One mass strikes and sticks to another mass atatched to a spring

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Homework Help Overview

The problem involves a block of mass sliding on a frictionless surface that collides inelastically with another mass attached to a spring. The goal is to determine how far the spring compresses after the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conservation of momentum during the inelastic collision and the subsequent conservation of energy to find the compression of the spring. There is uncertainty about the calculations and the interpretation of the problem.

Discussion Status

Some participants have provided guidance on using conservation laws, while others express confusion about the calculations and the accuracy of the problem statement. Multiple interpretations of the problem are being explored, particularly regarding the collision and energy conservation.

Contextual Notes

There is mention of a multiple-choice question, and participants are questioning whether the problem was stated correctly, as their calculated results do not match the provided options.

loganblacke
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Homework Statement


A block of mass 20.0 kg slides without friction at a speed of 6.0 m/s on a horizontal table surface until it strikes and sticks to a mass of 5.o kg attached to a horizontal spring (with a spring constant of k=8000.00 N), which in turn is attached to a wall. How far is the spring compressed before the masses come to rest?

Homework Equations


Kf-Ki = W
W = 1/2Kx^2


The Attempt at a Solution


Ki = 1/2(20kg)(6.0m/s)^2 = 360

I'm not sure how to account for the collision between the 20kg mass and the 5kg mass.
 
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Hi loganblacke...

See when its said that the 20kg mass sticks to the 5 kg mass its inelastic collision...
So all you have to do is conserve linear momentum(during the instant in which the collision takes place) ...You will get the initial velocity of the two masses(together as a mass of 25 kg)...Now conserve energy to get your answer...
 
I'm a little confused on the conserving momentum part, I did find the equation m1u1+m2u2 = (m1+m2)V.
 
I figured out the equations but my answer is still wrong, V2 = (m1V1)/(m1+m2) = 4.8 m/s. Then i used the equation .5(m1+m2)(V2^2) = .5kx^2 which when solved for x becomes x = sqrt(((m1+m2)*V2^2)/k) = .2683. I'm getting .2683 which is not one of the listed options on the multiple choice question
 
I don't see any mistake in what we did and even i am getting .2683...I have done similar problems and used the same logic...Are you sure this was the exact question?
 

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