Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompressed length at equilibrium. Block B of mass 2M is released from rest at a height h above block A . The two blocks collide and stick together. Express all answers in terms of d , h , M , and physical constants, as appropriate. Required:
a. Derive an expression for the spring constant k of the spring.
b. Derive an expression for the speed of block B just before it collides with block A.
c. Derive an expression for the speed of the blocks immediately after the collision.
d. Derive an expression for the maximum compression of the spring after the collision.
e. If the collision between the blocks was elastic, would the maximum compression of the spring be greater than, less than, or equal to that found in part (d)? Justify your answer.

a)  k = Mg / d , b)   v = √2gh , c)  v_{f} = ,  d)   x² + 6d x - dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

             ∑ F  = 0

             - W = 0

             k d = Mg

             k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

          Em₀ = U = m g h

lowest point. Right at the point of shock

          = K = ½ m v²2

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = √2gh

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

          p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

         p_{f} = (2M + M) v_{f}

 the moment is preserved

          p₀ = p_{f}

          2M v = 3M v_{f}

          v_{f} = ⅔ v

          v_{f} =

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

        Em₀ = K = ½ (3M)

        Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

        Em₀ = 4/3 M gh

final point. With the spring fully compressed

       Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

         Em₀ = Em_f

        4/3 M g h = ½ k x² + 3M g x

        ½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

          () x² + 3Mg x - Mgh = 0

          + 3 x - h = 0

to find the value of the spring compression, the second degree equation must be solved

          x² + 6d x - dh = 0

         x = [-6d ± ] / 2

         x = [-6d ± 6d  ]/2

         x = 3d ( -1±    )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

he kicked it 10 feet straight up into the air? i think this works.

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explanation: