before the 17 -m length of elastic bungee cord begins to stretch. The \(3 \mathrm{m}\) of rope above the elastic cord has no appreciable stretch. The man is observed to drop a total of \(44 \mathrm{m}\) before being projected upward. Neglect any energy loss and calculate \((a)\) the stiffness \(k\) of the bungee cord (increase in tension per meter of elongation) (b) the maximum velocity \(v_{\max }\) of the man during his fall, and \((c)\) his maximum acceleration \(a_{\max } .\) Treat the man as a particle located at the end of the bungee cord.

Freight car \(A\) of mass \(m_{A}\) is rolling to the right when it collides with freight car \(B\) of mass \(m_{B}\) initially at rest. If the two cars are coupled together at impact, show that the fractional loss of energy equals \(m_{B} /\left(m_{A}+m_{B}\right)\).

The nest of two springs is used to bring the 0.5 -kg plunger \(A\) to a stop from a speed of \(5 \mathrm{m} / \mathrm{s}\) and reverse its direction of motion. The inner spring increases the deceleration, and the adjustment of its position is used to control the exact point at which the reversal takes place. If this point is to correspond to a maximum deflection \(\delta=200 \mathrm{mm}\) for the outer spring, specify the adjustment of the inner spring by determining the distance \(s .\) The outer spring has a stiffness of \(300 \mathrm{N} / \mathrm{m}\) and the inner one a stiffness of \(150 \mathrm{N} / \mathrm{m}\).

A 90 -lb boy starts from rest at the bottom \(A\) of a 10-percent incline and increases his speed at a constant rate to \(5 \mathrm{mi} / \mathrm{hr}\) as he passes \(B, 50 \mathrm{ft}\) along the incline from \(A .\) Determine his power output as he approaches \(B\).

The quarter-circular slotted arm \(O A\) is rotating about a horizontal axis through point \(O\) with a constant counterclockwise angular velocity \(\Omega=\) 7 rad/sec. The 0.1 -lb particle \(P\) is epoxied to the arm at the position \(\beta=60^{\circ} .\) Determine the tangential force \(F\) parallel to the slot which the epoxy must support so that the particle does not move along the slot. The value of \(R=1.4 \mathrm{ft}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.