Help me please with this problem with collision!!!!!!!!!!!!!!BEG U!?

Question

Mass=4kg(M), which is immobile and is standing on the equillibrium point,is attached on a spring with k=1600N/m.We move the mass M from the equillibrium spending energy E=288 J and sometime we set it free to move.At the moment the mass M passes from the point D which is d1=0.3*sq.root3 m away from the equillibrium, it collides (headon collision) plastically with another mass m=12 Kg which is moving with u2 velocity oposite from the velocity of M.

Find
1)the velocity of the body (M+m) which is created,just after the collision.
2)the heat produced because of the collision and the energy of the oscillation.
3)the equation of the X of the body (M+m)from the equillibrium,assuming that the moment t=0 the collision happened.(the direction of m is the positive one).
4)When will the body(m+M) pass from the point D for the first time after the collision?

Hint:M is moving to the left,m is moving to the right.

Answer 1

This is exactly the same in principle to the one I answered the other day.
The velocity of M at equilibrium can be calculated using conservation of energy

288=.5*4*v^2
v=12

The energy to get to d1 is equal to the kinetic energy of the combined mass as
.5*k*d1^2=.5*16*vc^2
calculate vc as
100*.3*.3*3=vc^2
vc=3*sqrt(3)
now you have enough to calculate u2
12*4-12*u2=16*3*sqrt(3)

2) the loss of energy in the collision is
288+.5*12*u2^2-.5*16*vc^2

The energy of oscillation is
.5*16*vc^2-288-.5*12*u2^2
This should equal
.58k*d1^2 as a check

3)the equation of motion is found by first using
F=m*a
F=-k*x
so m*a=-k*x
or
a=-k*x/m
where m is the combined mass
keep in mind that
a=(dx/dt)(dx/dt)
using that at t=0, x=0 and dx/dt at t=0 is vc


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