Respuesta :
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 [tex]v_{b}^2[/tex]
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b = [tex]\sqrt{2gh}[/tex]
* 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 [tex]\sqrt{2gh}[/tex]
* 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’= [tex]\frac{v^2}{2g}[/tex]
h’ = [tex]\frac{1}{3^2} \ \frac{ 2gh}{2g}[/tex]
h’ = 1/9 h
