How Far Do Colliding Blocks Slide on a Spring?

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Homework Help Overview

The problem involves two blocks colliding after being released from a compressed spring. The first block has mass m and is initially at rest, while the second block has mass 2m and is positioned closer to the equilibrium position of the spring. The objective is to determine how far the blocks slide together after colliding inelastically.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to use conservation of energy to find the velocity at which the blocks collide and applies Newton's second law to determine acceleration. Some participants question the clarity of the original poster's explanation and seek further elaboration on the reasoning used.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the original poster's approach and reasoning. There is no explicit consensus on the correctness of the methods used, and some participants are exploring the connections to previous threads on similar topics.

Contextual Notes

There are indications of confusion regarding the application of conservation of momentum and the assumptions made about the masses involved in the collision. The original poster also expresses uncertainty about their approach.

naianator
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Homework Statement



problems_MIT_boriskor_BKimages_10-mass-spring-two-block-collision.png


A block of mass m is attached to a spring with a force constant k, as in the above diagram. Initially, the spring is compressed a distance x from the equilibrium and the block is held at rest. Another block, of mass 2m, is placed a distance x/2 from the equilibrium as shown. After the spring is released, the blocks collide inelastically and slide together. How far (Δx) would the blocks slide beyond the collision point? Neglect friction between the blocks and the horizontal surface.

Homework Equations


U_spring = 1/2*k*x^2

K = 1/2*m*v^2

F = m*a

v_f^2 = v_0^2+2*a*x

The Attempt at a Solution


E_i = 1/2*k*x^2 = E_f = 1/2*m*v^2 + 1/2*k*x_f^2

at x/2 where m collides with 2m:

1/2*k*x^2 = 1/2*3m*v^2 + 1/8*k*x^2

k*x^2 = 3m*v^2 + 1/4*k*x^2

3/4*k*x^2 = 3*m*v^2

yielding

v = sqrt(k*x^2/(4*m))

by Newtons second law a = -k*x/(3*m)

and using kinematics (v_f^2 = v_0^2+2*a*x)

0 = k*x^2/(4*m) - 2*(k*x/(3*m))*delta(x)

x/4 = 2/3*delta(x)

delta(x) = 3/8*x

Am I way off base with this approach?
 
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Could you possibly explain what you have written , in words ?
It is a bit difficult to understand what you have typed .
 
Qwertywerty said:
Could you possibly explain what you have written , in words ?
It is a bit difficult to understand what you have typed .
I used conservation of energy to find the velocity where the two blocks collide at x/2. I'm not sure about this though, because I don't think that conservation of momentum applies and I used 3m for the mass (assuming that the blocks had collided). Then I used N2L to find the acceleration based on the force (-kx) and mass (3m). I plugged that into the kinematics equation for final velocity and solved for delta x.
 
naianator said:
Yes, sorry, I wasn't having any luck with responses and it was due a half hour ago so I reposted it. Is there a way to delete this thread?
No, but you might be able to edit the title to something that will stop readers wasting their time delving into it.
A moderator could delete it.
 

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