A 100g block on a frictionless table is firmly attached to one end of a spring with k = 20N/m. ?

a) For an elastic, head-on collision, we know (from conservation of energy), that the relative velocity of approach = relative velocity of separation, or

5.0 m/s = u - v

where v is the post-collision velocity of the ball

and u is the post-collision velocity of the block

so

u = v + 5.0m/s

conserve momentum for the collision, substituting for u:

20g*5.0m/s = 20g*v + 100g*(v + 5.0m/s)

solves to

v = -3.33 m/s

and so the ball's speed is 3.33 m/s ◄

b) Then u = 1.67 m/s, and the block's KE gets converted into spring PE:

½ * 0.100kg * (1.67m/s)² = ½ * 20N/m * x²

solves to

x = 0.12 m ◄

c) Now the objects have a common post-collision velocity:

20g*5.0m/s = 120g*v

solves to

v = 0.83 m/s ◄

½*0.120kg*(0.83m/s)² = ½ * 20N/m * x²

solves to

x = 0.065 m ◄

Hope this helps!