Calculating Velocity and Distance for a Coupled Spring-Mass System

AI Thread Summary
The discussion focuses on calculating the velocity and distance for a coupled spring-mass system involving two masses, m1 and m2, with m1 in equilibrium and m2 compressing a spring. The maximum velocity of m2 after losing contact with m1 is determined to be 0.5 m/s. The participant calculates the energy and amplitude after m2 breaks free, finding an amplitude of 0.15 m, which raises questions about the logic of a decrease in amplitude. The time for m1 to reach this position is calculated as 0.47 seconds, and the distance traveled by m2 is derived to be 0.085 m. The participant seeks clarification on the energy loss and amplitude decrease when m2 separates from m1.
najd
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Homework Statement


Question statement:
An object of mass m1 = 9.00 kg is in equilibrium while connected to a light spring of constant s = 100 N/m that is fastened to a wall as shown in (figure a):
twoBlocksSpring01.png

A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.200 m (figure b). The system is then released, and both objects start moving to the right on the frictionless surface.

A- When m1 reaches the equilibrium point, m2 loses contact with m1 (figure c) and moves to the right with speed v. Determine the value of v.
B- How far apart are the objects when the spring is fully stretched for the first time (figure d)?

Homework Equations


T= 0.5mv²
E_tot= 0.5sA²
v_max= Aω

The Attempt at a Solution


Okay, so the amplitude for the coupled system is 0.2 metres. The angular frequency of them both is:
ω= √(s/(m1+m2)) = √(100/(16)) = 2.5/s

The velocity of the second mass would equal the maximum velocity reached by the two objects as they pass the equilibrium point, so:
v_max= Aω= 0.2*2.5 = 0.5m/s

So far so good, now the part which I'm unsure of:
To figure out the distance between them, first I calculated how far the spring will get stretched after m2 loses contact with m1.
1- I calculated T for m1:
T1= 0.5m1v² = 0.5*9*(0.5)^2= 1.125J

2- I used conservation of energy to calculate A:
E1_tot= T1 and so,
A= √(2T/s) = √(2*1.125/100) = 0.15m
[This turned out to be less than 0.2m, is that even logical? I mean, the amplitude decreased after m2 broke free. I imagine it should increase.. even though it looks smaller in the figure.]

3- I calculated the time needed for m1 to reach that position:
τ= 2∏/ω1 = 2∏√(m1/s) = 2∏√(9/100) = 1.88s for a whole cycle.
And so, the time for a quarter of a cycle would be:
1.88/4 = 0.47s

4- I calculated the distance traveled by m2:
d= vt = 0.5*0.47 = 0.235m

5- I subtracted the amplitude of m1 from the distance traveled by m2 to obtain D.
D= 0.235-0.15 = 0.085m.

Anything wrong with what I did?
 
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najd said:
[This turned out to be less than 0.2m, is that even logical? I mean, the amplitude decreased after m2 broke free. I imagine it should increase.. even though it looks smaller in the figure.]

I assume the amplitude decreased because the system lost part of its energy as m2 broke free, and so, it would make sense.. correct me if I'm wrong, please.
 
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