Solving for Compression of Unstretched Spring

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SUMMARY

The discussion focuses on solving a physics problem involving two blocks colliding and compressing a spring. Block 1 (mass 1.70 kg) traveling at 3.60 m/s collides with Block 2 (mass 1.10 kg) at rest, resulting in a spring with a spring constant of 144 N/m being compressed. The correct compression distance of the spring is determined to be 0.305 m. Key concepts include conservation of momentum and energy transfer to the spring during the collision.

PREREQUISITES
  • Understanding of conservation of momentum
  • Knowledge of kinetic energy calculations
  • Familiarity with Hooke's Law (Fsp = -kx)
  • Basic principles of elastic potential energy
NEXT STEPS
  • Study conservation of momentum in inelastic collisions
  • Learn about kinetic energy and its conversion to potential energy in springs
  • Explore Hooke's Law and its applications in mechanical systems
  • Investigate energy conservation principles in physics problems
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding mechanics, particularly inelastic collisions and spring dynamics.

DavidAp
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Block 2 (mass 1.10 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 144 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.70 kg), traveling at speed v1 = 3.60 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

Answer 0.305 m



Relavant Equations
Fsp = -kx
______________________________

How do I solve for this? I'm given k but what is Fsp? Is there any other way to solve for Fsp?

After scratching my head for a good while I tried something out of pure desperation, despite my gut feeling that it was wrong. I said,

m1v1^2 + m2v2^2 = m1u1^2 + m2u2^2
m1v1^2 = m2u2^2
u2^2 = (m1v1^2)/m2
u2 = sqrt((m1v1^2)/m2)
u2 = sqrt((1.7(3.6)^2)/1.10) = 4.4753

Being unsure about the relationship I said u2 = Fsp. Therefore,
Fsp = -kx
4.4753 = -144x
4.4753/-144 = x
-0.03107 = x

Now, I am way off with this attempt but I'm not quite sure how I'm suppose to approach this problem. Can somebody help explain to me what, and most importantly why, I have to do to solve this problem? I'm having difficulty with this section so baby steps is appreciated :).

Thank you for taking the time to review my question.
 
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DavidAp said:
Block 2 (mass 1.10 kg) is at rest on a frictionless surface and touching the end of an unstretched spring of spring constant 144 N/m. The other end of the spring is fixed to a wall. Block 1 (mass 1.70 kg), traveling at speed v1 = 3.60 m/s, collides with block 2, and the two blocks stick together. When the blocks momentarily stop, by what distance is the spring compressed?

Answer 0.305 m



Relavant Equations
Fsp = -kx
______________________________

How do I solve for this? I'm given k but what is Fsp? Is there any other way to solve for Fsp?

After scratching my head for a good while I tried something out of pure desperation, despite my gut feeling that it was wrong. I said,

m1v1^2 + m2v2^2 = m1u1^2 + m2u2^2
m1v1^2 = m2u2^2
u2^2 = (m1v1^2)/m2
u2 = sqrt((m1v1^2)/m2)
u2 = sqrt((1.7(3.6)^2)/1.10) = 4.4753

Being unsure about the relationship I said u2 = Fsp. Therefore,
Fsp = -kx
4.4753 = -144x
4.4753/-144 = x
-0.03107 = x

Now, I am way off with this attempt but I'm not quite sure how I'm suppose to approach this problem. Can somebody help explain to me what, and most importantly why, I have to do to solve this problem? I'm having difficulty with this section so baby steps is appreciated :).

Thank you for taking the time to review my question.

I think you need to use energy stored in the spring.

When those two masses collide, conservation of momentum will enable you to calculate the speed with which the blocks move after collision. From that velocity you can calculate the Kinetic energy of the blocks.
When they stop, that energy will be transferred to the spring..
 

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