/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 An object is tracked by a radar ... [FREE SOLUTION] | 91Ó°ÊÓ

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An object is tracked by a radar station and determined to have a position vector given by \(\vec{r}=(3500-160 t) \hat{i}+2700 \mathrm{j}+300 \hat{\mathrm{k}},\) with \(\vec{r}\) in meters and \(t\) in scconds. The radar station's \(x\) axis points cast. its \(y\) axis north, and its \(z\) axis vertically up. If the object is a \(250 \mathrm{~kg}\) meteorological missile, what are (a) its linear momentum, (b) its direction of motion, and (c) the net force on it?

Short Answer

Expert verified
(a) \( -40000\hat{i} \) kg m/s, (b) due east, (c) 0 N.

Step by step solution

01

Calculate the Velocity Vector

The velocity vector \( \vec{v} \) is obtained by differentiating the position vector \( \vec{r} \) with respect to time \( t \). \[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt} \left((3500 - 160t)\hat{i} + 2700\hat{j} + 300\hat{k}\right) \]This gives us:\[ \vec{v} = -160\hat{i} \] because the other terms are constants.
02

Calculate the Linear Momentum

The linear momentum \( \vec{p} \) is given by the product of the mass \( m \) and the velocity \( \vec{v} \).\[ \vec{p} = m \times \vec{v} = 250 \times (-160\hat{i}) \]Therefore:\[ \vec{p} = -40000\hat{i} \, \text{kg m/s} \]
03

Determine the Direction of Motion

Since the velocity vector \( \vec{v} \) is \(-160\hat{i}\), the direction of motion indicates movement in the negative \( x \)-direction only, which is due east.
04

Calculate the Net Force

The net force \( \vec{F} \) acting on the object is obtained by differentiating the linear momentum \( \vec{p} \) with respect to time. Since \( \vec{p} = -40000\hat{i} \), which is constant, \[ \vec{F} = \frac{d\vec{p}}{dt} = 0 \]Thus, the net force is zero.

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Key Concepts

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

Linear Momentum
Linear momentum is a fundamental concept in mechanics that describes the motion of an object. It is often denoted by the symbol \( \vec{p} \) and is defined as the product of an object's mass \( m \) and its velocity \( \vec{v} \). Mathematically, this is expressed as \( \vec{p} = m \times \vec{v} \). Linear momentum is a vector quantity, meaning it has both a magnitude and a direction. This is important because it indicates not just how fast something is moving, but also in which direction.

In mechanics, linear momentum can help us understand and predict outcomes in collisions and interactions between different objects. It is a conserved quantity, meaning in the absence of external forces, the total momentum of a closed system remains constant. In our exercise, the missile has a linear momentum of \(-40000\hat{i}\) kg m/s. This tells us that the momentum is directed along the negative x-axis, indicating a strong movement in that direction.
Velocity Vector
The velocity vector is an essential concept when analyzing the motion of objects. It defines the speed and direction in which an object is moving. You get the velocity vector by differentiating the position vector with respect to time. For our exercise, the velocity vector is calculated by taking the derivative of the position vector \( \vec{r} \) given as \((3500 - 160t)\hat{i} + 2700\hat{j} + 300\hat{k}\).

Upon differentiation, we find that the velocity vector \( \vec{v} \) is \(-160\hat{i}\). This result shows the object is moving solely in the negative \( x \)-direction with a constant speed of 160 m/s. No components in the \( \/ j \) or \( \hat{k} \) directions indicate that there is no movement north or vertically up, standard in three-dimensional motion coordinate systems. Velocity is crucial as it helps to establish subsequent calculations, such as force or linear momentum.
Net Force
Net force is the total force acting on an object and plays a vital role in determining changes in the motion of the object. According to Newton's Second Law, it is expressed mathematically as \( \vec{F} = \frac{d\vec{p}}{dt} \), which means it is the derivative of momentum with respect to time. If an object has constant momentum, as in our exercise, the net force is zero.

In practical terms, this means that there are no external influences causing the object to accelerate or change its velocity. In the example given, since \( \vec{p} = -40000\hat{i} \) is a constant value, taking the derivative results in \( \vec{F} = 0 \). This conclusion tells us the missile is moving at a constant velocity without being impacted by new forces, which is typically suggestive of a closed system in motion.
Direction of Motion
Direction of motion refers to the path that an object follows as it moves. Understanding this helps us predict the future trajectory of an object. In vector terms, the direction of motion aligns with the velocity vector of the object.

In our exercise, since the velocity vector \( \vec{v} \) is \(-160\hat{i}\), the missile is moving in the negative \( x \)-direction. This suggests the object is heading due east in the given spatial orientation. The absence of \( \/ j \) and \( \hat{k} \) components in the velocity vector implies that there are no movements to the north or vertically upward.

Knowing the direction tells engineers, pilots, or programmers exactly how an object will proceed, allowing for informed decisions and interventions if necessary. Recognizing the consistency and direction can also aid in anticipating future states of movement and planning resources accordingly.

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Most popular questions from this chapter

A rocket sled with a mass of \(2900 \mathrm{~kg}\) moves at \(250 \mathrm{~m} / \mathrm{s}\) on a set of rails. At a certain point, a scoop on the sled dips into a trough of water located between the tracks and scoops water into an empty tank on the sled. By applying the principle of conservation of linear momentum, determine the speed of the sled after \(920 \mathrm{~kg}\) of water has been scooped up. Ignore any retarding force on the scoop.

block \(A\) (mass \(1.6 \mathrm{~kg}\) ) slides into block \(B\) (mass \(2.4 \mathrm{~kg}\) ), along a frictionless surface. The directions of three velocities before \((i)\) and after \((f)\) the collision are indicated; the corresponding speeds are \(v_{A i}=5.5 \mathrm{~m} / \mathrm{s}\) \(v_{B i}=2.5 \mathrm{~m} / \mathrm{s},\) and \(v_{B f}=4.9 \mathrm{~m} / \mathrm{s} .\) What are the (a) speed and (b) direction (left or right) of velocity \(\vec{v}_{A f} ?\) (c) Is the collision elastic?

block 2 (mass \(1.0 \mathrm{~kg}\) ) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant \(200 \mathrm{~N} / \mathrm{m} .\) The other end of the spring is fixed to a wall. Block 1 (mass \(2.0 \mathrm{~kg}\) ), traveling at speed \(v_{1}=4.0 \mathrm{~m} / \mathrm{s}\), collides with block 2 , and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

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