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A two-stage rocket moves in space at a constant velocity of \(4900 \mathrm{m} / \mathrm{s} .\) The two stages are then separated by a small explosive charge placed between them. Immediately after the explosion the velocity of the \(1200-\mathrm{kg}\) upper stage is \(5700 \mathrm{m} / \mathrm{s}\) in the same direction as before the explosion. What is the velocity (magnitude and direction) of the \(2400-\mathrm{kg}\) lower stage after the explosion?

Short Answer

Expert verified
The velocity of the lower stage is \(4500 \mathrm{m/s}\) in the same direction as before.

Step by step solution

01

Understand the Principles

This problem involves the conservation of momentum, which states that in the absence of external forces, the total momentum before and after the explosion must be the same. This is because the forces involved in the explosion are internal.
02

Initial Conditions

Initially, the whole system (upper and lower stages) is moving at a constant velocity of \(4900\, \mathrm{m/s}\). The mass of the entire system is \(1200\, \mathrm{kg} + 2400\, \mathrm{kg} = 3600\, \mathrm{kg}\).
03

Calculate Initial Momentum

The initial momentum of the system can be calculated using the formula \( p = mv \). So, \( p_i = 3600\, \mathrm{kg} \times 4900\, \mathrm{m/s} = 17640000\, \mathrm{kg}\cdot\mathrm{m/s} \).
04

Apply Conservation of Momentum

After the explosion, the momentum of the system is the sum of the momentum of the individual stages: \( p_f = m_1v_1 + m_2v_2 \), where \( m_1 = 1200\, \mathrm{kg} \), \( v_1 = 5700\, \mathrm{m/s} \), and \( m_2 = 2400\, \mathrm{kg} \) is the velocity \( v_2 \) we need to find.
05

Input Known Values

Substitute the known values into the momentum conservation equation: \[ 17640000\, \mathrm{kg}\cdot\mathrm{m/s} = 1200\, \mathrm{kg} \times 5700\, \mathrm{m/s} + 2400\, \mathrm{kg} \times v_2 \] This simplifies to: \[ 17640000 = 6840000 + 2400v_2 \]
06

Solve for the Unknown

Rearrange the equation to solve for \( v_2 \): \[ 10800000 = 2400v_2 \] Dividing both sides by \(2400\) gives: \[ v_2 = \frac{10800000}{2400} = 4500\, \mathrm{m/s} \].
07

Determine the Direction

Since the problem states that the velocity of the upper stage is in the same direction as before the explosion and the resulting calculated velocity for the lower stage is positive, the direction of the lower stage's velocity is also the same as the initial direction.

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Key Concepts

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

Two-Stage Rocket
The concept of a two-stage rocket is fundamental in understanding how rockets separate during flight. Each stage of the rocket serves a specific purpose in different phases of the flight.
After using up its fuel, the first stage, also known as the lower stage, separates from the rest of the rocket to reduce weight, allowing the upper stage to continue onward with increased efficiency. This separation involves complex engineering principles, which include the use of explosive charges to initiate the separation.
In our exercise, the two-stage rocket's separation through an explosive charge is a key phenomenon that allows us to analyze momentum conservation, a critical principle in physics.
Velocity and Direction
Velocity is a vector quantity, which means it has both a magnitude (speed) and a direction. Understanding the changes in both these parameters is crucial when analyzing the motion of separate rocket stages.
Initially, the entire rocket moves at a uniform velocity of 4900 m/s. When the upper stage separates, it changes velocity due to an explosive force. It increases to 5700 m/s but maintains the same directional path.
In tasks involving rockets or any moving object, determining not just how fast something moves, but also in what direction, is vital to solving conservation of momentum problems, like in the rocket exercise. The direction remains consistent if forces act along the same line initially designated.
Momentum Calculation
The principle of momentum conservation is centered on the formula, \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity. It frames our understanding of how momentum before and after an event like a rocket stage separation need to equate, provided no external forces act on the system.
For the complete system initially, both the upper and lower stages, the initial momentum is calculated as \( 3600 \, \mathrm{kg} \times 4900 \, \mathrm{m/s} = 17640000 \, \mathrm{kg \cdot m/s} \). This sets a baseline as the momentum required post-explosion.
After the explosive charge separates the stages, individual momentums are summed to match this initial total. Understanding and applying these calculations are vital for figuring out the resulting motions of both rocket parts.
Explosion Dynamics
The dynamics involved in the explosion that separates rocket stages is a fascinating application of physics.
The explosion acts as an internal force, causing no change to the overall momentum of the larger system comprising both rocket stages.
This internal force changes the velocity of the individual stages. For the problem's context, it led to a new velocity for the upper stage and allowed calculations for the lower stage. It illustrates how explosions redistribute momentum across components within the closed system.
Analysis of explosion dynamics extends to understanding force impact, energy distribution, and eventual momentum readjustment post-event. It reinforces how vectors and magnitude play a role in directing these outcomes, maintaining momentum equilibrium.

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Most popular questions from this chapter

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