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(I) Calculate the force exerted on a rocket when the propelling gases are being expelled at a rate of \(1300 \mathrm{~kg} / \mathrm{s}\) with a speed of \(4.5 \times 10^{4} \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The force exerted on the rocket is 58,500,000 N (58.5 MN).

Step by step solution

01

Understand the concept of force

In this problem, we need to find the force exerted on a rocket due to the expulsion of propelling gases. According to Newton's second law of motion, force is defined as the rate of change of momentum.
02

Use the formula for force due to expelled gases

The force exerted by the expelled gases can be calculated using the formula: \[ F = u \frac{dm}{dt} \]where \( F \) is force, \( u \) is the speed of the gases, and \( \frac{dm}{dt} \) is the rate of mass expelled.
03

Assign the given values

From the problem, we have:- The rate of mass expelled, \( \frac{dm}{dt} = 1300 \text{ kg/s} \)- The speed of the expelled gases, \( u = 4.5 \times 10^4 \text{ m/s} \)
04

Calculate the force

Substitute the given values into the formula:\[ F = (4.5 \times 10^4) \times 1300 \]Now compute the multiplication:\[ F = 58,500,000 \text{ N} \]
05

Interpret the result

The calculated force is \( 58,500,000 \text{ N} \), which means the rocket experiences a thrust of 58.5 MegaNewtons as gases are expelled.

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Key Concepts

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

Force Calculation
In the world of rocket propulsion, understanding how to calculate force is essential. The force exerted by a rocket is directly related to how fast and in what quantity the exhaust gases are expelled. By employing the simple yet powerful formula \[ F = u \frac{dm}{dt} \]we gain an insight into how different factors contribute to rocket thrust.

Let's break it down:
  • \( F \): This is the force exerted by the expelled gases, essentially the thrust exerted on the rocket.
  • \( u \): This represents the speed at which the gases are expelled. Faster speeds mean more thrust.
  • \( \frac{dm}{dt} \): This is the rate of mass of the exhaust gases being expelled per second. A higher rate results in greater thrust.
When you multiply the speed of the exhaust (\( u \)) by the rate of gas expulsion (\( \frac{dm}{dt} \)), you get the total force applied to the rocket. Translating this into real-world terms, each action on the gases produces a reaction that propels the rocket forward. The larger the force, the faster the rocket can accelerate.
Newton's Second Law
Newton's Second Law of Motion is fundamental in understanding how force impacts movement. It states that force is the result of the change in motion, represented by the equation:

\[ F = ma \]
where \( F \) stands for force, \( m \) is the mass, and \( a \) is acceleration.

In the context of rockets, \( m \) refers to the changing mass of the rocket as it expels fuel, and \( a \) represents the acceleration achieved through thrusters.
To further grasp the concept:
  • The law emphasizes that the force applied to an object is equal to the mass of an object multiplied by its acceleration. Essentially, heavier objects require more force to achieve the same amount of acceleration.
  • When gases are expelled at high speeds, the rocket experiences a reinforced force in the opposite direction, validating Newton's third law of action and reaction within a context driven by the second law.
  • As a rocket burns fuel, its mass decreases, and thus for a consistent force, the acceleration increases as per the formula. This is why rockets speed up as they travel into space.
Momentum Change
Momentum is an essential concept in understanding rocket movement. It's defined as the product of mass and velocity, expressed as:\[ p = mv \]
Where \( p \) represents momentum, \( m \) stands for mass, and \( v \) is velocity.

In rockets, the change in momentum is key for calculating the force exerted by the engine on the rocket:
  • When gases are expelled, the rocket's momentum changes. This is a direct result of the momentum of the expelled gases moving in the opposite direction.
  • The rate of change of momentum is what produces thrust, providing the necessary power for the rocket to accelerate.
  • This change in momentum is used to calculate force using the previously mentioned formula \( F = u \frac{dm}{dt} \), as it embodies change over time.
Understanding momentum change helps solidify how rockets maneuver through space. The great synergy between fuel expulsion, speed, and force defines effective rocket propulsion.

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Most popular questions from this chapter

(III) ( \(a\) ) Calculate the impulse experienced when a \(65-\mathrm{kg}\) person lands on firm ground after jumping from a height of \(3.0 \mathrm{~m} .\) (b) Estimate the average force exerted on the person's feet by the ground if the landing is stiff-legged, and again \((c)\) with bent legs. With stiff legs, assume the body moves \(1.0 \mathrm{~cm}\) during impact, and when the legs are bent, about \(50 \mathrm{~cm} .\) [Hint: The average net force on her which is related to impulse, is the vector sum of gravity and the force exerted by the ground.

(II) A 0.280-kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball. (a) What is the mass of the second ball? (b) What fraction of the original kinetic energy \((\Delta K / K)\) gets transferred to the second ball?

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