Max Displacement of 6.0 kg Object: 0.27 m?

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Homework Help Overview

The problem involves a 2.0 kg object falling onto a 6.0 kg object supported by a vertical massless spring, leading to an oscillating mass/spring system. The goal is to determine the maximum displacement of the 6.0 kg object after the collision.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conservation of mechanical energy and momentum in the context of the inelastic collision. There are differing interpretations of the mass used in calculations, particularly regarding the total mass after the collision.

Discussion Status

Some participants are exploring the setup of energy conservation equations, while others express uncertainty about the problem's interpretation and the answer key's correctness. There is an acknowledgment of the complexity of the problem and a request for verification of the calculations.

Contextual Notes

Participants note potential discrepancies in the problem setup, including the treatment of gravitational potential energy and the initial conditions of the spring. There is also mention of varying interpretations among different individuals attempting to solve the problem.

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Homework Statement


. A 2.0 kg object falls from rest a distance of 5.0 meters onto a 6.0 kg object that is supported by a vertical massless spring with spring constant k = 72 N/m. The two objects stick together after the collision, which results in the mass/spring system oscillating. What is the maximum magnitude of the displacement of the 6.0 kg object from its original location before it is struck by the falling object?
(A) 0.27 m
(B) 1.1 m ← CORRECT ANSWER
(C) 2.5 m ← what I am getting
(D) 2.8 m
(E) 3.1 m

Homework Equations


Conservation of mechanical energy(spring, gravitational, and kinetic)
conservation of momentum for the inelastic collision

The Attempt at a Solution


I'm just copying the solution here because it's what I did, but I will point out where i did something different.

Conserve energy to find the speed of the 2.0 kg object at the instant before the collision:
m1gy = (2.0 kg)(10 m/s)(5.0 m) = 100 J

and this equals the kinetic energy, after plug and chug we get
v = 10.0 m/s

The collision is inelastic, so
vf = v1*m1/(m1+m2) = 2.5 m/s

Now conserve energy again
Kf = 1/2mtv2 = 1/2(8.0 kg)(2.5 m/s)2 = 25 J

This compresses the spring and changes the gravitational potential energy, so we must solve for x in the equation
1/2kx2 + mtgx = Kf
or
36x2 + 20x − 25 = 0 I used 8kg as the total mass instead of 2kg, why did they use 2kg?
which has solutions x = −1.15 m and x = 0.6 m.

solutions with mt=8kg: x = -2.5, 0.27.

I'm a bit reluctant to say this, but i think aapt might have made a mistake on the answer key. On top of my confidence that my solution is right, they already used mt=8kg in the previous step before apparently changing its value in the next step. Now, given I'm a mere high schooler, i would still like someone to double check my work so that i can ask aapt to make the small fix.
 
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I agree that the solution as given in the key is set up in an odd way, even though I think it gives the correct numerical answer.

Clearly both masses move downward after the collision, so both masses decrease their gravitational PE. Also, the spring already has some potential energy before the collision.

I suggest setting up the energy conservation using the total mass and including the initial PE of the spring. You will need to take care to distinguish between the initial and final compression of the spring and the distance traveled by the masses.
 
as in 1/2k(5/6-x)2+80x=25?
5/6 being the displacement from relaxed spring due to m2 alone.
 
I searched the problem up and seems like other people are understanding the problem differently as well. I think I'll stop trying to figure out this problem. Thanks for the help.
 

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