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

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

It's not clear if U is the impact speed or the launch speed but let's refer to the impact speed as U2. If U = 7 m/s is defined as launch speed, not impact speed U2, compute U2 from U^2 - U2^2 = 2as, where a = -9.81 m/s^2. The collision conserves momentum so the combined velocity of the system comprising m & M = U2/2. Energy is also conserved so the separation velocity vM - vm = U2, vm = U2/2 - U2/2 = 0 and vM = U2/2 + U2/2 = U2 m/s.
1. You can compute M's momentum from that.
2. m is now in free fall and reaches the floor at t = sqrt(2s/g) (where s = 2 m) = 0.4515 s, which is also when M passes through the spring-mass equilibrium point at half the vibration period T/2. The natural frequency F of the system = 1/T and the radian frequency w = 2pi*F. Then k=Mw^2.
3. The oscillation amplitude = U2/w. Add twice this to 2 m to find the max. height M reaches.