/*! 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 44 A \(15.0 \mathrm{~kg}\) block is... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(15.0 \mathrm{~kg}\) block is attached to a very light horizontal spring of force constant \(500.0 \mathrm{~N} / \mathrm{m}\) and is resting on a frictionless horizontal table (Fig. \(\mathrm{LX.44}\) ). Suddenly it is struck by a \(3.00 \mathrm{~kg}\) stone traveling horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) to the right, whereupon the stone rebounds at \(2.00 \mathrm{~m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

Short Answer

Expert verified
To solve this exercise, apply conservation of momentum to find the final velocity of the block and then conservation of energy to find the maximum compression of the spring.

Step by step solution

01

Calculate the initial and final momentum

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Initially, only the stone is moving, so the initial momentum is the mass of the stone \(m_s = 3.00kg\) times its initial velocity \(v_{si} = 8.00m/s\). The final momentum after the collision is the sum of the momentum of the stone and the block. The stone's momentum is its mass times its final velocity \(v_{sf} = -2.00m/s\) (the negative sign indicates the stone moves to the left), and the block's momentum is the mass of the block \(m_b = 15.0kg\) times its final velocity \(v_{b}\). So we have: \(m_s*v_{si} = m_s*v_{sf} + m_b*v_{b}\). This allows us to solve for \(v_{b}\).
02

Use energy conservation to find spring compression

After the collision, the block moves and compresses the spring. At the maximum compression of the spring, all the kinetic energy of the block would have been transferred into potential energy stored in the spring. The kinetic energy of the block is \(\frac{1} {2} m_b * v_{b}^2\), and the spring potential energy is \(\frac{1} {2} k * x^2\), where \(x\) is the compression of the spring and \(k = 500.0N/m\) is the spring constant. Using energy conservation, we equate these two expressions: \(\frac{1} {2} m_b * v_{b}^2 = \frac{1} {2} k * x^2\). We can now solve for \(x\).
03

Substitute the values

Now we just substitute all the known values in our formulas from steps 1 and 2 to solve for \(v_{b}\) and \(x\). Keep in mind to check if the units are consistent and be careful with the signs: positive for right motion and negative for left motion.

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Key Concepts

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

Spring Force Constant
When we talk about springs in physics, one of the key parameters is the spring force constant, often denoted as \(k\). This constant characterizes how stiff the spring is, and is measured in Newtons per meter (N/m). A higher spring constant means the spring is stiffer and requires more force to compress or extend it by a unit length.

In the given problem, the spring attached to the block has a spring constant of \(500.0 \mathrm{~N/m}\). This rigidity implies that substantial force is needed to compress the spring. The spring force constant links directly to Hooke's Law, expressed as \(F = kx\), where \(F\) is the force applied to the spring and \(x\) is the displacement (compression or elongation) of the spring from its equilibrium position.
  • The larger the spring constant, the more force is required for the same displacement.
  • This concept is crucial in determining how much the spring compresses when the block moves after the collision, as in the problem.
Energy Conservation
The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In the context of our problem, after the collision, the kinetic energy from the block is eventually transformed into potential energy stored in the spring.

This transformation is governed by the equation: \( \text{Kinetic Energy of Block} = \text{Potential Energy in Spring} \)This translates into the mathematical expression: \[ \frac{1}{2} m_b v_{b}^2 = \frac{1}{2} k x^2 \]Here's a breakdown of the terms:
  • \(\frac{1}{2} m_b v_{b}^2\) is the kinetic energy of the block, with \(m_b\) and \(v_b\) representing the mass and velocity of the block.
  • \(\frac{1}{2} k x^2\) is the potential energy stored in the compressed spring, where \(k\) is the spring force constant and \(x\) is the compression distance.
In problems like these, energy conservation helps us determine how far the spring compresses after absorbing the block's energy.
Kinetic and Potential Energy
Kinetic energy and potential energy are two fundamental forms of mechanical energy that often convert into one another.
  • **Kinetic energy** is the energy of motion, calculated as \(\frac{1}{2} mv^2\), where \(m\) is mass and \(v\) is velocity. In our scenario, when the block is struck by the stone, it moves and possesses kinetic energy.
  • **Potential energy** is stored energy dependent on an object's position or arrangement. For a spring, this is given by \(\frac{1}{2} k x^2\), where \(k\) is the spring constant and \(x\) is the displacement from equilibrium.
As the problem illustrates, initially, the block has kinetic energy from its motion. Once it hits the spring, as the spring compresses, the kinetic energy transforms into potential energy, stored within the spring's coils.

Understanding these energy types and their conversion is crucial for problems involving conservation of energy, such as calculating the spring's maximum compression.

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