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

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SUMMARY

The problem involves a 20.0 kg block sliding at 6.0 m/s colliding inelastically with a 5.0 kg mass attached to a spring with a spring constant of 8000.00 N. The initial kinetic energy of the system is calculated as 360 J. After applying conservation of momentum, the combined mass moves at 4.8 m/s post-collision. The spring compression is determined using energy conservation, yielding a compression of 0.2683 meters, which does not match the provided multiple-choice options.

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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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