/*! 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 14 A glider of mass \(m\) is free t... [FREE SOLUTION] | 91Ó°ÊÓ

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A glider of mass \(m\) is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Model the launcher as a light spring of force constant \(k\) compressed by a distance \(x\). The glider is released from rest. (a) Show that the glider attains a speed of \(v=x(k / m)^{1 / 2} .\) (b) Does a glider of large or of small mass attain a greater speed? (c) Show that the impulse imparted to the glider is given by the expression \(x(k m)^{1 / 2} .\) (d) Is a greater impulse injected into a large or a small mass? (e) Is more work done on a large or a small mass?

Short Answer

Expert verified
The glider attains a speed of \(v=\sqrt{\frac{k}{m}}x\), smaller mass gliders attain greater speeds, the impulse imparted is \(\sqrt{km}x\), larger masses get larger impulses, and equal work is done on large and small masses.

Step by step solution

01

Identify the necessary physics principles

For this problem, the key physics principles are the conservation of energy (kinetic energy + potential energy = constant) and the Impulse-Momentum theorem (impulse equals the change in momentum). We can use these two principles to solve the different parts of the problem.
02

Solve for the maximum speed

As per conservation of energy, the potential energy of the compressed spring (given by the equation \(\frac{1}{2}kx^2\)) is converted into kinetic energy of the glider (given by the equation \(\frac{1}{2}mv^2\)). Set these two equivalent and solve for \(v\), giving a speed of \(v=\sqrt{\frac{k}{m}}x\).
03

Determine the impact of mass on speed

From the derived formula for \(v\), it's clear that the speed attained by the glider is inversely proportional to the square root of its mass, meaning a glider of smaller mass will attain a greater speed.
04

Calculate the impulse imparted to the glider

Using the Impulse-Momentum theorem (Impulse = Change in momentum = Final momentum - Initial momentum), and knowing that the initial momentum was 0 (glider at rest), the impulse is equal to the final momentum, which is \(mv\). Substituting \(v=\sqrt{\frac{k}{m}}x\) into the equation for impulse gives \(Impulse = m * \sqrt{\frac{k}{m}}x = \sqrt{km}x\).
05

Evaluate the impact of mass on impulse and work

Again, from the derived formula it can be seen that the impulse is proportional to the square root of mass, meaning a greater impulse is injected into a glider of larger mass. As for the work done, this is determined by the formula Work = Change in kinetic energy = \(\frac{1}{2}mv^2 - \frac{1}{2}mu^2\). Given that \(u=0\) and using the earlier formula for \(v\), the work is \(\frac{1}{2}m(\sqrt{\frac{k}{m}}x)^2\), which just simplifies to \(\frac{1}{2}kx^2\), making it independent of the mass of the glider, meaning equal work is done on both large and small masses.

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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 key concept in physics that relates the impulse exerted on an object to its change in momentum. Impulse refers to the effect of a force acting over a period of time and is calculated using the formula:\[\text{Impulse} = F \cdot \Delta t\]Here, \(F\) is the force applied, and \(\Delta t\) is the time duration over which the force acts. Momentum, on the other hand, is a measure of the quantity of motion an object has and is determined using mass and velocity:\[\text{Momentum} = m \cdot v\]In the context of the glider problem, we are essentially looking at the change in momentum, calculated by the final momentum minus the initial momentum. Given that the glider starts from rest, the initial momentum is zero, simplifying the calculation to:\[\text{Impulse} = m \cdot v\]By substituting the derived velocity expression \(v = \sqrt{\frac{k}{m}}x\), we arrive at the impulse imparted to the glider:\[\text{Impulse} = \sqrt{km}x\]This shows that the impulse depends on both \(\sqrt{k}\), the spring constant, \(m\), the mass of the glider, and the compression distance \(x\). A larger mass receives a greater impulse due to the square root dependency on mass.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It is one of the fundamental types of energy in physics, especially when analyzing dynamics like the glider example. The equation for kinetic energy \(KE\) is:\[KE = \frac{1}{2}mv^2\]For the glider, initially at rest, all potential energy stored in the compressed spring gets converted entirely into kinetic energy as the spring releases. This concept supports the conservation of energy principle, where the total energy in a closed system remains constant:
  • Initial potential energy in the spring = Final kinetic energy of the glider
The potential energy stored in the spring is \(\frac{1}{2}kx^2\), so equating it with the kinetic energy gives us:\[\frac{1}{2}kx^2 = \frac{1}{2}mv^2\]Solving this for velocity \(v\), we find:\[v = \sqrt{\frac{k}{m}}x\]This formula shows the relationship between mass and velocity: a smaller mass will achieve a higher velocity, resulting in more significant kinetic energy, emphasizing how mass influences dynamics in motion scenarios.
Potential Energy
Potential energy is the energy held by an object due to its position relative to other objects or forces. In this glider exercise, we're dealing with elastic potential energy in a compressed spring, defined by:\[\text{Potential Energy} = \frac{1}{2}kx^2\]Here, \(k\) is the spring constant—a measure of the spring's stiffness—and \(x\) is the distance the spring is compressed. When the spring is released, this potential energy converts to kinetic energy, propelling the glider along the track.The main connection of potential energy to this exercise lies in its conversion to kinetic energy as the spring expands. The idea is straightforward:
  • When the spring is fully compressed, energy stored is at its maximum.
  • When the spring returns to its original length, potential energy is zero, and all of it turns into kinetic energy.
Potential energy is pivotal because it powerfully impacts the dynamics and speeds which the glider can attain. Notably, potential energy here is independent of the glider's mass, and instead purely dependent on the spring's force constant and compression—showing how energy transformations hinge on system configurations.

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Most popular questions from this chapter

There are (one can say) three coequal theories of motion: Newton's second law, stating that the total force on an object causes its acceleration; the work- kinetic energy theorem, stating that the total work on an object causes its change in kinetic energy; and the impulse-momentum theorem, stating that the total impulse on an object causes its change in momentum. In this problem, you compare predictions of the three theories in one particular case. A 3.00-kg object has velocity 7.00 \(\hat{\mathbf{j}}\) m/s. Then, a total force \(12.0 \hat{\mathbf{i}} \mathrm{N}\) acts on the object for 5.00 s. (a) Calculate the object's final velocity, using the impulse-momentum theorem. (b) Calculate its acceleration from \(\mathbf{a}=\left(\mathbf{v}_{f}-\mathbf{v}_{i}\right) / \Delta t\) (c) Calculate its acceleration from \(\mathbf{a}=\Sigma \mathbf{F} / m\) (d) Find the object's vector displacement from \(\Delta \mathbf{r}=\mathbf{v}_{i} t+\frac{1}{2} \mathbf{a} t^{2}\).(e) Find the work done on the object from \(W=\mathbf{F} \cdot \Delta \mathbf{r}\).(f) Find the final kinetic energy from \(\frac{1}{2} m v_{f}\space^{2}$$=\frac{1}{2} m \mathbf{v}_{f} \cdot \mathbf{v}_{f}\).(g) Find the final kinetic energy from \(\frac{1}{2} m v_{i}\space^{2}+W\).

A 2.00 -kg particle has a velocity \((2.00 \hat{\mathbf{i}}-3.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s},\) and a \(3.00-\mathrm{kg}\) particle has a velocity \((1.00 \hat{\mathbf{i}}+6.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) Find (a) the velocity of the center of mass and (b) the total momentum of the system.

An object of mass \(3.00 \mathrm{kg},\) moving with an initial velocity of \(5.00 \mathrm{i} \mathrm{m} / \mathrm{s},\) collides with and sticks to an object of mass \(2.00 \mathrm{kg}\) with an initial velocity of \(-3.00 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s} .\) Find the final velocity of the composite object.

Sand from a stationary hopper falls onto a moving conveyor belt at the rate of \(5.00 \mathrm{kg} / \mathrm{s}\) as in Figure \(\mathrm{P} 9.72 .\) The conveyor belt is supported by frictionless rollers and moves at a constant speed of \(0.750 \mathrm{m} / \mathrm{s}\) under the action of a constant horizontal external force \(\mathbf{F}_{\text {ext }}\) supplied by the motor that drives the belt. Find (a) the sand's rate of change of momentum in the horizontal direction, (b) the force of friction exerted by the belt on the sand, (c) the external force \(\mathbf{F}_{\mathrm{ext}},\) (d) the work done by \(\mathbf{F}_{\mathrm{ext}}\) in \(1 \mathrm{s},\) and \((\mathrm{e})\) the kinetic energy acquired by the falling sand each second due to the change in its horizontal motion. (f) Why are the answers to \((\mathrm{d})\) and \((\mathrm{e})\) different?

A friend claims that, as long as he has his seat belt on, he can hold on to a 12.0 -kg child in a \(60.0 \mathrm{mi} / \mathrm{h}\) head-on collision with a brick wall in which the car passenger compartment comes to a stop in 0.050 0 s. Show that the violent force during the collision will tear the child from his arms. A child should always be in a toddler seat secured with a seat belt in the back seat of a car.

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