Fill in the graph for 50 points
2 answers:
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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
Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.