Respuesta :
Answer:
v₁₀ = 1.90 m / s
Explanation:
In this exercise we are given the maximum height data, with energy we can know how fast the body came out
Final mechanical energy, maximum height
[tex]Em_{f}[/tex] = U = m g h
Initial mechanical energy, in the lower part of the track
Em₀ = K = ½ m v²
Em= [tex]Em_{f}[/tex]
½ m v² = m g h
v = √ 2gh
Now we can use the moment to find the speed with which objects collide
The large object has a mass M = 5.41 kg a velocity starts v₁₀, the small object has a mass m = 1.68 kg an initial velocity of zero v₂₀ = 0 and final velocity v
Initial before the crash
p₀ = M v₁₀ + 0
Final after the crash
[tex]p_{f}[/tex] = M v1f + m v
p₀ = [tex]p_{f}[/tex]
M v₁₀ = M [tex]v_{1f}[/tex]+ m v
As the shock is elastic the kinetic energy is conserved
K₀ = [tex]K_{f}[/tex]
½ M v₁₀² = ½ M [tex]v_{1f}[/tex]² + ½ m v²
Let's write the system of equations
M v₁₀ = M [tex]v_{1f}[/tex] + m v
M v1₁₀² = M [tex]v_{1f}[/tex]² + m v²
We cleared v1f in the first we replaced in the second
[tex]v_{1f}[/tex] = (M v₁₀ - mv) / M
M v₁₀² = M (M v₁₀ - mv)² / M² + m v²
M v₁₀² = 1 / M (M² v₁₀² - 2mM v v₁₀ + m² v²) +m v²
v₁₀² (M - M) + 2 m v v₁₀ - v² (m2 + m) / M = 0
2 m v₁₀ - v (m + 1) m/ M = 0
v₁₀ = v (m +1) / (2M)
Let's substitute the value of v
v1₁₀= √ (2gh) (m +1) / (2M)
Let's calculate
v₁₀ = √ (2 9.8 3) (1+ 1.68) / (2 5.41)
V₁₀ = 7.668 (2.68) / 10.82
v₁₀ = 1.90 m / s
