/*! 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 59 block 1 (mass \(2,0 \mathrm{~kg}... [FREE SOLUTION] | 91Ó°ÊÓ

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block 1 (mass \(2,0 \mathrm{~kg}\) ) is moving rightward at \(10 \mathrm{~m} / \mathrm{s}\) and block 2 (mass \(5.0 \mathrm{~kg}\) ) is moving rightward at \(3.0 \mathrm{~m} / \mathrm{s}\) The surface is frictionless, and a spring with a spring constant of \(1120 \mathrm{~N} / \mathrm{m}\) is fixed to block 2 . When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.

Short Answer

Expert verified
The maximum compression of the spring is 0.25 meters.

Step by step solution

01

Analyze the Collision

We first need to understand that the collision between the two blocks is perfectly inelastic at maximum compression. At this point, both blocks move with the same velocity. This occurs because energy in the form of kinetic energy is transferred into potential energy stored in the spring. Since the blocks continue to move together, the momentum before and after collision is conserved.
02

Apply Conservation of Momentum

Using the conservation of momentum principle, calculate the velocity of the blocks at maximum spring compression.Let the final velocity be\( v_f \).Initial momentum \( p_{initial} \) is given by:\[ p_{initial} = m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = 2.0 \cdot 10 + 5.0 \cdot 3 \]\[ p_{initial} = 20 + 15 = 35 \, \text{kg} \cdot \text{m/s} \]Since momentum is conserved:\[ m_1 \cdot v_f + m_2 \cdot v_f = 35 \]\[ (m_1 + m_2) \cdot v_f = 35 \]\[ (2.0 + 5.0) \cdot v_f = 35 \]Solving for \( v_f \):\[ v_f = \frac{35}{7} = 5 \, \text{m/s} \]
03

Calculate Initial Kinetic Energy

Calculate the initial total kinetic energy of the system before the blocks collide.\[ KE_{initial} = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 \]\[ KE_{initial} = \frac{1}{2} \cdot 2 \cdot 10^2 + \frac{1}{2} \cdot 5 \cdot 3^2 \]\[ KE_{initial} = 100 + 22.5 = 122.5 \, ext{J} \]
04

Calculate Final Kinetic Energy

Calculate the kinetic energy of the blocks when they move together at the velocity \( v_f \).\[ KE_{final} = \frac{1}{2} (m_1 + m_2) v_f^2 \]\[ KE_{final} = \frac{1}{2} \cdot 7 \cdot 5^2 \]\[ KE_{final} = \frac{1}{2} \cdot 7 \cdot 25 \]\[ KE_{final} = 87.5 \, ext{J} \]
05

Calculate Potential Energy Stored in the Spring

The difference in initial and final kinetic energy is stored as potential energy in the spring when compression is maximum:\[ PE_{spring} = KE_{initial} - KE_{final} \]\[ PE_{spring} = 122.5 - 87.5 = 35 \, ext{J} \]Equate this to spring potential energy:\[ \frac{1}{2} k x^2 = 35 \]where \( x \) is the maximum compression. Substitute \( k = 1120 \; \text{N/m} \) and solve for \( x \):
06

Solve for Compression x

Set up the equation for spring potential energy:\[ \frac{1}{2} \times 1120 \times x^2 = 35 \]Solving for \(x^2\):\[ 560 x^2 = 35 \]\[ x^2 = \frac{35}{560} \]\[ x^2 = \frac{1}{16} \]\[ x = \sqrt{\frac{1}{16}} \]\[ x = \frac{1}{4} = 0.25 \, ext{m} \]
07

Final Answer

The maximum compression of the spring when both blocks have the same velocity is 0.25 meters.

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Key Concepts

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

Inelastic Collision
An inelastic collision is a type of collision where the colliding objects stick together after the impact. In this scenario, momentum is conserved but kinetic energy is not necessarily conserved.
\[ m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f \]
In the exercise, when two blocks collide, they move together at the maximum compression of the spring. This means the collision is perfectly inelastic as the blocks have a shared final velocity.

It's essential to recognize that during an inelastic collision:
  • Momentum remains conserved throughout the process.
  • Some kinetic energy transforms into other forms of energy, such as heat or, in this case, potential energy stored in the spring.
Understanding how momentum transfers between the objects helps us solve problems where energy transformations occur.
Spring Potential Energy
Spring potential energy is the energy stored in a spring when it is compressed or stretched. Hooke's Law describes the force exerted by a spring as proportional to its displacement:
\[ F = -kx \]
where \( k \) is the spring constant, and \( x \) is the displacement from the spring's rest length.

In our exercise, as the moving blocks collide and compress the spring, their kinetic energy transforms into spring potential energy:
\[ PE_{spring} = \frac{1}{2} k x^2 \]
Here, the work done on the spring through compression becomes stored potential energy. It's crucial for students to distinguish this transformation as it allows energy conservation analysis.

Key points to understand:
  • The spring constant \( k \) determines how resistant the spring is to compression or stretching.
  • The maximum compression represents the point where kinetic energy is fully converted to potential energy.
This conversion plays a pivotal role in analyzing mechanical systems involving springs and collisions.
Kinetic Energy Transformation
Kinetic energy transformation involves converting a system's kinetic energy to other forms. In the exercise, the initial total kinetic energy reduces as the blocks collide and compress the spring.
Initial kinetic energy is given by:
\[ KE_{initial} = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 \]
After the collision, as the blocks move together at velocity \( v_f \), their combined kinetic energy is:
\[ KE_{final} = \frac{1}{2} (m_1 + m_2) v_f^2 \]
The difference between initial and final kinetic energies converts to potential energy in the spring:
\[ KE_{initial} - KE_{final} = PE_{spring} \]

Understanding this transformation is fundamental in analyzing how energy enters or leaves a system:
  • In inelastic collisions, kinetic energy decreases because some energy transforms to potential energy or another form.
  • Accounting for this transformation ensures accurate calculations of energy changes during a collision.
Recognizing energy conservation principles can simplify complex physical scenarios significantly.

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