Answer:
vₐ₀ = 29.56 m / s
Explanation:
In this exercise the initial velocity of car A is asked, to solve it we must work in parts
* The first with the conservation of the moment
* the second using energy conservation
let's start with the second part
we must use the relationship between work and kinetic energy
W = ΔK (1)
for this part the mass is
M = mₐ + m_b
the final velocity is zero, the initial velocity is v
friction force work is
W = - fr x
the negative sign e because the friction forces always oppose the movement
we write Newton's second law for the y-axis
N -W = 0
N = W = Mg
friction forces have the expression
fr =μ N
fr = μ M g
we substitute in 1
-μ M g x = 0 - ½ M v²
v² = 2 μ g x
let's calculate
v² = 2 0.750 9.8 6.00
v = ra 88.5
v = 9.39 m / s
Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.
Initial instant. Before the crash
p₀ = + mₐ vₐ₀ - m_b v_{bo}
instant fianl. Right after the crash, but the cars are still not moving
p_f = (mₐ + m_b) v
p₀ = p_f
+ mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v
mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}
let's reduce to the SI system
v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s
let's calculate
660 vₐ₀ = (660 +490) 9.39 + 490 17.778
vₐ₀ = 19509.72 / 660
vₐ₀ = 29.56 m / s
we can see that car A goes much faster than vehicle B
