For a freely falling object dropped from
1 answer:
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Answer:
F = 99 v₂₀
v₂₀ = 1 m / s, F = 99 N
Explanation:
In this exercise it is asked to find the force during the collision, for this we use the relationship between the momentum and the momentum of car 1
I = Δp
F t = p_f- p₀
F t = m (v_f -v₀) (1)
We must find the final speed of car 1, for this we define a system formed by the two cars, in this case the forces during the collision are internal and the moment is conserved
initial instant. Before the crash
p₀ = 0 + m₂ v₂₀
final instant. After the crash
p_f = m₁ v₁ + m₂ v_{2f}
the moment is preserved
p₀ = p_f
m₂ v₂₀ = m₁ v_{1f} + m₂ v_{2f} (2)
m₂ (v₂₀ - v_2f}) = m₁ v_{1f}
as the collision is elastic the kinetic energy is also conserved
K₀ = K_f
½ m₂ v₂₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²
m₂ (v₂₀² -v_{2f}²) = m₁ v_{1f}²
let's write our system of equations, using
a² - b² = (a + b) (a-b)
m₂ (v₂₀ - v_{2f}) = m₁ v_{1f}
m₂ (v₂₀ -v_{2f}) (v₂₀ + v_{2f}) = m₁ v_{1f}²
to solve we divide the equations
v₂₀ + v_{2f} = v_{1f}
with this we substitute in equation 2 and find the speed of each car, in this case we need the speed of car 1
m₂ v₂₀ = m₁ v_{1f} + m₂ (v_{1f}-v₂₀)
2m₂ v₂₀ = (m₁ + m₂) v_{1f}
v_{1f} =
We substitute in the drive ratio of car 1
F t = m (v_f -v₀)
F = m₁ (\frac{2m_2}{m_1+m_2} v_{2o} - 0) / t
F =
the mass of each car is the mass of the car plus the mass of the boy
m₁ = 50 +40 = 90 kg
m₂ = 50 +60 = 110 kg
time is t = 1
we substitute the values
F = 2 90 100/90 + 110 vo2 / 1
F = 99 v₂₀
The value of the initial velocity of car 2 is not indicated in the problem, if this velocity is known it can be included and the force value is obtained, suppose that the initial velocity v₂₀ = 1 m / s
F = 99 N