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Chapter 9: Problem 58

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?

Short Answer

Expert verified
The spring is compressed by approximately 0.327 meters.

Step by step solution

01

Understand Conservation of Momentum

Since Block 1 and Block 2 stick together after the collision, we use the law of conservation of momentum to find their velocity just before they stop moving: \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \). Here, \(v_2 = 0\) as Block 2 is initially at rest.
02

Apply Known Values

Substitute the known values into the momentum equation: \( 2.0 \, \text{kg} \cdot 4.0 \, \text{m/s} + 1.0 \, \text{kg} \cdot 0 = (2.0 \, \text{kg} + 1.0 \, \text{kg})v_f \). This simplifies to \( 8.0 \, \text{kg} \cdot \text{m/s} = 3.0 \, \text{kg} \cdot v_f \).
03

Solve for Final Velocity

Rearrange the equation to solve for \(v_f\): \( v_f = \frac{8.0 \, \text{kg} \cdot \text{m/s}}{3.0 \, \text{kg}} = 2.67 \, \text{m/s} \).
04

Relate Kinetic Energy to Spring Compression

The kinetic energy of the blocks is converted to the potential energy stored in the spring when they stop. Use \( KE = \frac{1}{2}(m_1 + m_2)v_f^2 = \frac{1}{2}kx^2 \).
05

Substitute and Simplify

Substitute known values into the energy equation: \( \frac{1}{2}(3.0 \, \text{kg})(2.67 \, \text{m/s})^2 = \frac{1}{2}(200 \, \text{N/m})x^2 \). Simplify to find \(10.67 \, \text{J} = 100x^2 \).
06

Solve for Spring Compression

Rearrange to solve for \(x\): \(x^2 = \frac{10.67}{100} = 0.1067\), so \( x = \sqrt{0.1067} \approx 0.327 \, \text{m} \).

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Key Concepts

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

Spring Constant
When trying to understand the behavior of springs, the spring constant represents how stiff the spring is. It's usually denoted by the letter "k".
  • A large spring constant means the spring is very stiff and difficult to compress or stretch.
  • A small spring constant implies a more flexible spring.
For our exercise, the given spring constant is 200 N/m.
This means that to compress the spring by 1 meter, you'd need to apply 200 Newtons of force.
The spring constant helps us calculate potential energy stored in the spring when it's compressed, or stretched using the formula: \[ PE = \frac{1}{2} k x^2 \] where "PE" stands for potential energy and "x" is the displacement from the equilibrium position (how much the spring is stretched or compressed). This was crucial in our example because we needed to know how much energy was stored after the collision of the blocks.
Kinetic Energy
Kinetic energy is the energy an object has because of its motion. In our problem, this applies when the blocks move after the collision.
  • It's calculated using the formula: \[ KE = \frac{1}{2} mv^2 \]
  • "m" stands for mass, and "v" represents velocity.
For instance, just before the blocks stop, they have a combined kinetic energy due to their movement, with a mass of 3 kg (sum of both blocks) traveling at a velocity found from the conservation of momentum equation.
This energy then transforms into potential energy stored in the compressed spring. Understanding this conversion is essential to determine how far the spring compresses. Thus, kinetic energy plays a pivotal role because it's what eventually compresses the spring.
Potential Energy
Potential energy, in the context of springs, is the energy stored in the spring when it is compressed or stretched. The primary formula used to find this energy was: \[ PE = \frac{1}{2} k x^2 \]
  • "k" is the spring constant, indicating spring stiffness.
  • "x" is the distance by which the spring is compressed or stretched.
In our exercise, when the blocks collide and stick together, their kinetic energy is converted into potential energy. This stored energy determines how much the spring compresses.
It's also important to remember that potential energy increases with both the stiffness of the spring and the amount of compression or extension.Utilizing the known values of kinetic energy and spring constant, we systematically solved for "x", which signifies the distance by which the spring was compressed.

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Most popular questions from this chapter

A uniform soda can of mass \(0.140 \mathrm{~kg}\) is \(12.0 \mathrm{~cm}\) tall and filled with \(0.354 \mathrm{~kg}\) of soda (Fig. \(9-41\) ). Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. What is the height \(h\) of the com of the can and contents (a) initially and (b) after the can loses all the soda? (c) What happens to \(h\) as the soda drains out? (d) If \(x\) is the height of the remaining soda at any given instant, find \(x\) when the com rcaches its lowest point.

A \(0.70 \mathrm{~kg}\) ball moving horizontally at \(5.0 \mathrm{~m} / \mathrm{s}\) strikes a vertical wall and rebounds with speed \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in its linear momentum?

A \(6100 \mathrm{~kg}\) rocket is set for vertical firing from the ground. If the exhaust speed is \(1200 \mathrm{~m} / \mathrm{s}\), how much gas must be cjected cach second if the thrust (a) is to equal the magnitude of the gravitational force on the rocket and (b) is to give the rocket an initial upward acceleration of \(21 \mathrm{~m} / \mathrm{s}^{2} ?\)

block \(L\) of mass \(m_{L}=1.00 \mathrm{~kg}\) and block \(R\) of mass \(m_{R}=0.500 \mathrm{~kg}\) are held in place with a compressed spring between them. When the blocks are released, the spring sends them sliding across a frictionless floor. (The spring has negligible mass and falls to the floor after the blocks leave it.) (a) If the spring gives block \(L\) a release speed of \(1.20 \mathrm{~m} / \mathrm{s}\) relative to the floor, how far does block \(R\) travel in the next \(0.800 \mathrm{~s} ?\) (b) If, instead, the spring gives block \(L\) a release speed of \(1.20 \mathrm{~m} / \mathrm{s}\) relative to the velocity that the spring gives block \(R,\) how far does block \(R\) travel in the next \(0.800 \mathrm{~s} ?\)

Two particles are launched from the origin of the coordinate system at time \(t=0 .\) Particle 1 of mass \(m_{1}=5.00 \mathrm{~g}\) is shot directly along the \(x\) axis on a frictionless floor, with constant speed \(10.0 \mathrm{~m} / \mathrm{s} .\) Particle 2 of mass \(m_{2}=3.00 \mathrm{~g}\) is shot with a velocity of magnitude \(20.0 \mathrm{~m} / \mathrm{s},\) at an upward angle such that it always stays directly above particle \(1 .\) (a) What is the maximum height \(H_{\max }\) reached by the com of the two-particle system? In unit- vector notation, what are the (b) velocity and (c) acceleration of the com when the com reaches \(H_{\max } ?\)
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