Answer:
v = 91.8 Km / h
Explanation:
We must start this exercise at the end, let's look for the lighter car acceleration (B), for this we use Newton's second law
fr = m a
a = fr / m
fr = μ N
N-W = 0
Let's replace
a = μ m g / m
a = μ g
a = 0.8 9.8
a = 7.84 m / s²
As the car B after the crash reached an initial velocity vo₀₂ and at the end of the fine speed zero, let us use kinematics
v² = v₀₂² - 2 a x
0 = v₀₂² - 2 a x
v₀₂ = 2 a x₂
v₀₂ = √ (2 7.84 26)
v₀₂ = 20.19 m / s
Let's perform the same procedure for car A, the acceleration is the same as it does not depend on the mass of the vehicles
v₀₁ = √ 2 a x₁
v₀₁ = √ (2 7.84 19)
v₀₂ = 17.36 m / s
Now let's use moment conservation, where the system is the two vehicles
Initial before crash.
p₀ = M v₁ + 0
After the crash
[tex]p_{f}[/tex] = M v₀₁ + m v₀₂
p₀ =[tex]p_{f}[/tex]
M v₁ = M v₀₁ + m v₀₂
v₁ = v₀₁ + m / M v₀₂
v₁ = 17.36 + 1841/3000 20.19
v₁ = 20.75 m / s
This is the speed of car 1 (A) just before the crash, now let's look for the speed when I apply the brakes the initial speed (v)
v₁² = v² - 2 a x₁
v = √ (v₁² + 2 a x₁)
v = √ (20.75² + 2 7.84 14)
v = 25.50 m / s
v = 25.50 m / s (1km / 1000m) (3600s / 1h)
v = 91.8 Km / h