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Chapter 8: Problem 63

A C6-5 model rocket engine has an impulse of 10.0 \(\mathrm{N} \cdot \mathrm{s}\) while burning 0.0125 \(\mathrm{kg}\) of propellant in 1.70 s. It has a maximum thrust of 13.3 \(\mathrm{N} .\) The initial mass of the engine plus propellant is 0.0258 \(\mathrm{kg} .\) (a) What fraction of the maximum thrust is the average thrust? (b) Calculate the relative speed of the exhaust gases, assuming it is constant. (c) Assuming that the relative speed of the exhaust gases is constant, find the final speed of the engine if it was attached to a very light frame and fired from rest in gravity-free outer space.

Short Answer

Expert verified
(a) 0.442, (b) 800 m/s, (c) 555.2 m/s.

Step by step solution

01

Calculate Average Thrust

The average thrust \( F_{avg} \) can be calculated using the formula \( F_{avg} = \frac{I}{t} \), where \( I \) is the impulse of 10.0 Ns and \( t \) is the burn time of 1.70 s.\[ F_{avg} = \frac{10.0 \, \text{N} \cdot \text{s}}{1.70 \, \text{s}} = 5.88 \, \text{N} \]
02

Calculate Fraction of Maximum Thrust

The fraction of the average thrust compared to the maximum thrust is given by \( \frac{F_{avg}}{F_{max}} \), where the maximum thrust \( F_{max} \) is 13.3 N. Thus:\[ \frac{F_{avg}}{F_{max}} = \frac{5.88}{13.3} \approx 0.442 \]
03

Calculate Relative Speed of Exhaust Gases

The relative speed of the exhaust gases \( v_e \) can be calculated using the formula \( v_e = \frac{I}{m_b} \), where \( I = 10.0 \, \text{N} \cdot \text{s} \) and \( m_b = 0.0125 \, \text{kg} \).\[ v_e = \frac{10.0 \, \text{N} \cdot \text{s}}{0.0125 \, \text{kg}} = 800 \, \text{m/s} \]
04

Calculate Final Speed of the Engine

The final speed \( v_f \) can be calculated using the formula \( v_f = v_e \ln \left( \frac{m_i}{m_f} \right) \), assuming constant relative speed of the exhaust gases. The initial mass is \( m_i = 0.0258 \, \text{kg} \) and the final mass \( m_f = m_i - m_b = 0.0258 \, \text{kg} - 0.0125 \, \text{kg} = 0.0133 \, \text{kg} \).\[ v_f = 800 \, \text{m/s} \times \ln \left( \frac{0.0258}{0.0133} \right) = 800 \, \text{m/s} \times 0.694 = 555.2 \, \text{m/s} \]

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Key Concepts

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

Average Thrust
In rocket propulsion, average thrust is a critical parameter. It indicates the steady force that a rocket engine produces over the burning period. Calculating average thrust involves dividing the total impulse by the duration of the burn time.
For instance, if a rocket engine has an impulse of 10.0 Ns and burns for 1.70 seconds, the average thrust can be found as follows:
  • Use the formula: \( F_{avg} = \frac{I}{t} \)
  • Substitute the values: \( F_{avg} = \frac{10.0 \, \text{N} \cdot \text{s}}{1.70 \, \text{s}} \)
  • Solve to get: \( F_{avg} = 5.88 \, \text{N} \)
This means throughout its burn, the engine exerted a constant average force of 5.88 N.
Impulse
Impulse is the product of force applied and the time duration over which it acts. It is a key concept in understanding how rockets generate thrust. Impulse provides the push that propels the rocket forward.
In the context of a C6-5 model rocket engine, the impulse is given as 10.0 Ns. This value represents the total momentum change imparted by the exhaust gases. A higher impulse means more "push," allowing rockets to achieve greater speeds or carry heavier payloads.
  • Impulse is calculated with the formula \( I = F \times t \).
  • The measurement unit for impulse is Newton-seconds (Ns).
Understanding impulse helps us grasp the effectiveness of rocket engines in various applications.
Exhaust Velocity
Exhaust velocity reflects the speed at which exhaust gases exit a rocket engine's nozzle. This speed is a vital factor for rocket performance. High exhaust velocity translates to efficient thrust generation.
For the engine in question, the relative exhaust velocity is computed with the formula \( v_e = \frac{I}{m_b} \), where \( m_b \) is the mass of the propellant. Substituting into the equation:
  • Impulse \( I = 10.0 \, \text{N} \cdot \text{s} \)
  • Burn mass \( m_b = 0.0125 \, \text{kg} \)
  • Calculate \( v_e = \frac{10.0}{0.0125} = 800 \, \text{m/s} \)
A high exhaust velocity of 800 m/s shows the engine鈥檚 capability to provide effective thrust.
Maximum Thrust
The maximum thrust is the peak force exerted by the rocket engine during its operation. It represents the engine's full potential to propel the rocket. In our example, the C6-5 engine has a maximum thrust of 13.3 N.
To understand how this compares with average thrust, we calculate the fraction of maximum thrust used as average thrust by taking the ratio \( \frac{F_{avg}}{F_{max}} \). From the previous calculation:
  • \( F_{avg} = 5.88 \, \text{N} \)
  • \( F_{max} = 13.3 \, \text{N} \)
  • Calculate \( \frac{5.88}{13.3} \approx 0.442 \)
This shows that the average thrust is about 44.2% of the maximum thrust, illustrating how the engine operates at less than its peak most of the time.
Relative Speed
Relative speed in rocketry pertains to how quickly the exhaust moves away relative to the rocket itself. It鈥檚 pivotal for determining how effective the propellant is in gaining momentum for the rocket.
The relative speed of the exhaust gases for the C6-5 engine is calculated as 800 m/s, meaning the gases are expelled at this velocity. This high relative speed contributes significantly to thrust and vehicle acceleration.
Rocketry often involves concepts like the Tsiolkovsky rocket equation, which encompasses relative speed, to predict final speeds of rockets post-burn. The larger the relative speed, the better the performance of the rocket in terms of speed and payload delivery.
Outer Space Physics
Outer space physics provides a unique environment for studying rocket propulsion. Unlike within Earth's atmosphere, there is no air resistance or gravity affecting the vehicle directly. This environment allows for straightforward application of physical laws to achieve maximum performance.
In outer space, rockets behave strictly according to Newton's laws of motion. The absence of external forces means that calculations such as determining final speed from thrust and propellant mass become precise.
  • Final speed is determined by the rocket's initial mass and the mass of propellant burned.
  • This is where concepts like the Tsiolkovsky rocket equation shine, relating exhaust speed to final velocity.
These calculations show how efficiently propulsion systems work when free from external terrestrial influences.

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Most popular questions from this chapter

You are at the controls of a particle accelerator, sending a beam of \(1.50 \times 10^{7} \mathrm{m} / \mathrm{s}\) protons (mass \(m )\) at a gas target of an unknown element. Your detector tells you that some protons bounce straight back after a collision with one of the nuclei of the unknown element. All such protons rebound with a speed of \(1.20 \times 10^{7} \mathrm{m} / \mathrm{s}\) . Assume that the initial speed of the target nucleus is negligible and the collision is elastic. (a) Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass \(m .\) (b) What is the speed of the unknown nucleus immediately after such a collision?

A blue puck with mass \(0.0400 \mathrm{kg},\) sliding with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass \(m,\) initially at rest. After the collision, the velocity of the blue puck is 0.050 \(\mathrm{m} / \mathrm{s}\) in the same direction as its initial velocity. Find (a) the velocity (magnitude and direction) of the red puck after the collision and (b) the mass \(m\) of the red puck.

One 110 -kg football lineman is running to the right at 2.75 \(\mathrm{m} / \mathrm{s}\) while another 125 -kg lineman is running directly toward him at 2.60 \(\mathrm{m} / \mathrm{s}\) . What are (a) the magnitude and direction of the net momentum of these two athletes, and (b) their total kinetic energy?

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass \(70.0 \mathrm{kg},\) is moving to the right at \(2.00 \mathrm{m} / \mathrm{s},\) while the other, of mass \(65.0 \mathrm{kg},\) is moving to the left at 2.50 m/s. What are the magnitude and direction of the velocity of these skaters just after they collide?

On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 \(\mathrm{kg}\) is moving toward puck \(B\) (with mass \(0.350 \mathrm{kg} ),\) which is initially at rest. After the collision, puck \(A\) has a velocity of 0.120 \(\mathrm{m} / \mathrm{s}\) to the left, and puck \(B\) has a velocity of 0.650 \(\mathrm{m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.
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