Question

Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has mass 2mm and is hanging at rest with its rope vertical. Sphere BB has mass mm and is held so that its rope makes an angle with the vertical that puts BB a vertical height HH above AA. Sphere BB is released from rest and swings down, collides with sphere AA, and sticks to it.
In terms of H,H, what is the maximum height above the original position of A reached by the combined spheres after their collision?

Answer 1

Answer:

  h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

         Em₀ = U = m g h

final point. Lower, just before the crash

         Em_f = K = ½ m $v_{b}^2$

energy is conserved

         Em₀ = Em_f

         m g h = ½ m v²

         v_b = $\sqrt{2gh}$

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

         p₀ = 2m 0 + m v_b

final instant. Right after the crash

         p_f = (2m + m) v

       

the moment is preserved

         p₀ = p_f

         m v_b = 3m v

         v = v_b / 3

         

          v = ⅓ $\sqrt{2gh}$

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

          Em₀ = K = ½ 3m v²

Final point. Higher

          Em_f = U = (3m) g h'

          Em₀ = Em_f

          ½ 3m v² = 3m g h’

           

we substitute

         h’=  $\frac{v^2}{2g}$

         h’ =  $\frac{1}{3^2} \ \frac{ 2gh}{2g}$

         h’ = 1/9 h