Answer:
a) h₂ = (m₁ / (m₁ + m₂))² h₁
b) h₂ = (2m₁ / (m₁ + m₂))² h₁
Explanation:
Let's analyze this exercise, we have an energy that is given to marble 1, then it collides with canine 2 and they rise to a new height; therefore the problem has to be solved in parts.
Let's start by using the energy concepts for the canine 1
Starting point. Highest point
Em₀ = U = m₁ g h
Final point. Lowest point, just before touching the other marble
= K = ½ m₁ v²
Emo = Em_{f}
m₁ g h = ½ m₁ v²
v₁ = √ 2gh
now let's analyze the clash of the two marbles
We define a system formed by the two marbles, so that the outside during the shock have been internal and the moment is preserved
initial. Just before the crash
p₀ = m₁ v₁ + 0
finsl. Right after the crash
= (m₁ + m₂) v
This case inelastic collisions
p₀ = p_{f}
m₁ v₁ = (m₁ + m₂) v
v = m₁ / (m₁ + m₂) v₁
this is the speed of the set before starting to climb. Let's use energy conservation for these two marbles
Starting point. Right after the crash
Em₀ = K = ½ (m₁ + m₂) v
final point. At the highest point of the set
Em_f = U = (m₁ + m₂) h₂
Em₀ = Em_f
½ (m₁ + m₂) v² = (m₁ + m₂) gh
we substitute
½ v² = gh
h = v² / 2g
we substitute the equation for speed
h = (m₁ / (m₁ + m₂))² (2gh₁) / 2g
h₂ = (m₁ / (m₁ + m₂))² h₁
b) In the case of elastic collision
in this case the conservation of the moment changes
initial p₀ = m₁ v₁
End p_f = m₁ v₁ ’+ m₂ v₂’
p₀ = p_f
m₁ v₁ = m₁ v₁ ’+ m₂ v₂’
also kinetic energy is conserved
K₀ = K_f
½ m₁ v₁² = ½ m₁ v₁’² + ½ m₂ v₂²
we write the two equations
m₁ (v₁ - v₁ ’) = m₂ v₂²
m1 (v₁² - v₁’²) = m₂ v₂²
solving this system of equations we are left with
v₁ ’= (m₁-m₂) / (m₁ + m₂) v₁
v₂ = 2m₁ / (m₁ + m₂) v₁
with this result marble 2 rises to height, let's use conservation energy
Em₀ = Em_f
½ m₂ v₂² = m₂ g h₂
h₂ = v₂² / 2g
h₂ = (2m₁ / (m₁ + m₂))² 2gh₁ / 2g
h₂ = (2m₁ / (m₁ + m₂))² h₁
C) in the elastic case there is no heat in the collision
in the inelastic case Q = ΔK