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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):

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?