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At time \(t=0\) a \(2150 \mathrm{~kg}\) rocket in outer space fires an engine that excrts an increasing force on it in the \(+x\) -direction. This force ohcys the equation \(F_{x}=A t^{2}\), where \(t\) is time, and his a magritude of \(781.25 \mathrm{~N}\) when \(t=1.25 \mathrm{~s}\). (a) Find the \(\mathrm{Sl}\) value of the constant \(\mathrm{A},\) including its units. (b) What impulse does the engine exert on the rocket during the \(1.50 \mathrm{~s}\) intcrval starting \(200 \mathrm{~s}\) after the cngine is firod? (c) By how much docs the rockel's velocity change during this interval? Assume constant mass.

Short Answer

Expert verified
The constant \(A\) is approximately \(500 N/s^{2}\). The impulse exerted by the engine is approximately \(0.125 N.s\), and the change in the rocket's velocity is approximately \(0.000058 m/s\).

Step by step solution

01

Finding the value of constant A

Based on the given information, the equation of the force exerted on the rocket is \(F_{x}=A t^{2}\). At \(t=1.25\) s, \(F_{x}=781.25\) N. Substituting these values into the equation, we can solve for A. Let's do that: \(781.25 = A \cdot (1.25)^{2}\). Solving for \(A\) gives \(A = 781.25 / (1.25)^{2}\).
02

Calculating the impulse

The impulse \(J\) exerted on an object is the integral of the force \(F\) over time \(t\), in this case, from \(t=200s\) to \(t=201.50s\). Here is how to calculate it: \(J = \int_{200}^{201.50} F_{x} dt = \int_{200}^{201.50} A t^{2} dt = A \cdot \int_{200}^{201.50} t^{2} dt = A \cdot [\frac{1}{3} t^{3}]_{200}^{201.50}\). Substituting the previously found value of \(A\) and evaluating the integral yields the impulse.
03

Calculating the velocity change

The change in velocity \(\Delta v\) of the rocket is equal to the impulse divided by the rocket's mass (\(m = 2150 kg\)). That comes from the definition of momentum (\(p = mv\)), which is the change in momentum divided by the mass. Here's the calculation: \(\Delta v = \frac{J}{m}\). Substituting the previously calculated value of \(J\) and the given mass of the rocket gives the change in velocity.

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Key Concepts

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

Impulse-Momentum Theorem
The impulse-momentum theorem is a fundamental principle in rocket physics that connects impulse, the force applied over a period of time, with the change in momentum of a rocket.

An impulse is produced when a force acts on an object over a period of time. The theorem states that the impulse on an object is equal to the change in the object's momentum. Mathematically, this is represented as:
\[\begin{equation}J = \triangle pewline \triangle p = m \triangle vewline \therefore J = m \triangle v ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline ewline \end{equation}\]For rockets in space, the change in momentum is particularly important because it affects the rocket's velocity. In the textbook problem, the rocket experiences an impulse from the engine over a period of 1.50 seconds. By calculating this impulse and knowing the rocket's mass, we can use the impulse-momentum theorem to determine how much the rocket's velocity changes.
  • The theorem is integral in understanding how forces over time can accelerate rockets.
  • It illustrates the relation between a time-varying force, impulse, and the resulting change in velocity.
  • It applies to objects like rockets that might be subject to forces over varying durations.
Force-Time Relationship
In rocket physics, the force-time relationship plays a crucial role in understanding how forces affect the motion of rockets. According to Newton's second law of motion, the net force acting on an object is equal to the rate of change of its momentum.

This relationship is particularly vital when examining variable forces, such as the increasing force exerted by a rocket engine described in the textbook problem. The engine's force on the rocket is a function of time, given by the equation: \[\begin{equation}F_{x}=A t^{2}ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline ewline ewline onumberewline ewline ewline ewline onumberewline ewline onumberewline ewline ewline ewline onumberewline ewline ewline ewline onumberewline ewline onumberewline ewline ewline ewline onumberewline ewline ewline ewline onumberewline ewline onumberewline ewline ewline ewline onumber\end{equation}\],where 'A' is a constant that determines how the force changes over time. The force-time graph for this would show a parabolic curve, indicating increasing force as time progresses.
  • The force-time relationship helps us predict future motion by understanding how force varies with time.
  • It serves as a foundation for calculating impulse, since impulse is the integral (or the area under the curve) of the force-time graph.
  • Different engines can generate different force-time profiles, affecting how the rocket accelerates.
Velocity Change
Velocity change is a direct consequence of the forces applied to a rocket, as described by the impulse-momentum theorem and force-time relationship. It signifies how much the speed and direction of the rocket's motion are altered over the period during which forces act.

In our textbook scenario, the velocity change is the result of the impulse exerted on the rocket by the engine's force over a specified time interval. Once the impulse has been calculated, the change in the rocket's velocity can be determined by dividing the impulse by the rocket's mass. The equation that describes this is: \[\begin{equation}\triangle v = \frac{J}{m}ewline onumberewline ewline onumberewline ewline onumberewline ewline onumberewline ewline \end{equation}\],where
  • \[\begin{equation}Jewline onumberewline \end{equation}\] is the impulse measured in Newton-seconds (Ns),
  • \[\begin{equation}mewline onumberewline \end{equation}\] is the rocket's mass in kilograms (kg), and
  • \[\begin{equation}\triangle vewline onumberewline \end{equation}\] is the change in velocity in meters per second (m/s).
The velocity change gives us practical information about the rocket's performance, such as how quickly it can reach a desired speed or alter its orbit. It is a critical factor in mission planning and maneuvering in space.

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Most popular questions from this chapter

A Variable-Mass Raindrop. In a rocket-propulsion problem the mass is variable. Another such problem is a raindrop falling through a cloud of small water droplets. Some of these small droplets adhere to the raindrop, thereby increasing its mass as it falls. The force on the raindrop is $$ F_{\mathrm{ext}}=\frac{d p}{d t}=m \frac{d v}{d t}+v \frac{d m}{d t} $$ Suppose the mass of the raindrop depends on the distance \(x\) that it has fallen. Then \(m=k x,\) where \(k\) is a constant, and \(d m / d t=k u .\) This gives, since \(F_{\text {cut }}=m g_{*}\) $$ m g=m \frac{d v}{d t}+v(k v) $$ Or, dividing by \(k_{*}\) $$ x_{B}=x \frac{d v}{d t}+v^{2} $$ This is a differential equation that has a solution of the form \(v=a t\) where \(a\) is the acceleration and is constant. Take the initial velocity of the raindrop to be zero. (a) Using the proposed solution for \(v\). find the acccleration \(a\). (b) Find the distance the raindrop has fallen in \(t=3.00 \mathrm{~s} .(\mathrm{c})\) Given that \(k=2.00 \mathrm{~g} / \mathrm{m},\) find the mass of the raindrop at \(t=3.00 \mathrm{~s}\). (For many more intriguing aspects of this problem, see K. S. Krane, American Jourmal of Physics, Vol. 49(1981) . pp. \(113-117 .\) )

When cars are equipped with flexible bumpers, they will bounce off cach other during low-specd collisions, thus causing less damage, In one such accident, a \(1750 \mathrm{~kg}\) car traveling to the right at \(1.50 \mathrm{~m} / \mathrm{s}\) collides with a \(1450 \mathrm{~kg}\) car going to the left at \(1.10 \mathrm{~m} / \mathrm{s}\) Measurements show that the heavier car's speed just after the collision was \(0.250 \mathrm{~m} / \mathrm{s}\) in its original direction. Ignore any road friction during the collision. (a) What was the speed of the lighter car just after the collision? (b) Calculate the change in the combined kinetic eneryy of the two-car system during this collision.

You are standing on a large shect of frictionless ice and holding a large rock. In orver to get off the ice. you throw the rock so it has velocity \(12.0 \mathrm{~m} / \mathrm{s}\) relutive to the eurth at an angle of \(35.0^{\circ}\) above the horizontal. If your mass is \(70.0 \mathrm{~kg}\) and the rock's mass is \(3.00 \mathrm{~kg}\), what is your speed after you throw the rock? (See Discussion Question \(Q 8.7 .)\)

A system consists of two particles. \(\mathrm{At} t=0\) one particle is at the origin; the other, which has a mass of \(0.50 \mathrm{~kg}\). is on the \(y\) -axis at \(y=6,0 \mathrm{~m} .\) At \(t=0\) the center of mass of the system is on the \(y\) -axis at \(y=2.4 \mathrm{~m} .\) The velocity of the center of mass is given by \(\left(0.75 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2} \hat{i}\) (a) Find the total mass of the system. (b) Find the acceleration of the center of mass at any time \(t\). (c) Find the net external force acting on the system at \(t=3.0 \mathrm{~s}\)

Two figure skaters, one weighing \(625 \mathrm{~N}\) and the other \(725 \mathrm{~N}\) push off against cach other on frictionless ice. (a) If the heavicr skater travels at \(1.50 \mathrm{~m} / \mathrm{s}\), how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this cnergy come from?

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