Help me please with this problem with collision!!!!I am desperate!!!?

Question

Mass=1kg(M), which is immobile and is standing on the equillibrium point,is attached on a spring with k=400N/m.We move the mass M to a point D which is d=0.2 m away from the equillibrium and then we set it free to move.At the moment t=0 the mass M passes from the equillibrium for the first time and it collides (headon collision) plastically with another mass m=3 Kg which is moving with u2 velocity.The new (M+m) body just after the collision has a velocity with the same direction of the velocity of mass M just before the collision and it immobilizes for a moment for the first time after its creation to a point E which is d1=0,25m away from D.
Find
1) the Energy of the oscillation of the body (M+m)
2)the u2.
3)the percentage of the kinetic energy of the wo masses beforethe collision,which turnd into heat because of the collision.

Hint:m1 is moving to the left,m2 is moving to the right.

Answer 1

1) At the point d1, all energy of the combined system is stored in the spring, which is .05 m from the equilibrium point. Assuming no friction. The system will oscillate with this energy.
.5*k*x^2
in this case
.5*400*.05^2
0.5 Joules


2. To find u2, consider the momentum of the system and note that it will be conserved.
The velocity of M at equilibrium can be found by looking at energy
.5*k*x^2=.5*m*v^2
v=sqrt(400*.2^2)
v= 4 m/s

now to compute the velocity post collision, using the same technique, the kinetic energy of the combined mass will get converted to the energy in the spring,'so
.5=.5*4*v^2
v=.5 m/s

Now set up the conservation of momentum equations
1*4+3*u2=4*.5
u2= -2/3 the negative sign indicates that it was moving in the opposite direction of the mass on the spring at the moment of collision.

3. The kinetic energy post collision has already been computed as .5 Joules

The kinetic energy pre collision is
.5*1*16+.5*3*4/9
8.7 Joules

the percentage lost is
((8.7-.5)/8.7)*100
over 94%

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