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
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b =
* 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 
* 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’=
h’ =
h’ = 1/9 h
Answer:
F₁ = 4,120.2 N
F₂ = 3,924N
Explanation:
1) Balance of angular momentum around the end where F₁ is applied.
F₂ × 0.5m - F₁ × 0 = mass × g × 1m
⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m
F₂ = 196.2 Nm / 0.5m = 3,924 N
2) Balance of forces
F₁ - F₂ = mg
F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N
Answer:
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Explanation: