Math Problem Statement

A block with mass m1 slides with velocity v1 on a frictionless surface towards another block with mass m2. A spring with spring constant k is attached to m2 and compresses when the two blocks interact. Determine: a) Whether the collision is elastic or not. b) The maximum compression of the spring. c) The final velocities of the blocks after the collision.

Solution

This task involves analyzing a collision between two blocks, m1m_1 and m2m_2, with a spring (with spring constant kk) attached to m2m_2. Here's how to approach the questions:


Analysis and Solutions

a) Is the collision elastic or not?
To determine if the collision is elastic:

  1. Elastic collision implies both kinetic energy and momentum are conserved.
  2. Inelastic collision involves momentum conservation, but not necessarily kinetic energy conservation.

Here, as the spring compresses, potential energy is stored in the spring and then released, implying no energy loss to heat or other factors. Thus, the collision is elastic, assuming an ideal spring and no energy loss.


b) Maximum compression of the spring (xmaxx_{\text{max}}):
The maximum compression occurs when the relative velocity of the blocks equals zero momentarily, meaning both blocks move together as one unit.

  1. Initial kinetic energy of the system: Ekin, initial=12m1v12E_{\text{kin, initial}} = \frac{1}{2} m_1 v_1^2

  2. At maximum compression, all the kinetic energy of m1m_1 is converted into the spring's potential energy: Espring=12kxmax2E_{\text{spring}} = \frac{1}{2} k x_{\text{max}}^2

  3. By conservation of energy: 12m1v12=12kxmax2\frac{1}{2} m_1 v_1^2 = \frac{1}{2} k x_{\text{max}}^2

  4. Solve for xmaxx_{\text{max}}: xmax=m1v12kx_{\text{max}} = \sqrt{\frac{m_1 v_1^2}{k}}


c) Final velocities of the blocks (v1v_1' and v2v_2'):
By conservation of momentum and the fact that the collision is elastic, we use these equations:

  1. Momentum conservation: m1v1=m1v1+m2v2m_1 v_1 = m_1 v_1' + m_2 v_2'

  2. Velocity relationship in elastic collisions: v1v2=(v1v2)v_1 - v_2 = -(v_1' - v_2')

Combine these equations to solve for v1v_1' and v2v_2': v1=m1m2m1+m2v1v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1 v2=2m1m1+m2v1v_2' = \frac{2 m_1}{m_1 + m_2} v_1


Summary of Results

  • a) The collision is elastic.
  • b) Maximum compression of the spring: xmax=m1v12kx_{\text{max}} = \sqrt{\frac{m_1 v_1^2}{k}}
  • c) Final velocities of the blocks: v1=m1m2m1+m2v1,v2=2m1m1+m2v1v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1, \quad v_2' = \frac{2 m_1}{m_1 + m_2} v_1

Let me know if you'd like further details!
Follow-up questions:

  1. How does the spring constant kk affect xmaxx_{\text{max}}?
  2. What happens if m1=m2m_1 = m_2?
  3. How is energy shared between the blocks after the collision?
  4. Can this method handle a non-ideal spring?
  5. How would damping forces affect the analysis?

Tip: Always verify whether energy losses occur when analyzing real-world systems!

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Math Problem Analysis

Mathematical Concepts

Elastic Collisions
Conservation of Momentum
Conservation of Energy
Hooke's Law

Formulas

Conservation of Momentum: m1 * v1 = m1 * v1' + m2 * v2'
Conservation of Kinetic Energy: (1/2) * m1 * v1^2 = (1/2) * k * x_max^2
Spring Compression: x_max = sqrt((m1 * v1^2) / k)
Velocity Relationships: v1 - v2 = -(v1' - v2')

Theorems

Conservation of Momentum
Conservation of Energy in Elastic Collisions
Hooke's Law

Suitable Grade Level

Undergraduate Physics or Advanced High School Physics