Answer:
a) v₁ = - ½ v₂, K₁ / K ₂ = ½,)
Step-by-step explanation:
A) Let's use the conservation of the moment, for this we define the system formed by the spring and the two cars, so the force during the separation is internal, therefore the moment is conserved
initial instant. Before releasing the carts
p₀ = 0
final instant. After jumping the cars
p_f = M v₁ + m v₂₂
how the moment was preserved
p₀ = p_f
0 = M v₁ + m v₂
v₁ = - m / M v₂
indicate that M = 2m
v₁ = - ½ v₂
the kinetic energy of each car is
K₁ = ½ M v₁²
K₁ = ½ 2m (v₂/ 2) ²
K₁ = m ₂v₂² / 4
K₂ = ½ m v₂²
the relationship between the kinetic energies is
K₁ / K₂ = ½
B) If the much greater than the mass of car 1 is mass of car 2
v = - m / M v₂
In this case, the speed of car 1 is very small, so the car 1 practitioner does not backtrack and car 2 leaves with a lot of speed.
The energy is still conserved, where almost all the energy has it is car 2