An object of mass m 1 = 9.40 kg is in equilibrium while connected to a light spr

Question

An object of mass m1 = 9.40 kg is in equilibrium while connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in Figure P15.52a. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.260 m, (see Fig. P15.52b). 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 (see Fig. P15.52c) 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 (D in Fig. P15.52d)? (Suggestion: First determine the period of oscillation and the amplitude of the m1-spring system after m2 loses contact with m1.)

An object of mass m1 = 9.40 kg is in equilibrium while connected to a light spring of constant k = 100 N/m that is fastened to a wall as shown in Figure P15.52a. A second object, m2 = 7.00 kg, is slowly pushed up against m1, compressing the spring by the amount A = 0.260 m, (see Fig. P15.52b). The system is then released and both objects start moving to the right on the frictionless surface. When m1 reaches the equilibrium point, m2 loses contact with m1 (see Fig. P15.52c) and moves to the right with speed v. Determine the value of v. How far apart are the objects when the spring is fully stretched for the first time (D in Fig. P15.52d)?

Explanation / Answer

Okay in order to find out their distance, you will need to know the amplitude of the m1 to find the distance from equilibrium for m1 after the m2 left m1 and the time that has passed for m2 so you can multiply speed and time for the distance of m2 and then subtract them to get the distance. So here's how it goes:
For amplitude, since the amplitude would change since it lost mass, I needed energy formula for oscillation to help me get the new amplitude. So the system originally had the total energy:
(1/2)kA^2=E
(1/2)(100)(.16)^2=1.28J
Since m2 has a velocity of its own and took some energy, then I can use the kinetic energy formula:
(1/2)(m2)(v)^2=K
(1/2)(7)(.4103)^2=.589J
Therefore the remaining energy in m1 is:
1.28J-.589J=.691J
Then I can use this reuse the energy formula of oscillation for the new amplitude:
(1/2)kA^2=E
(1/2)(100)A^2=.691
Solve for A which is .1176m which is the distance m1 will be from equilibrium when spring is fully stretched.

Now to find the distance of m2 from equilibrium...I will need to use T or period of m1 to help me find out how much time has passed since m2 left from equilibrium to find distance. So:
T=2(pi)* square root of (m/k)
T=2(3.14)(square root of (8.2/100))=1.7983 seconds
But beware...the time found is considered the period of the whole time from m1 which includes before equilibrium which we aren't interested since we had m2 attached before equilibrium. So we must divide the time we found by four because the time between equilibrium and maxed stretched is one fourth of the entire period of oscillation. So:
T/4=1.7983/4=.4496 seconds.
And finally we find the distance of m2 from equilibrium: speed times time= .4496*.4103=.1845m

And now we can tell how far apart they are...
.1845m-.1176m=.0669m

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