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

Question

Mass=4kg(M), which is immobile and is standing on the equillibrium point,is attached on a spring and it is h=2 m away from the floor.We shoot from the floor another mass m=4 kg to the mass M with a velocity U= 7 m/s.The two masses collide (headon collision) elastically.After the collision the mass m returns to the floor the moment the mass M passes from the equilibrium point for the first time after the collision.

Find:
1.the momentum of the mass M hardly after the collision.
2.the spring constant k
3.the maximum height from the floor of the mass M during the oscillation

Answer 1

I will use g=10 m/s^2 for this solution
Lets look at the speed at which the second mass strikes the hanging mass at 2m above the floor

Using conservation of energy
.5*m*v^2=.5*m*v0^2-m*g*h

v^2=v0^2-2*g*h
in this case
v^2=7^2-2*10*2
v=3 m/s

a) After collision, momentum of the system is maintained, so setting up as positive
3*4=v1*4+v2*4
or
3=v1+v2
I will assume that the v1 is negative. Since there is symmetry in the system, the v2=-v1
and v1=-1.5

b) Next, let's look at the time it takes for the mass to reach the floor:
2=1.5*t+.5*10*t^2
0=5*t^2+1.5*t-2

The roots are .5 and -.08

That means the the hanging mass compresses the spring and returns back to the equilibrium point in 0.5
Set y as the maximum compression of the spring
using conservation of energy
.5*4*1.5^2=.5*k*y^2-4*10*(4*10/k+y)

Also
a=(-k*y-g) for
y=-4*10/k to y=0

and
a=-(g+k*y)
for
y=0 to y=ymax

so
a(t)=-(g+k*y)
this is d^2y/dt^2

BTW: That is a partial differential equation.

Solving it gives
f=sqrt(k/m)
where f is the frequency of oscillation

In this case, the oscillator went through 1/2 of a period, so
the period is one
and
1=sqrt(k/4)
k=4



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