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Chapter 8: Problem 11

CALC At time \(t=0,\) a 2150 -kg rocket in outer space fires an engine that exerts an increasing force on it in the \(+x\) -direction. This force obeys the equation \(F_{x}=A t^{2},\) where \(t\) is time, and has a magnitude of 781.25 \(\mathrm{N}\) when \(t=1.25 \mathrm{s}\) . (a) Find the SI value of the constant \(A,\) including its units. (b) What impulse does the engine exert on the rocket during the 1.50 -s interval starting 2.00 \(\mathrm{s}\) after the engine is fired? (c) By how much does the rocket's velocity change during this interval?

Short Answer

Expert verified
(a) A = 500.0 N/s^2. (b) Impulse = 5812.5 Ns. (c) Velocity change = 2.703 m/s.

Step by step solution

01

Determine Constant A

Given \(F_x = A t^2\) and \(F_x = 781.25 \, \text{N}\) at \(t = 1.25 \, \text{s}\), substitute into the formula:\[781.25 = A (1.25)^2\]Solve for \(A\):\[A = \frac{781.25}{1.25^2} = \frac{781.25}{1.5625} = 500.0 \, \text{N/s}^2\]Thus, the constant \(A = 500.0 \, \text{N/s}^2\).
02

Calculate the Impulse

Impulse \(J\) is the integral of force over time:\[J = \int_{t_1}^{t_2} F(t) \, dt = \int_{2.00}^{3.50} A t^2 \, dt\]Substitute \(A = 500.0\):\[J = \int_{2.00}^{3.50} 500 t^2 \, dt\]Compute the integral:\[J = 500 \left[ \frac{t^3}{3} \right]_{2.00}^{3.50} = 500 \left( \frac{3.50^3}{3} - \frac{2.00^3}{3} \right)\]\[J = 500 \left( \frac{42.875}{3} - \frac{8.000}{3} \right) = 500 \left( 14.2917 - 2.6667 \right)\]\[J = 500 \times 11.625 = 5812.5 \, \text{Ns}\]
03

Calculate Change in Velocity

The change in velocity \(\Delta v\) can be found using the relationship \(J = m \Delta v\):\[\Delta v = \frac{J}{m} = \frac{5812.5}{2150} \]\[\Delta v = 2.703 \, \text{m/s}\]

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Key Concepts

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

Newton's Second Law of Motion
Newton's Second Law of Motion is a fundamental principle of classical physics, guiding our understanding of how forces affect the motion of an object. Simply put, this law states that an object accelerates when a force is applied to it. The relationship can be expressed mathematically as \[ F = ma \].
Here, \( F \) represents the force applied, \( m \) stands for the object's mass, and \( a \) is the acceleration produced.

In the context of rocket propulsion, this means that the engine's force is directly related to the change in motion of the rocket.
For example, if a 2150-kg rocket experiences an increasing force as described by \( F_x = At^2 \), this force dictates the acceleration the rocket undergoes.
Since the force depends on time, the acceleration and thus the motion of the rocket will also vary with time.
  • The bigger the force, the greater the acceleration.
  • An increase in mass requires more force to achieve the same acceleration.
Consequently, understanding Newton's Second Law is crucial for determining how changes in force impact the motion of rockets and other objects.
Impulse-Momentum Theorem
The Impulse-Momentum Theorem bridges the concept of impulse, the product of force and time, with momentum changes in an object. It's represented by the equation: \[ J = ext{Impulse} = F imes ext{time} = ext{Change in Momentum} \]
Impulse is essentially the "oomph" that a force gives to an object over a period of time.
In simpler terms, when a force acts on an object for a certain time, it changes the object's momentum.
Momentum, denoted by \( p \), can be defined as \( p = mv \), where \( m \) is mass and \( v \) is velocity.
Therefore, a change in momentum implies a velocity change:
\[ J = riangle p = m imes riangle v \]
  • Greater impulse results in a larger momentum change.
  • A prolonged application of force (long time) can lead to significant velocity changes.
In the rocket scenario, the impulse delivered by the engine over 1.5 seconds changes the rocket's momentum, measured by how much the velocity is altered."
Calculus in Physics
Calculus plays an essential role in understanding the dynamics of physics, where quantities change continuously over time. Concepts such as differentiation and integration allow us to describe and predict these changes accurately.

Integration in Calculus

When we talk about forces that change over time, like the rocket's force given by \( F_x = At^2 \), calculus helps us determine total effects such as impulse.
The impulse can be calculated as the integral of force with respect to time, i.e., \[ J = \int F(t) \, dt \]This calculates the total impulse applied by the varying force over a specified time interval.
  • This integration tells us how changes in force accumulate over time.
  • It reveals the overall impact of time-varying forces on an object's motion.

Finding Velocity

Similarly, calculus allows us to find velocity changes when the force is not constant. By integrating force over time and knowing the object's mass, we can determine how velocity changes.
  • The interplay of force, time, and mass is crucial for predicting real-world motion effectively.
In essence, calculus provides the tools necessary to tackle complex physics problems involving variability and time-dependence.

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Most popular questions from this chapter

A fireworks rocket is fired vertically upward. At its maximum height of \(80.0 \mathrm{m},\) it explodes and breaks into two pieces: one with mass 1.40 \(\mathrm{kg}\) and the other with mass 0.28 \(\mathrm{kg} .\) In the explosion, 860 \(\mathrm{J}\) of chemical energy is converted to kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

A single-stage rocket is fired from rest from a deep-space platform, where gravity is negligible. If the rocket burns its fuel in 50.0 s and the relative speed of the exhaust gas is \(v_{\text { ex }}=2100 \mathrm{m} / \mathrm{s}\) what must the mass ratio \(m_{0} / m\) be for a final speed \(v\) of 8.00 \(\mathrm{km} / \mathrm{s}\) (about equal to the orbital speed of an earth satellite)?

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 \(d m / d t=k v\) . This gives, since \(F_{\text { ext }}=m g\) , $$m g=m \frac{d v}{d t}+v(k v)$$ Or, dividing by \(k\) $$x g=x \frac{d v}{d t}+v^{2}$$ This is a differential equation that has a solution of the form \(v=\) at, 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 acceleration \(a\) . (b) Find the distance the raindrop has fallen in \(t=3.00\) s. (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 \(\mathrm{K} .\) S. Krane, American Journal of Physics, Vol. 49 \((1981),\) pp. \(113-117.)\)

Accident Analysis. Two cars collide at an intersection. Car \(A,\) with a mass of 2000 \(\mathrm{kg}\) , is going from west to east, while car \(B,\) of mass \(1500 \mathrm{kg},\) is going from north to south at 15 \(\mathrm{m} / \mathrm{s}\) . As a result of this collision, the two cars become enmeshed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of \(65^{\circ}\) south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car \(A\) going just before the collision?

CALC A radio-controlled model airplane has a momentum given by \(\left[\left(-0.75 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}^{3}\right) t^{2}+(3.0 \mathrm{kg} \cdot \mathrm{m} / \mathrm{s})\right] \hat{\imath}+(0.25 \mathrm{kg} \cdot \) \(\mathrm{m} / \mathrm{s}^{2} ) t \hat{J} .\) What are the \(x-, y-\) and \(z\) -components of the net force on the airplane?
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