Respuesta :
Answer:
h = [tex](\frac{m}{m+M} )^2 \ \frac{v_o^2}{2g}[/tex]
Explanation:
To solve this problem, let's work in parts, let's start with the conservation of the moment, for this we define a system formed by the block and the bullet, in such a way that the forces during the collision have been internal and the moment is conserved.
initial instant. Before the crash
p₀ = m v₀
final instant. Right after the crash, but before the pendulum started to climb
m_f = (m + M) v
the moment is preserved
p₀ = p_f
m v₀ = (m + M) v
v = [tex]\frac{m}{m+M} \ v_o[/tex]
Now we work the pendulum system with embedded block, we use the concept of conservation of energy
starting point. Lower
Em₀ = K = ½ (m + M) v²
final point. higher, when it stops
Em_f = U = (m + M) g h
as there is no friction, energy is conserved
Em₀ = Em_f
½ (m + M) v² = (m + M) g h
h = [tex]\frac{v^2}{2g}[/tex]
we substitute the speed value of the block plus bullet set
h = [tex]( \frac{m}{ m+M} \ v_o )^2 \ \frac{1}{2g}[/tex]
h = [tex](\frac{m}{m+M} )^2 \ \frac{v_o^2}{2g}[/tex]
