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Chapter 7: Problem 5

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

Short Answer

Expert verified
The thrust (force exerted) on the rocket is \(5.85 \times 10^7\, \text{N}\).

Step by step solution

01

Understand the Concept

The force exerted on a rocket is calculated using the principle of momentum and the concept of thrust. Thrust, in this context, refers to the force generated by expelling gases at a high speed from the rocket's engines.
02

Use Newton's Second Law of Motion

Newton's Second Law of Motion states that the rate of change of momentum of an object is equal to the net force acting on the object. In this scenario, the force exerted by the rocket is given by the rate of change of momentum of the gases being expelled.
03

Identify Given Data

The following data is given in the problem:- Rate of mass expulsion (\dot{m}) = 1300 kg/s- Speed of expelled gases (\(v_e\)) = \(4.5 \times 10^4 m/s\)
04

Apply the Thrust Formula

The formula to calculate the thrust (force, F) exerted by the rocket is given by:\[ F = \dot{m} \times v_e \]Where \(\dot{m}\) is the mass flow rate and \(v_e\) is the velocity of the expelled gases.
05

Substitute the Values and Calculate

Substitute the values into the formula:\[ F = 1300\, \text{kg/s} \times 4.5 \times 10^4\, \text{m/s} \]Calculate:\[ F = 5.85 \times 10^7\, \text{N} \]
06

Interpret the Result

The calculated force, \(5.85 \times 10^7\, \text{N}\), represents the thrust generated by the rocket as a reaction to the expulsion of gases.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that describes the quantity of motion an object has. It is calculated as the product of an object's mass and its velocity. In simpler terms, momentum tells us how hard it will be to stop an object in motion. The formula for momentum (p) is given by: \[ p = m imes v \] where \( m \) is the mass of the object, and \( v \) is its velocity.
  • Momentum is a vector quantity, meaning it has both magnitude and direction.
  • In rocket mechanics, the momentum of expelled gases plays a crucial role.
When the rocket expels gases at high speed, they carry away momentum. By Newton's third law, the rocket experiences an equal and opposite momentum change, which propels it forward.
Newton's Second Law of Motion
Newton's Second Law of Motion is central to understanding how rockets work. This law states that the force acting on an object is equal to the rate of change of its momentum over time. Mathematically, it can be expressed as: \[ F = rac{dp}{dt} \] Where \( F \) is force, \( dp \) represents a change in momentum, and \( dt \) is a change in time. In the context of a rocket:
  • The force exerted by the rocket engines is what alters the momentum of the expelled gas.
  • As gas leaves the rocket, it changes the rocket's momentum, allowing it to accelerate.
For constant mass flow rates and speeds, this simplifies to the thrust formula: \( F = ext{mass flow rate} imes ext{velocity of expelled gas} \).
Thrust Formula
The thrust formula is a critical equation used to calculate the force (or thrust) exerted by a rocket. Thrust is the mechanical force generated by the engine due to the rapid expulsion of gas. This formula comes directly from Newton's Second Law of Motion and is expressed as: \[ F = ext{mass flow rate} imes ext{velocity of expelled gas} \]
  • The mass flow rate (\( \dot{m} \)) is the amount of mass expelled per unit of time.
  • The velocity (\( v_e \)) is the speed at which this mass exits the rocket.
The thrust formula helps engineers and scientists design rockets by predicting how much force a rocket engine will produce. This is essential for determining if the rocket can overcome Earth's gravity and reach the desired altitude.
Mass Flow Rate
Mass flow rate is the amount of mass that passes through a given surface or system per unit of time. In rockets, it refers to how quickly fuel and oxidizer are consumed and expelled. This rate is crucial for calculating thrust because it directly influences how much force the rocket engine generates.
  • It is denoted as \( \dot{m} \).
  • In our scenario, the mass flow rate is \( 1300 \) kg/s, meaning the rocket expels this much mass each second.
A higher mass flow rate usually means more reactive force or thrust, assuming the expulsion speed remains constant. The interplay between mass flow rate and exhaust velocity is what allows a rocket to perform maneuvers and maintain flight path stability during its journey.

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Most popular questions from this chapter

(II) Rain is falling at the rate of 2.5 cm/h and accumulates in a pan. If the raindrops hit at 8.0 m/s, estimate the force on the bottom of a \(1.0-m^2\) pan due to the impacting rain which we assume does not rebound.Water has a mass of \(1.00 \times 10^3 kg\) per \(m^3.\)

(II) A 28-g rifle bullet traveling 190 m/s embeds itself in a 3.1-kg pendulum hanging on a 2.8-m-long string, which makes the pendulum swing upward in an arc. Determine the vertical and horizontal components of the pendulum's maximum displacement.

A massless spring with spring constant \(k\) is placed between a block of mass \(m\) and a block of mass \(3m\). Initially the blocks are at rest on a frictionless surface and they are held together so that the spring between them is compressed by an amount \(D\) from its equilibrium length. The blocks are then released and the spring pushes them off in opposite directions. Find the speeds of the two blocks when they detach from the spring.

(III) A neon atom (\(m = 20.0\) u) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a 55.6\(^{\circ}\) angle from its original direction and the unknown atom travels away at a -50.0\(^{\circ}\) angle. What is the mass (in u) of the unknown atom? [\(Hint\): You could use the law of sines.]

(II) A wooden block is cut into two pieces, one with three times the mass of the other. A depression is made in both faces of the cut, so that a firecracker can be placed in it with the block reassembled. The reassembled block is set on a rough-surfaced table, and the fuse is lit.When the firecracker explodes inside, the two blocks separate and slide apart. What is the ratio of distances each block travels?
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