/*! 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 66 You are given the equation used ... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given the equation used to solve a problem. For each of these, you are to a. Write a realistic problem for which this is the correct equation. b. Draw the before-and-after pictorial representation. c. Finish the solution of the problem. $$\begin{array}{l} (0.10 \mathrm{kg}+0.20 \mathrm{kg}) v_{\mathrm{lx}}=(0.10 \mathrm{kg})(3.0 \mathrm{m} / \mathrm{s}) \\ \frac{1}{2}(0.30 \mathrm{kg})(0 \mathrm{m} / \mathrm{s})^{2}+\frac{1}{2}(3.0 \mathrm{N} / \mathrm{m})\left(\Delta x_{2}\right)^{2} \\ =\frac{1}{2}(0.30 \mathrm{kg})\left(v_{1 x}\right)^{2}+\frac{1}{2}(3.0 \mathrm{N} / \mathrm{m})(0 \mathrm{m})^{2} \end{array}$$

Short Answer

Expert verified
The blocks have a speed of 1.0 m/s right after the collision, and the spring is compressed by a displacement of 0.39 m when the blocks stop.

Step by step solution

01

Understand the equations and write a realistic problem

In order to write a realistic problem for these equations, first it must be understood what they represent. The first one represents a collision between two objects of different masses. The second equation represents the potential energy of a mass attached to a spring in a system where no friction is present, and the third is the conservation of energy. Based on this, a possible scenario could be: A 0.10 kg block is moving with a speed of 3.0 m/s on a frictionless surface, where it hits and sticks to a stationary 0.20 kg block. Both blocks are attached to a spring with a spring constant of 3.0 N/m that was initially unstretched. Find the speed of the combined mass immediately after the collision and the displacement of the spring when the blocks stop.
02

Draw the before-and-after pictorial representation

To represent the situation, you can draw two diagrams. The before diagram shows the 0.10 kg block moving by itself and the 0.20 kg block being still, with an unstretched spring attached to it. The after diagram shows both blocks attached and moving together right after the collision, and the spring being compressed when they stop.
03

Finish the solution of the problem

Using the momentum conservation equation \((0.10 kg + 0.20 kg) v_{lx} = (0.10 kg)(3.0 m/s)\), the left side represents the total momentum after the collision and the right side is the initial momentum before it. Solving for \( v_{lx} \) gives a value of \( v_{lx} = 1.0 m/s \). Then, using the conservation of energy equation \(\frac{1}{2} (0.30 kg) (0 m/s)^2 + \frac{1}{2} (3.0 N/m) (\Delta x_{2})^2 = \frac{1}{2} (0.30 kg) (v_{1x})^2 + \frac{1}{2} (3.0N/m) (0m)^2 \), where the left side of the equation represents the final combined kinetic and potential energy, and the right side represents the initial energies. Substituting \( v_{1x} = 1.0 m/s \) you can solve for \( \Delta x_{2} \) and get \( \Delta x_{2} = 0.39 m \).

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Key Concepts

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

Elastic Collision
An elastic collision is an event where two objects collide and bounce off each other without enduring any deformation or heat generation. In such collisions, both momentum and kinetic energy are conserved. Imagine two balls of different masses colliding on a frictionless surface. If they bounce back without changing their shape and with no loss of energy in other forms like heat or sound, that’s an elastic collision.

In the realistic problem provided, the 0.10 kg block hits a stationary 0.20 kg block and sticks to it, which suggests the collision is perfectly inelastic rather than elastic. However, understanding elastic collisions helps us appreciate that in such encounters, if they had bounced off, we could have applied the same momentum conservation principle but also would have had to consider kinetic energy conservation to fully describe the behavior of the blocks post-collision.
Conservation of Energy
The conservation of energy principle states that energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of physics problems, such as the one involving the blocks and spring, this means that the total energy at one point in time must equal the total energy at another, provided no energy is lost to the surroundings.

Initially, the moving block has kinetic energy due to its motion and the spring has potential energy since it’s capable of doing work when compressed. After the collision, as the blocks come to a rest, the kinetic energy of the system is converted into the potential energy of the spring. The conservation of energy equation is used to determine how much energy is stored in the spring (spring potential energy) when the blocks are momentarily at rest.
Momentum Conservation Equation
The momentum conservation equation is a fundamental law of physics which expresses the conservation of momentum in a system. Momentum, the product of an object's mass and its velocity, remains constant in isolated systems. This equation is pivotal when analyzing collisions.

In the step-by-step solution to the problem provided, the momentum conservation equation \(0.10 \text{ kg} + 0.20 \text{ kg}) v_{\text{1x}} = (0.10 \text{ kg})(3.0 \text{ m/s})\) is utilized to determine the velocity of the blocks immediately after they stick together in the collision. This equation illustrates that the initial momentum (mass times velocity of the moving block) equals the final momentum (combined mass times their shared velocity) because there are no external forces acting on the system.
Spring Potential Energy
Spring potential energy is the energy stored within a spring when it is either compressed or stretched from its natural length. According to Hooke's Law, the potential energy stored in a spring is proportional to the square of its displacement (change in length) when a force is applied to it. The equation for spring potential energy is \(\frac{1}{2} k (\Delta x)^2\), where \(k\) is the spring constant and \(\Delta x\) is the displacement.

In our problem, after the combined blocks stop moving due to the collision, they compress the spring. We use the conservation of energy principle to equate the kinetic energy immediately after the collision to the spring potential energy at the point of maximum compression. The spring's potential energy can then be calculated, telling us how much the spring was compressed by the impact of the blocks.

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Most popular questions from this chapter

It's your birthday, and to celebrate you're going to make your first bungee jump. You stand on a bridge 100 m above a raging river and attach a 30 -m-long bungee cord to your harness. A bungee cord, for practical purposes, is just a long spring, and this cord has a spring constant of 40 N/m. Assume that your mass is 80 kg. After a long hesitation, you dive off the bridge. How far are you above the water when the cord reaches its maximum elongation?

You're driving at \(35 \mathrm{km} / \mathrm{hr}\) when the road suddenly descends \(15 \mathrm{m}\) into a valley. You take your foot off the accelerator and coast down the hill. Just as you reach the bottom you see the policeman hiding behind the speed limit sign that reads 770 km/hr." Are you going to get a speeding ticket?

Ball \(1,\) with a mass of \(100 \mathrm{g}\) and traveling at \(10 \mathrm{m} / \mathrm{s},\) collides head-on with ball \(2,\) which has a mass of \(300 \mathrm{g}\) and is initially at rest. What is the final velocity of each ball if the collision is (a) perfectly clastic? (b) perfectly inclastic?

An oxygen atom is four times as massive as a helium atom. In an experiment, a helium atom and an oxygen atom have the same kinetic energy. What is the ratio \(v_{\mathrm{H}} / v_{\mathrm{O}}\) of their speeds?

You have been asked to design a "ballistic spring system" to measure the speed of bullets. A spring whose spring constant is \(k\) is suspended from the ceiling. A block of mass \(M\) hangs from the spring. A bullet of mass \(m\) is fired vertically upward into the bottom of the block. The spring's maximum compression \(d\) is measured. a. Find an expression for the bullet's speed \(v_{\mathrm{B}}\) in terms of \(m, M\) \(k,\) and \(d\) b. What was the speed of a \(10 \mathrm{g}\) bullet if the block's mass is \(2.0 \mathrm{kg}\) and if the spring, with \(k=50 \mathrm{N} / \mathrm{m},\) was compressed by \(45 \mathrm{cm} ?\)

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