/*! 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 106 The spring in the muzzle of a ch... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The spring in the muzzle of a child's spring gun has a spring constant of \(700 \mathrm{~N} / \mathrm{m}\). To shoot a ball from the gun, first the spring is compressed and then the ball is placed on it. The gun's trigger then releases the spring, which pushes the ball through the muzzle. The ball leaves the spring just as it leaves the outer end of the muzzle. When the gun is inclined upward by \(30^{\circ}\) to the horizontal, a \(57 \mathrm{~g}\) ball is shot to a maximum height of \(1.83 \mathrm{~m}\) above the gun's muzzle. Assume air drag on the ball is negligible. (a) At what speed does the spring launch the ball? (b) Assuming that friction on the ball within the gun can be neglected, find the spring's initial compression distance.

Short Answer

Expert verified
(a) The launch speed is 11.97 m/s. (b) The spring's initial compression is 0.0341 m.

Step by step solution

01

Understanding Launch Speed Calculation

To find the speed at which the ball is launched, we use the fact that the maximum height reached by the ball can help us determine the initial vertical speed. Using the equation for motion under gravity, we have \(v_y^2 = u_y^2 + 2as\) where \(v_y = 0\) at maximum height, \(a = -g = -9.8 \ m/s^2\), and \(s = 1.83 \ m\). This simplifies to \(0 = u_y^2 - 2 \cdot 9.8 \cdot 1.83\). Solve for \(u_y\).
02

Solving for Initial Vertical Speed

Using the equation from Step 1, \(u_y^2 = 2 \cdot 9.8 \cdot 1.83\). Calculating this gives \(u_y^2 = 35.868\). Thus, \(u_y = \sqrt{35.868} = 5.985 \ m/s\).
03

Finding Total Launch Speed

The initial launch angle is \(30^{\circ}\). The vertical component of speed is \(u_y = v_0 \sin 30^{\circ}\) and the horizontal component is \(v_0 \cos 30^{\circ}\). To find \(v_0\), use \(u_y = v_0 \sin 30^{\circ} = \frac{v_0}{2}\). Solving \(v_0 = 2 \times 5.985 = 11.97 \ m/s\).
04

Initial Compression Distance Calculation

Use the conservation of energy. The potential energy in the compressed spring when released converts to the kinetic energy of the ball. Spring energy is \(\frac{1}{2}kx^2\) and kinetic energy is \(\frac{1}{2}mv_0^2\). Set them equal: \(\frac{1}{2}kx^2 = \frac{1}{2}mv_0^2\). Solve for \(x\).
05

Solving for Spring Compression

Substitute known values: spring constant \(k = 700 \ N/m\), mass \(m = 0.057 \ kg\), and velocity \(v_0 = 11.97 \ m/s\) into the equation from Step 4: \(700x^2 = 0.057 \cdot 11.97^2\). Calculate to find \(x\).
06

Calculating Compression Distance

Substituting and simplifying gives \(700x^2 = 0.81338\). Solving for \(x\), we get \(x^2 = \frac{0.81338}{700} = 0.001162\). Therefore, \(x = \sqrt{0.001162} = 0.0341 \ m\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Projectile Motion
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. In this exercise, the ball is launched from a spring gun at an angle of \(30^{\circ}\) above the horizontal. This angle affects how far and how high the ball travels.
The launch speed determines the horizontal and vertical components of the velocity:
  • The horizontal component is described by \(v_0 \cos \theta\), where \(\theta\) is the launch angle.
  • The vertical component is \(v_0 \sin \theta\).
The key here is that the vertical component is what determines how high the projectile will go. In this particular problem, when the ball reaches the maximum height, its vertical velocity becomes zero. This allows us to use motion equations to find the initial vertical speed, which plays a crucial role in solving for the overall launch speed. Understanding these components help solve any projectile problem efficiently.
Conservation of Energy
The principle of conservation of energy is pivotal in this exercise. It tells us that energy cannot be created or destroyed, only transformed from one form to another. Here, it's used to link the energy from the compressed spring to the motion of the ball.
When the spring is compressed, it stores potential energy given by:\[\text{Potential Energy} = \frac{1}{2} k x^2\]where \(k\) is the spring constant, and \(x\) is the compression distance. When released, this energy converts entirely into the kinetic energy of the ball, described by:\[\text{Kinetic Energy} = \frac{1}{2} m v_0^2\]where \(m\) is the mass of the ball, and \(v_0\) is the velocity of the ball just as it leaves the spring. By equating the potential energy in the spring to the kinetic energy of the ball, we can solve for the unknown compression distance, thus applying the conservation of energy to find out how far the spring had to be compressed initially. This highlights the seamless conversion between stored and motion energy.
Compression Distance
Compression distance is how much the spring is compressed initially to launch the ball. It's a vital step in linking the initial state of the spring with the launch speed of the ball. This measurement is crucial because the amount of compression determines how much potential energy is initially stored in the spring.
The relationship is expressed through the equation for spring potential energy:\[\frac{1}{2} k x^2 = \frac{1}{2} m v_0^2\]Solving this equation with known parameters such as the spring constant \(k\), the mass of the ball \(m\), and the initial launch speed \(v_0\) allows us to find \(x\), the compression distance.Once the calculations are carried out, we get the measure of how much the spring was compressed, confirming the required energy to project the ball at the desired speed. This concept binds together the mechanics of springs, energy transformations, and projectile motion into a single problem-solving strategy.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

To form a pendulum, a \(0.092 \mathrm{~kg}\) ball is attached to one end of a rod of length \(0.62 \mathrm{~m}\) and negligible mass, and the other end of the rod is mounted on a pivot. The rod is rotated until it is straight up, and then it is released from rest so that it swings down around the pivot. When the ball reaches its lowest point, what are (a) its speed and (b) the tension in the rod? Next, the rod is rotated until it is horizontal, and then it is again released from rest. (c) At what angle from the vertical does the tension in the rod equal the weight of the ball? (d) If the mass of the ball is increased, does the answer to (c) increase, decrease, or remain the same?

A metal tool is sharpened by being held against the rim of a wheel on a grinding machine by a force of \(180 \mathrm{~N}\). The frictional forces between the rim and the tool grind off small pieces of the tool. The wheel has a radius of \(20.0 \mathrm{~cm}\) and rotates at \(2.50 \mathrm{rev} / \mathrm{s}\) The coefficient of kinetic friction between the wheel and the tool is \(0.320 .\) At what rate is energy being transferred from the motor driving the wheel to the thermal energy of the wheel and tool and to the kinetic energy of the material thrown from the tool?

The magnitude of the gravitational force between a particle of mass \(m_{1}\) and one of mass \(m_{2}\) is given by $$ F(x)=G \frac{m_{1} m_{2}}{x^{2}} $$ where \(G\) is a constant and \(x\) is the distance between the particles. (a) What is the corresponding potential energy function \(U(x) ?\) Assume that \(U(x) \rightarrow 0\) as \(x \rightarrow \infty\) and that \(x\) is positive. (b) How much work is required to increase the separation of the particles from \(x=x_{1}\) to \(x=x_{1}+d ?\)

A \(1.50 \mathrm{~kg}\) water balloon is shot straight up with an initial speed of \(3.00 \mathrm{~m} / \mathrm{s}\). (a) What is the kinetic energy of the balloon just as it is launched? (b) How much work does the gravitational force do on the balloon during the balloon's full ascent? (c) What is the change in the gravitational potential energy of the balloon-Earth system during the full ascent? (d) If the gravitational potential energy is taken to be zero at the launch point, what is its value when the balloon reaches its maximum height? (e) If, instead, the gravitational potential energy is taken to be zero at the maximum height, what is its value at the launch point? (f) What is the maximum height?

A \(0.50 \mathrm{~kg}\) banana is thrown directly upward with an initial speed of \(4.00 \mathrm{~m} / \mathrm{s}\) and reaches a maximum height of \(0.80 \mathrm{~m} .\) What change does air drag cause in the mechanical energy of the bananaEarth system during the ascent?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.